Class 10 Optional Mathematics Coordinate Geometry
Exercise 8.1: Two-Point Form • Exercise 8.2: Angle Between Two Lines • Exercise 8.3: Conic Sections • Exercise 8.4: Equation of a Circle
135 solved questions (Exercises 8.1 to 8.4) with English and Nepali questions, step by step solutions, SVG graphs and frequently asked questions for the Nepal Grade 10 SEE.
1. Class 10 Optional Mathematics Coordinate Geometry: Unit Overview and Key Formulas
Unit 8: Coordinate Geometry (Optional Mathematics, Grade 10, SEE, Nepal) • Exercise 8.1 Equation of a Straight Line in Two-Point Form • Exercise 8.2 Angle Between Two Lines, Parallel and Perpendicular Lines • Exercise 8.3 Conic Sections • Exercise 8.4 Equation of a Circle
Class 10 Optional Mathematics Coordinate Geometry is the eighth unit of the course prescribed by the Curriculum Development Centre (CDC), Nepal. This page solves every question of Exercise 8.1, Exercise 8.2, Exercise 8.3 and Exercise 8.4. Each question is written in English with the Nepali question in brackets. Multiple choice solutions also have Nepali in brackets, and the other solutions are in English because the steps and symbols are the same in both languages.
- Two-point form: find the slope from the two points, then use the point-slope form.
- Collinear points: three points are collinear when the slopes of \(AB\) and \(BC\) are equal.
- Triangles: the midpoint of a side, the length of a median and the equation of a median.
Key formulas to remember
Figure: The two-point form: the slope of the line is (y₂ − y₁) / (x₂ − x₁).
For more chapters, visit the complete SEE Class 10 Optional Mathematics notes. You can also read about the equation of a line for extra understanding.
2. Exercise 8.1 Equation of a Straight Line in Two-Point Form: Class 10 Optional Mathematics Coordinate Geometry
Exercise 8.1: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)
Question 1(A). Which one is the equation of the straight line in two-point form? (दुई बिन्दु भएर जाने रेखाको समीकरण कुन हो ?)
- a) \(y-y_1=\frac{x_2-x_1}{y_2-y_1}(x-x_1)\)
- b) \(y-y_1=\frac{y_2+y_1}{x_2+x_1}(x-x_1)\)
- c) \(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x+x_1)\)
- d) \(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)
Correct Answer: d) \(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)
Solution:
- The slope of the line joining \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\dfrac{y_2-y_1}{x_2-x_1}\).
- Put it in the point-slope form \(y-y_1=m(x-x_1)\).
(रेखाको ढाल \(m=\frac{y_2-y_1}{x_2-x_1}\) लाई \(y-y_1=m(x-x_1)\) मा राख्दा दुई बिन्दु रूपको समीकरण प्राप्त हुन्छ ।)
Question 1(B). Which one is the equation of the line passing through the points \((2,3)\) and \((-1,0)\) from the given below? (बिन्दुहरू \((2,3)\) र \((-1,0)\) भएर जाने रेखाको समीकरण तलका मध्ये कुन हो ?)
- a) \(x-y+1=0\)
- b) \(x+y+1=0\)
- c) \(x-y-1=0\)
- d) \(-x-y-1=0\)
Correct Answer: a) \(x-y+1=0\)
Solution:
Here, \((x_1,y_1)=(2,3)\) and \((x_2,y_2)=(-1,0)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y-3 &= \frac{0-3}{(-1)-2}(x-2) \\ &= \frac{-3}{-3}(x-2) \\ &= x-2\end{aligned}\]\[x-y+1=0\](दुई बिन्दु रूपको सूत्र प्रयोग गर्दा \(x-y+1=0\) प्राप्त हुन्छ ।)
Question 1(C). Which is the equation of the line passing through \((a,0)\) and \((0,b)\)? (बिन्दुहरू \((a,0)\) र \((0,b)\) भएर जाने रेखाको समीकरण कुन हो ?)
- a) \(\frac xb+\frac ya=1\)
- b) \(\frac xa+\frac yb=1\)
- c) \(bx+ay=0\)
- d) \(ax+by=0\)
Correct Answer: b) \(\frac xa+\frac yb=1\)
Solution:
Here, \((x_1,y_1)=(a,0)\) and \((x_2,y_2)=(0,b)\).
\[\begin{aligned}& y-0=\frac{b-0}{0-a}(x-a) \\ &\implies -ay=b(x-a) \\ &\implies bx+ay=ab\end{aligned}\]Divide both sides by \(ab\):
\[\frac{bx}{ab}+\frac{ay}{ab}=1 \implies \frac xa+\frac yb=1\](दुवै पक्षलाई \(ab\) ले भाग गर्दा \(\frac xa+\frac yb=1\) प्राप्त हुन्छ ।)
Question 1(D). Which one of the following represents the equation of the line passing through the points \((-4,-2)\) and \((-3,5)\)? (बिन्दुहरू \((-4,-2)\) र \((-3,5)\) भएर जाने रेखाको समीकरण तलका मध्ये कुन हो ?)
- a) \(y=7x+26\)
- b) \(y=-7x+26\)
- c) \(y=-7x-26\)
- d) \(y=\frac17x+26\)
Correct Answer: a) \(y=7x+26\)
Solution:
Here, \((x_1,y_1)=(-4,-2)\) and \((x_2,y_2)=(-3,5)\).
\[\begin{aligned}& y+2=\frac{5+2}{-3+4}(x+4)=7(x+4) \\ &\implies y+2=7x+28 \\ &\implies y=7x+26\end{aligned}\](\(m=7\) भएकाले \(y+2=7(x+4)\) बाट \(y=7x+26\) आउँछ ।)
Exercise 8.1: Written Questions (लेख्ने प्रश्न)
Let the two given points be \((x_1,y_1)\) and \((x_2,y_2)\).
- Find the slope: \(m=\dfrac{y_2-y_1}{x_2-x_1}\).
- Put the slope and one of the points in the point-slope form: \(y-y_1=m(x-x_1)\).
- Combining the two steps gives the two-point form:
- Simplify it into the general form \(Ax+By+C=0\) or the slope-intercept form \(y=mx+c\).
Figure: Graph for Question 3(a): the line passes through (−5, 0) and (4, 3).
From the graph, the line passes through the points \((-5,0)\) and \((4,3)\).
Here, \((x_1,y_1)=(-5,0)\) and \((x_2,y_2)=(4,3)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y &= \frac{3-0}{4-(-5)}(x+5) \\ &= \frac{3}{9}(x+5) \\ &= \frac{1}{3}(x+5)\end{aligned}\]\[3y=x+5\]\[x-3y+5=0\]Figure: Graph for Question 3(b): the line passes through (−1, −2) and (5, 0).
From the graph, the line passes through the points \((-1,-2)\) and \((5,0)\).
Here, \((x_1,y_1)=(-1,-2)\) and \((x_2,y_2)=(5,0)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y+2 &= \frac{0-(-2)}{5-(-1)}(x+1) \\ &= \frac{2}{6}(x+1) \\ &= \frac{1}{3}(x+1)\end{aligned}\]\[3(y+2)=x+1\]\[x-3y-5=0\]Figure: Graph for Question 3(c): the line passes through (0, 5) and (6, 1).
From the graph, the line passes through the points \((0,5)\) and \((6,1)\).
Here, \((x_1,y_1)=(0,5)\) and \((x_2,y_2)=(6,1)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y-5 &= \frac{1-5}{6-0}x \\ &= \frac{-4}{6}x \\ &= -\frac{2}{3}x\end{aligned}\]\[3(y-5)=-2x\]\[2x+3y-15=0\]Note
The graph was not supplied, so the two points are read from the textbook answer key. If your graph shows other points, use the same method.
Figure: Graph for Question 3(d): the line passes through (−4, 0) and (0, −2).
From the graph, the line passes through the points \((-4,0)\) and \((0,-2)\).
Here, \((x_1,y_1)=(-4,0)\) and \((x_2,y_2)=(0,-2)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y &= \frac{(-2)-0}{0-(-4)}(x+4) \\ &= \frac{-2}{4}(x+4) \\ &= -\frac{1}{2}(x+4)\end{aligned}\]\[2y=-(x+4)\]\[x+2y+4=0\]Here, \((x_1,y_1)=(2,3)\) and \((x_2,y_2)=(-1,0)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y-3 &= \frac{0-3}{(-1)-2}(x-2) \\ &= \frac{-3}{-3}(x-2) \\ &= x-2\end{aligned}\]\[x-y+1=0\]Here, \((x_1,y_1)=(4,3)\) and \((x_2,y_2)=(-2,4)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y-3 &= \frac{4-3}{(-2)-4}(x-4) \\ &= \frac{1}{-6}(x-4) \\ &= -\frac{1}{6}(x-4)\end{aligned}\]\[6(y-3)=-(x-4)\]\[x+6y-22=0\]Here, \((x_1,y_1)=(6,-3)\) and \((x_2,y_2)=(-1,-5)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y+3 &= \frac{(-5)-(-3)}{(-1)-6}(x-6) \\ &= \frac{-2}{-7}(x-6) \\ &= \frac{2}{7}(x-6)\end{aligned}\]\[7(y+3)=2(x-6)\]\[2x-7y-33=0\]Here, \((x_1,y_1)=(0,-a)\) and \((x_2,y_2)=(b,0)\).
\[y+a=\frac{0+a}{b-0}(x-0)=\frac abx\]Multiply both sides by \(b\):
\[\begin{aligned}& b(y+a)=ax \\ &\implies by+ab=ax \\ &\implies ax-by-ab=0\end{aligned}\]Here, \((x_1,y_1)=(p,0)\) and \((x_2,y_2)=(0,q)\).
\[\begin{aligned}y-0 &= \frac{q-0}{0-p}(x-p) \\ &= -\frac qp(x-p)\end{aligned}\]Multiply both sides by \(-p\):
\[\begin{aligned}& -py=q(x-p) \\ &\implies -py=qx-pq \\ &\implies qx+py-pq=0\end{aligned}\]Here, \((x_1,y_1)=(4,3)\) and \((x_2,y_2)=(5,6)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y-3 &= \frac{6-3}{5-4}(x-4) \\ &= \frac{3}{1}(x-4)\end{aligned}\]\[3x-y-9=0\]Let \(A=(1,-1)\), \(B=(5,2)\) and \(C=(9,5)\). Three points are collinear if the slopes of \(AB\) and \(BC\) are equal.
\[\begin{aligned}m_{AB} &= \frac{2-(-1)}{5-1} \\ &= \frac{3}{4} \\ &= \frac{3}{4}\end{aligned}\]\[\begin{aligned}m_{BC} &= \frac{5-2}{9-5} \\ &= \frac{3}{4} \\ &= \frac{3}{4}\end{aligned}\]Since \(m_{AB}=m_{BC}=\frac{3}{4}\) and \(B\) is common to both segments, \(A\), \(B\) and \(C\) lie on the same straight line.
Let \(A=(-3,0)\), \(B=(0,-6)\) and \(C=(-1,-4)\). Three points are collinear if the slopes of \(AB\) and \(BC\) are equal.
\[\begin{aligned}m_{AB} &= \frac{(-6)-0}{0-(-3)} \\ &= \frac{-6}{3} \\ &= -2\end{aligned}\]\[\begin{aligned}m_{BC} &= \frac{(-4)-(-6)}{(-1)-0} \\ &= \frac{2}{-1} \\ &= -2\end{aligned}\]Since \(m_{AB}=m_{BC}=-2\) and \(B\) is common to both segments, \(A\), \(B\) and \(C\) lie on the same straight line.
Let \(A=(2,4)\), \(B=(4,6)\) and \(C=(6,8)\). Three points are collinear if the slopes of \(AB\) and \(BC\) are equal.
\[m_{AB}=\frac{6-4}{4-2}=\frac{2}{2}=1\]\[m_{BC}=\frac{8-6}{6-4}=\frac{2}{2}=1\]Since \(m_{AB}=m_{BC}=1\) and \(B\) is common to both segments, \(A\), \(B\) and \(C\) lie on the same straight line.
Let \(A=(1,-1)\), \(B=(2,1)\) and \(C=(4,5)\). Three points are collinear if the slopes of \(AB\) and \(BC\) are equal.
\[\begin{aligned}m_{AB} &= \frac{1-(-1)}{2-1} \\ &= \frac{2}{1} \\ &= 2\end{aligned}\]\[m_{BC}=\frac{5-1}{4-2}=\frac{4}{2}=2\]Since \(m_{AB}=m_{BC}=2\) and \(B\) is common to both segments, \(A\), \(B\) and \(C\) lie on the same straight line.
Let \(A=(7,-2)\), \(B=(2,3)\) and \(C=(-1,6)\). Three points are collinear if the slopes of \(AB\) and \(BC\) are equal.
\[\begin{aligned}m_{AB} &= \frac{3-(-2)}{2-7} \\ &= \frac{5}{-5} \\ &= -1\end{aligned}\]\[\begin{aligned}m_{BC} &= \frac{6-3}{(-1)-2} \\ &= \frac{3}{-3} \\ &= -1\end{aligned}\]Since \(m_{AB}=m_{BC}=-1\) and \(B\) is common to both segments, \(A\), \(B\) and \(C\) lie on the same straight line.
Let \(A=(0,4)\), \(B=(3,-2)\) and \(C=(6,-8)\). Three points are collinear if the slopes of \(AB\) and \(BC\) are equal.
\[\begin{aligned}m_{AB} &= \frac{(-2)-4}{3-0} \\ &= \frac{-6}{3} \\ &= -2\end{aligned}\]\[\begin{aligned}m_{BC} &= \frac{(-8)-(-2)}{6-3} \\ &= \frac{-6}{3} \\ &= -2\end{aligned}\]Since \(m_{AB}=m_{BC}=-2\) and \(B\) is common to both segments, \(A\), \(B\) and \(C\) lie on the same straight line.
The midpoint of \(AC\) is
\[\left(\frac{b+3}{2},\frac{0+3}{2}\right)=\left(\frac{b+3}{2},\frac32\right)\]For \(B(0,c)\) to be this midpoint we need \(\frac{b+3}{2}=0\) and \(c=\frac32\), that is \(b=-3\) and \(c=\frac32\).
The problem only says that the three points are collinear, so \(b\) and \(c\) are not fixed. The midpoint condition holds only for the special values \(b=-3\), \(c=\frac32\) (these do satisfy \(\frac1b+\frac1c=\frac13\) of part (b)).
Step 1: equation of the line through \(A(b,0)\) and \(B(0,c)\)
\[\begin{aligned}& y-0=\frac{c-0}{0-b}(x-b) \\ &\implies -by=c(x-b) \\ &\implies cx+by=bc\end{aligned}\]Divide both sides by \(bc\):
\[\frac xb+\frac yc=1 \quad (1)\]Step 2: use the third point \(C(3,3)\)
Since \(C\) lies on the line, put \(x=3\) and \(y=3\) in (1):
\[\begin{aligned}& \frac3b+\frac3c=1 \\ &\implies 3\left(\frac1b+\frac1c\right)=1 \\ &\implies \frac1b+\frac1c=\frac13\end{aligned}\]Three or more points are called collinear if all of them lie on one and the same straight line. Then the slope of the segment joining any two of these points is the same.
Step 1: equation of the line through \(A(p,0)\) and \(B(0,q)\)
\[\begin{aligned}& y-0=\frac{q-0}{0-p}(x-p) \\ &\implies -py=q(x-p) \\ &\implies qx+py=pq\end{aligned}\]Divide both sides by \(pq\):
\[\frac xp+\frac yq=1 \quad (1)\]Step 2: use the third point \(C(5,5)\)
Put \(x=5\) and \(y=5\) in (1):
\[\begin{aligned}& \frac5p+\frac5q=1 \\ &\implies 5\left(\frac1p+\frac1q\right)=1 \\ &\implies \frac1p+\frac1q=\frac15\end{aligned}\]Figure: Triangle PQR with P(1, 0), Q(0, 1), R(2, 3), the midpoint M(1, 2) of QR and the median PM.
- Plot \(P(1,0)\) on the \(x\)-axis and \(Q(0,1)\) on the \(y\)-axis.
- Plot \(R(2,3)\) in the first quadrant: \(2\) units right and \(3\) units up.
- Join \(P\) to \(Q\), \(Q\) to \(R\) and \(R\) to \(P\) with straight line segments.
Here, \(Q(x_1,y_1)=(0,1)\) and \(R(x_2,y_2)=(2,3)\). The midpoint is \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
\[\begin{aligned}M &= \left(\frac{0+2}{2},\frac{1+3}{2}\right) \\ &= \left(\frac22,\frac42\right) \\ &= (1,2)\end{aligned}\]Here, \(P(1,0)\) and the midpoint \(M(1,2)\) (part (b)). Both points have the same \(x\)-coordinate, \(x_1=x_2=1\).
So the denominator \(x_2-x_1=0\) and the slope is not defined: the line is vertical, parallel to the \(y\)-axis.
\[\begin{aligned}& y-0=\frac{2-0}{1-1}(x-1) \\ &\implies (1-1)y=2(x-1) \\ &\implies x-1=0\end{aligned}\]Figure: Triangle ABC with A(2, 2), B(2, 8), C(−6, 2), the midpoint M(−2, 5) of BC and the median AM.
Here, \(B(2,8)\) and \(C(-6,2)\).
\[\begin{aligned}M &= \left(\frac{2+(-6)}{2},\frac{8+2}{2}\right) \\ &= \left(\frac{-4}{2},\frac{10}{2}\right) \\ &= (-2,5)\end{aligned}\]Here, \(A(2,2)\) and \(M(-2,5)\). Use the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
\[\begin{aligned}AM &= \sqrt{(-2-2)^2+(5-2)^2} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5\end{aligned}\]Here, \(A(2,2)\) and \(M(-2,5)\).
\[\begin{aligned}y-2 &= \frac{5-2}{-2-2}(x-2) \\ &= -\frac34(x-2)\end{aligned}\]Multiply both sides by \(4\):
\[\begin{aligned}& 4(y-2)=-3(x-2) \\ &\implies 4y-8=-3x+6 \\ &\implies 3x+4y=14\end{aligned}\]Figure: Triangle XYZ with X(−1, 3), Y(1, −1), Z(5, 1), the midpoint D(3, 0) of YZ and the median XD.
Step 1: midpoint \(D\) of \(YZ\)
\[\begin{aligned}D &= \left(\frac{1+5}{2},\frac{-1+1}{2}\right) \\ &= (3,0)\end{aligned}\]Step 2: length of \(XD\)
\[\begin{aligned}XD &= \sqrt{(3+1)^2+(0-3)^2} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5\end{aligned}\]Here, \(X(-1,3)\) and \(D(3,0)\).
\[y-3=\frac{0-3}{3+1}(x+1)=-\frac34(x+1)\]Multiply both sides by \(4\):
\[\begin{aligned}& 4(y-3)=-3(x+1) \\ &\implies 4y-12=-3x-3 \\ &\implies 3x+4y-9=0\end{aligned}\]Since \(A(a,b)\) lies on \(6x-y=1\):
\[6a-b=1 \implies b=6a-1 \quad (1)\]Since \(B(b,a)\) lies on \(2x-5y=5\):
\[2b-5a=5 \quad (2)\]Substitute (1) in (2):
\[\begin{aligned}& 2(6a-1)-5a=5 \\ &\implies 12a-2-5a=5 \\ &\implies 7a=7 \\ &\implies a=1\end{aligned}\]From (1), \(b=6(1)-1=5\).
From part (a), \(a=1\) and \(b=5\).
\[A=(a,b)=(1,5),\qquad B=(b,a)=(5,1)\]Here, \(A(1,5)\) and \(B(5,1)\).
\[\begin{aligned}AB &= \sqrt{(5-1)^2+(1-5)^2} \\ &= \sqrt{16+16} \\ &= \sqrt{32} \\ &= 4\sqrt2\end{aligned}\]Here, \((x_1,y_1)=(1,5)\) and \((x_2,y_2)=(5,1)\).
\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]\[\begin{aligned}y-5 &= \frac{1-5}{5-1}(x-1) \\ &= \frac{-4}{4}(x-1) \\ &= -(x-1)\end{aligned}\]\[x+y-6=0\]1. A. d B. a C. b D. a
3. a. \(x-3y+5=0\) b. \(x-3y-5=0\) c. \(2x+3y-15=0\) d. \(x+2y+4=0\)
4. a. \(x-y+1=0\) b. \(x+6y-22=0\) c. \(2x-7y-33=0\) d. \(ax-by-ab=0\) e. \(qx+py-pq=0\) f. \(3x-y-9=0\)
5. All collinear 8. b. \((1,2)\) c. \(x=1\)
9. a. \((-2,5)\) b. \(5\) 10. a. \(5\) b. \(3x+4y-9=0\)
11. a. \(a=1,\ b=5\) b. \(A(1,5),\ B(5,1)\) c. \(4\sqrt2\) d. \(x+y-6=0\)
3. Exercise 8.2 Angle Between Two Lines, Parallel and Perpendicular Lines: Class 10 Optional Mathematics Coordinate Geometry
Key Formulas for Exercise 8.2
Exercise 8.2: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)
Question 1(A). Which of the following is the formula to find the angle between two straight lines \(y=m_1x+c_1\) and \(y=m_2x+c_2\)? (दुई सरल रेखा \(y=m_1x+c_1\) र \(y=m_2x+c_2\) बिचको कोण पत्ता लगाउने सूत्र तलका मध्ये कुन हो ?)
- a) \(\theta=\tan^{-1}\pm\left(\dfrac{m_1+m_2}{1-m_1m_2}\right)\)
- b) \(\theta=\tan^{-1}\pm\left(\dfrac{1-m_1m_2}{m_1-m_2}\right)\)
- c) \(\theta=\tan^{-1}\pm\left(\dfrac{m_1-m_2}{1+m_1m_2}\right)\)
- d) \(\theta=\tan^{-1}\pm\left(\dfrac{1+m_1m_2}{m_1-m_2}\right)\)
Correct Answer: c) \(\theta=\tan^{-1}\pm\left(\dfrac{m_1-m_2}{1+m_1m_2}\right)\)
Solution:
Let \(\theta\) be the angle between two straight lines having slopes \(m_1\) and \(m_2\).
The tangent of the angle between the two lines is
\[\tan\theta=\pm\left(\frac{m_1-m_2}{1+m_1m_2}\right)\]Taking the inverse tangent on both sides:
\[\theta=\tan^{-1}\pm\left(\frac{m_1-m_2}{1+m_1m_2}\right)\](दुई रेखाबिचको कोणको लागि \(\tan\theta=\pm\frac{m_1-m_2}{1+m_1m_2}\) हुन्छ, त्यसैले विकल्प c सही हो ।)
Question 1(B). Which of the following is the condition for two straight lines to be parallel? (दुई सरल रेखाहरू समानान्तर हुने अवस्था तलका मध्ये कुन हो ?)
- a) \(m_1-m_2=1\)
- b) \(m_1m_2=-1\)
- c) \(m_1+m_2=-1\)
- d) \(m_1-m_2=0\)
Correct Answer: d) \(m_1-m_2=0\)
Solution:
Two non-vertical straight lines are parallel if and only if their slopes are equal, \(m_1=m_2\).
Subtracting \(m_2\) from both sides:
\[m_1-m_2=0\](समानान्तर रेखाहरूको ढाल बराबर हुन्छ, अर्थात् \(m_1=m_2\) वा \(m_1-m_2=0\) ।)
Question 1(C). Which of the following is the condition for two straight lines to be perpendicular? (दुई सरल रेखाहरू लम्ब हुने अवस्था तलका मध्ये कुन हो ?)
- a) \(m_1m_2=1\)
- b) \(m_1m_2=-1\)
- c) \(m_1+m_2=-1\)
- d) \(m_1-m_2=0\)
Correct Answer: b) \(m_1m_2=-1\)
Solution:
Two straight lines are perpendicular if the product of their slopes is \(-1\).
\[m_1m_2=-1\](लम्ब रेखाहरूका ढालहरूको गुणनफल \(-1\) हुन्छ ।)
Question 1(D). What is the angle between the straight lines \(3x-y-7=0\) and \(x+3y-6=0\)? (सरल रेखाहरू \(3x-y-7=0\) र \(x+3y-6=0\) बिचको कोण कति हुन्छ ?)
- a) \(0^\circ\)
- b) \(30^\circ\)
- c) \(45^\circ\)
- d) \(90^\circ\)
Correct Answer: d) \(90^\circ\)
Solution:
Slope of a line \(ax+by+c=0\) is \(m=-\dfrac ab\).
\[m_1=-\frac{3}{-1}=3,\qquad m_2=-\frac13\]\[m_1m_2=3\times\left(-\frac13\right)=-1\]Since the product of the slopes is \(-1\), the lines are perpendicular. So the angle between them is \(90^\circ\).
(\(m_1m_2=-1\) भएकाले रेखाहरू लम्ब छन् र कोण \(90^\circ\) हो ।)
Question 1(E). What is the acute angle between the straight lines \(x-3y-2=0\) and \(2x-y-1=0\)? (सरल रेखाहरू \(x-3y-2=0\) र \(2x-y-1=0\) बिचको न्यूनकोण कति हुन्छ ?)
- a) \(0^\circ\)
- b) \(30^\circ\)
- c) \(45^\circ\)
- d) \(60^\circ\)
Correct Answer: c) \(45^\circ\)
Solution:
Here, \(m_1=-\dfrac{1}{-3}=\dfrac13\) and \(m_2=-\dfrac{2}{-1}=2\).
\[\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\left|\frac{\frac13-2}{1+\frac13\times2}\right|=\left|\frac{-\frac53}{\frac53}\right|=1\]\[\theta=\tan^{-1}(1)=45^\circ\](\(\tan\theta=1\) भएकाले न्यूनकोण \(45^\circ\) हुन्छ ।)
Question 1(F). What is the obtuse angle between the straight lines \(\sqrt3x-y-5=0\) and \(y+10=0\)? (सरल रेखाहरू \(\sqrt3x-y-5=0\) र \(y+10=0\) बिचको अधिककोण कति हुन्छ ?)
- a) \(120^\circ\)
- b) \(135^\circ\)
- c) \(150^\circ\)
- d) \(160^\circ\)
Correct Answer: a) \(120^\circ\)
Solution:
Here, \(m_1=-\dfrac{\sqrt3}{-1}=\sqrt3\) and \(m_2=0\) (the line \(y=-10\) is horizontal).
\[\tan\theta_1=\left|\frac{\sqrt3-0}{1+\sqrt3\times0}\right|=\sqrt3\implies\theta_1=60^\circ\]The obtuse angle is
\[\theta_2=180^\circ-60^\circ=120^\circ\](न्यूनकोण \(60^\circ\) भएकाले अधिककोण \(180^\circ-60^\circ=120^\circ\) हुन्छ ।)
Exercise 8.2: Written Questions (लेख्ने प्रश्न)
Line 1: \(-3x+y-4=0\implies m_1=-\dfrac{-3}{1}=3\).
Line 2: \(y=-2x+1\implies 2x+y-1=0\implies m_2=-2\).
\[\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\left|\frac{3-(-2)}{1+3(-2)}\right|=\left|\frac{5}{-5}\right|=1\]\[\theta=\tan^{-1}(1)=45^\circ\]Line 1: \(y=3x+2\implies m_1=3\).
Line 2: \(x+2y-2=0\implies m_2=-\dfrac12\).
\[\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\left|\frac{3+\frac12}{1-\frac32}\right|=\left|\frac{\frac72}{-\frac12}\right|=|-7|=7\]\[\theta=\tan^{-1}(7)\]Here, \(m_1=3\) and \(m_2=-2\). First find the acute angle \(\alpha\).
\[\tan\alpha=\left|\frac{3-(-2)}{1+3(-2)}\right|=\left|\frac{5}{-5}\right|=1\implies\alpha=45^\circ\]The obtuse angle is
\[\theta=180^\circ-45^\circ=135^\circ\]Here, \(m_1=-\dfrac23\) and \(m_2=-\dfrac12\). First find the acute angle \(\alpha\).
\[\tan\alpha=\left|\frac{-\frac23+\frac12}{1+\frac23\times\frac12}\right|=\left|\frac{-\frac16}{\frac43}\right|=\frac18=0.125\]\[\alpha=\tan^{-1}(0.125)\approx7.13^\circ\]The obtuse angle is
\[\theta=180^\circ-7.13^\circ=172.87^\circ\]Here, \(m_1=-3\) and \(m_2=-\dfrac13\).
Acute angle:
\[\tan\theta=\left|\frac{-3+\frac13}{1+(-3)\left(-\frac13\right)}\right|=\left|\frac{-\frac83}{2}\right|=\frac43\]\[\theta=\tan^{-1}\left(\frac43\right)\approx53.13^\circ\]Obtuse angle:
\[180^\circ-53.13^\circ=126.87^\circ\]Here, \(m_1=2\) and \(m_2=-\dfrac12\).
\[m_1m_2=2\times\left(-\frac12\right)=-1\]Since the product of the slopes is \(-1\), the lines are perpendicular.
Line 1: \(2x-3y-5=0\). Line 2: \(2x-3y-12=0\).
\[m_1=-\frac{2}{-3}=\frac23,\qquad m_2=-\frac{2}{-3}=\frac23\]Since \(m_1=m_2=\dfrac23\), the lines are parallel.
Line 1: \(x-5y-4=0\). Line 2: \(3x-15y+12=0\).
\[m_1=-\frac{1}{-5}=\frac15,\qquad m_2=-\frac{3}{-15}=\frac15\]Since \(m_1=m_2=\dfrac15\), the lines are parallel.
Line 1: \(\sqrt3x+y+4=0\implies m_1=-\sqrt3\).
Line 2: \(x-\sqrt3y-5=0\implies m_2=-\dfrac{1}{-\sqrt3}=\dfrac1{\sqrt3}\).
\[m_1m_2=(-\sqrt3)\times\frac1{\sqrt3}=-1\]Since \(m_1m_2=-1\), the lines are perpendicular.
Line 1: \(ax+by+c=0\implies m_1=-\dfrac ab\).
Line 2: \(bx-ay+c=0\implies m_2=-\dfrac{b}{-a}=\dfrac ba\).
\[m_1m_2=\left(-\frac ab\right)\left(\frac ba\right)=-1\]Since \(m_1m_2=-1\), the lines are perpendicular.
Line 1: \(m_1=-\dfrac p3\). Line 2: \(-3x+4y+7=0\implies m_2=\dfrac34\).
For parallel lines, \(m_1=m_2\).
\[-\frac p3=\frac34\implies-4p=9\implies p=-\frac94\]Line 1: \(m_1=-\dfrac{2}{-8}=\dfrac14\). Line 2: \(qx-12y-7=0\implies m_2=-\dfrac{q}{-12}=\dfrac q{12}\).
For parallel lines, \(m_1=m_2\).
\[\frac14=\frac q{12}\implies4q=12\implies q=3\]Line 1: \(m_1=-\dfrac a3\). Line 2: \(m_2=-\dfrac39=-\dfrac13\).
For parallel lines, \(m_1=m_2\).
\[-\frac a3=-\frac13\implies a=1\]Line 1: \(3x+my-5=0\implies m_1=-\dfrac3m\).
Line 2: \(\dfrac x2+\dfrac y3=1\implies3x+2y-6=0\implies m_2=-\dfrac32\).
For parallel lines, \(m_1=m_2\).
\[-\frac3m=-\frac32\implies m=2\]Line 1: \(m_1=-\dfrac p3\). Line 2: \(-3x+4y+7=0\implies m_2=\dfrac34\).
For perpendicular lines, \(m_1m_2=-1\).
\[\left(-\frac p3\right)\left(\frac34\right)=-1\implies-\frac p4=-1\implies p=4\]Line 1: \(m_1=-\dfrac23\). Line 2: \(m_2=-\dfrac{m}{-5}=\dfrac m5\).
For perpendicular lines, \(m_1m_2=-1\).
\[\left(-\frac23\right)\left(\frac m5\right)=-1\implies-\frac{2m}{15}=-1\implies m=\frac{15}2\]Line 1: \(m_1=-\dfrac43\). Line 2: \(m_2=-\dfrac3q\).
For perpendicular lines, \(m_1m_2=-1\).
\[\left(-\frac43\right)\left(-\frac3q\right)=-1\implies\frac4q=-1\implies q=-4\]Line 1: \(m_1=-\dfrac a5\). Line 2: \(m_2=-\dfrac6{10}=-\dfrac35\).
For perpendicular lines, \(m_1m_2=-1\).
\[\left(-\frac a5\right)\left(-\frac35\right)=-1\implies\frac{3a}{25}=-1\implies a=-\frac{25}3\]The slope of \(ax+by+c=0\) is \(m=-\dfrac{\text{coefficient of }x}{\text{coefficient of }y}\).
\[m_1=-\frac pq\]Line 1: \(m_1=-\dfrac pq\). Line 2: \(m_2=-\dfrac{l}{-m}=\dfrac lm\).
Since the lines are perpendicular, \(m_1m_2=-1\).
\[\left(-\frac pq\right)\left(\frac lm\right)=-1\]\[-\frac{pl}{qm}=-1\implies pl=qm\]The equation of any line parallel to \(ax+by+c=0\) is \(ax+by+k=0\). So the required line is
\[3x+4y+k=0\qquad\cdots(1)\]It passes through \((3,4)\):
\[3(3)+4(4)+k=0\implies25+k=0\implies k=-25\]Put \(k=-25\) in (1):
Slope of the line joining \(A(2,3)\) and \(B(3,-1)\):
\[m=\frac{-1-3}{3-2}=-4\]The required line is parallel to \(AB\), so its slope is also \(-4\). It passes through \((3,1)\):
\[y-1=-4(x-3)\]\[y-1=-4x+12\]The equation of any line perpendicular to \(ax+by+c=0\) is \(bx-ay+k=0\). So the required line is
\[5x-2y+k=0\qquad\cdots(1)\]It passes through \((2,5)\):
\[5(2)-2(5)+k=0\implies k=0\]Put \(k=0\) in (1):
Slope of the line joining \(A(4,7)\) and \(B(2,-3)\):
\[m_1=\frac{-3-7}{2-4}=\frac{-10}{-2}=5\]Slope of the perpendicular line:
\[5\times m_2=-1\implies m_2=-\frac15\]The line passes through \((3,1)\):
\[y-1=-\frac15(x-3)\]\[5y-5=-x+3\](i) Solve the two equations together:
\[3x+4y=7\qquad\cdots(1)\]\[5x-2y=3\qquad\cdots(2)\]Multiply (2) by 2: \(10x-4y=6\) \(\cdots(3)\). Add (1) and (3):
\[13x=13\implies x=1\]Put \(x=1\) in (1): \(3+4y=7\implies y=1\). So \(Z=(1,1)\).
(ii) A line perpendicular to \(2x+3y-5=0\) has the form \(3x-2y+k=0\). It passes through \(Z(1,1)\):
\[3(1)-2(1)+k=0\implies k=-1\]Step 1: Intersection point. From \(x+y=3\), \(y=3-x\). Put it in \(2x-3y=1\):
\[2x-3(3-x)=1\implies5x=10\implies x=2,\quad y=1\]The intersection point is \((2,1)\).
Step 2: The slope of \(x-y-6=0\) is \(-\dfrac1{-1}=1\), so the required line also has slope \(1\).
Step 3: Line through \((2,1)\) with slope \(1\):
\[y-1=1(x-2)\implies x-y=1\implies\frac x1+\frac y{-1}=1\]Comparing with \(\dfrac xa+\dfrac yb=1\):
Let \(A(-2,3)\) and \(B(4,7)\). Midpoint of \(AB\):
\[M=\left(\frac{-2+4}2,\frac{3+7}2\right)=(1,5)\]Slope of \(AB\) and of the perpendicular bisector:
\[m_1=\frac{7-3}{4+2}=\frac23,\qquad m_2=-\frac32\]Equation through \(M(1,5)\):
\[y-5=-\frac32(x-1)\]\[2y-10=-3x+3\]Let \(A(4,-2)\) and \(B(-8,9)\). Midpoint of \(AB\):
\[M=\left(\frac{4-8}2,\frac{-2+9}2\right)=\left(-2,\frac72\right)\]Slope of \(AB\) and of the perpendicular bisector:
\[m_1=\frac{9+2}{-8-4}=-\frac{11}{12},\qquad m_2=\frac{12}{11}\]Equation through \(M\left(-2,\dfrac72\right)\):
\[y-\frac72=\frac{12}{11}(x+2)\]\[11(2y-7)=24(x+2)\]\[22y-77=24x+48\]Let \(A(6,-5)\) and \(B(-8,9)\). Midpoint of \(AB\):
\[M=\left(\frac{6-8}2,\frac{-5+9}2\right)=(-1,2)\]Slope of \(AB\) and of the perpendicular bisector:
\[m_1=\frac{9+5}{-8-6}=-1,\qquad m_2=1\]Equation through \(M(-1,2)\):
\[y-2=1(x+1)\]The slope of the given line is \(m_1=\dfrac23\). Let the slope of the required line be \(m\).
\[\tan45^\circ=\pm\left(\frac{\frac23-m}{1+\frac23m}\right)\]\[1=\pm\left(\frac{2-3m}{3+2m}\right)\implies3+2m=\pm(2-3m)\]Taking (+) sign:
\[3+2m=2-3m\implies5m=-1\implies m=-\frac15\]\[y+2=-\frac15(x-3)\implies5y+10=-x+3\implies x+5y+7=0\]Taking (−) sign:
\[3+2m=-2+3m\implies m=5\]\[y+2=5(x-3)\implies5x-y-17=0\]The slope of the given line is \(m_1=-1\). Let the slope of the required line be \(m\).
\[\tan60^\circ=\pm\left(\frac{-1-m}{1-m}\right)\]\[\sqrt3(1-m)=\pm(-1-m)\]Taking (+) sign:
\[\sqrt3-\sqrt3m=-1-m\implies m(\sqrt3-1)=\sqrt3+1\]\[m=\frac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3\]Line through the origin: \(y=(2+\sqrt3)x\), i.e. \((2+\sqrt3)x-y=0\).
Taking (−) sign:
\[\sqrt3-\sqrt3m=1+m\implies m(\sqrt3+1)=\sqrt3-1\]\[m=\frac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3\]Line through the origin: \(y=(2-\sqrt3)x\), i.e. \((2-\sqrt3)x-y=0\).
1. A. c B. d C. b D. d E. c F. a
2. a. \(45^\circ\) b. \(\tan^{-1}7\) 3. a. \(135^\circ\) b. \(172.87^\circ\)
4. a. \(53.13^\circ\) and \(126.87^\circ\) b. \(90^\circ\)
7. a. \(p=-\frac94\) b. \(q=3\) c. \(a=1\) d. \(m=2\)
8. a. \(p=4\) b. \(m=\frac{15}2\) c. \(q=-4\) d. \(a=-\frac{25}3\)
9. a. \(-\frac pq\)
10. a. \(3x+4y-25=0\) b. \(4x+y-13=0\) 11. a. \(5x-2y=0\) b. \(x+5y-8=0\)
12. a. \(Z(1,1),\ 3x-2y-1=0\) b. \(a=1,\ b=-1\)
13. a. \(3x+2y-13=0\) b. \(24x-22y+125=0\) c. \(x-y+3=0\)
14. a. \(x+5y+7=0,\ 5x-y-17=0\) b. \((2+\sqrt3)x-y=0,\ (2-\sqrt3)x-y=0\)
4. Exercise 8.3 Conic Sections: Class 10 Optional Mathematics Coordinate Geometry
Key Facts for Exercise 8.3
Exercise 8.3: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)
Question 1(A). When a plane surface cuts a cone parallel to its base, which conic section is formed? (कुनै समतल सतहले एउटा सोलीलाई आधारसँग समानान्तर हुने गरी काट्दा कुन साङ्ङ्किक क्षेत्र बन्छ ?)
- a) Parabola (पाराबोला)
- b) Ellipse (दीर्घवृत्त)
- c) Circle (वृत्त)
- d) Hyperbola (हाइपरबोला)
Correct Answer: c) Circle (वृत्त)
Solution:
A plane parallel to the circular base is perpendicular to the axis of the cone, so \(\theta=90^\circ\). Every point of the section is at the same distance from the axis.
(आधारसँग समानान्तर समतलले अक्षसँग \(90^\circ\) कोण बनाउँछ, त्यसैले वृत्त बन्छ ।)
Question 1(B). When a plane surface cuts a cone perpendicular to its axis, not passing through the vertex, which conic section is formed? (एउटा समतल सतहले सोलीको अक्षसँग लम्ब हुने गरी र सोलीको शीर्षविन्दुबाट नजाने गरी प्रतिच्छेदित गर्दा बन्ने साङ्ङ्किक क्षेत्रलाई के भनिन्छ ?)
- a) Parabola (पाराबोला)
- b) Ellipse (दीर्घवृत्त)
- c) Circle (वृत्त)
- d) Hyperbola (हाइपरबोला)
Correct Answer: c) Circle (वृत्त)
Solution:
The plane is perpendicular to the axis, so \(\theta=90^\circ\). When \(\theta=90^\circ\), the section is a circle.
(\(\theta=90^\circ\) हुँदा वृत्त बन्छ ।)
Question 1(C). When a plane surface cuts a right circular cone such that the angle it makes with the axis is less than the semi-vertical angle, which conic section is formed? (एउटा सतहले समकोणी सोलीलाई सोलीको अक्षसँग बनाएको कोण अर्धशीर्षकोणभन्दा सानो हुने गरी प्रतिच्छेदित गर्दा बन्ने साङ्ङ्किक क्षेत्रलाई के भनिन्छ ?)
- a) Parabola (पाराबोला)
- b) Ellipse (दीर्घवृत्त)
- c) Circle (वृत्त)
- d) Hyperbola (हाइपरबोला)
Correct Answer: d) Hyperbola (हाइपरबोला)
Solution:
Let \(\alpha\) be the semi-vertical angle of the cone and \(\theta\) the angle between the cutting plane and the axis.
When \(\theta<\alpha\), the plane cuts both nappes (upper and lower halves) of the double cone and gives two open curves, a hyperbola.
(\(\theta<\alpha\) हुँदा समतलले दुवै भाग काट्छ र हाइपरबोला बन्छ ।)
Question 1(D). When a plane surface cuts a cone parallel to its generator, which conic section is formed? (एउटा सोलीलाई समतलीय सतहले सोलीको जेनेरेटरसँग समानान्तर हुने गरी काट्दा बन्ने साङ्ङ्किक क्षेत्रलाई के भनिन्छ ?)
- a) Parabola (पाराबोला)
- b) Ellipse (दीर्घवृत्त)
- c) Circle (वृत्त)
- d) Hyperbola (हाइपरबोला)
Correct Answer: a) Parabola (पाराबोला)
Solution:
Let \(\alpha\) be the semi-vertical angle of the cone and \(\theta\) the angle between the cutting plane and the axis.
A plane parallel to a generator makes \(\theta=\alpha\). The section is an open, single curve, a parabola.
(जेनेरेटरसँग समानान्तर हुँदा \(\theta=\alpha\) हुन्छ र पाराबोला बन्छ ।)
Question 1(E). If the semi-vertical angle of a cone is \(\alpha\), the angle made by the cutting plane with the axis is \(\theta\) and \(\alpha<\theta<90^\circ\), which conic section is formed? (सोलीको अर्धशीर्षकोण \(\alpha\) र अक्षसँग समतल सतहले बनाएको कोण \(\theta\) छ । यदि \(\alpha<\theta<90^\circ\) भए बन्ने साङ्ङ्किक क्षेत्रलाई के भनिन्छ ?)
- a) Parabola (पाराबोला)
- b) Ellipse (दीर्घवृत्त)
- c) Circle (वृत्त)
- d) Hyperbola (हाइपरबोला)
Correct Answer: b) Ellipse (दीर्घवृत्त)
Solution:
When \(\alpha<\theta<90^\circ\), the plane cuts all the generators of one nappe. The curve is a closed oval, an ellipse.
(\(\alpha<\theta<90^\circ\) हुँदा बन्द अण्डाकार वक्र, दीर्घवृत्त बन्छ ।)
Question 1(F). If the semi-vertical angle of a cone is \(\alpha\), the angle made by the cutting plane with the axis is \(\theta\) and \(\theta=\alpha\), which conic section is formed? (सोलीको अर्धशीर्षकोण \(\alpha\) र अक्षसँग समतल सतहले बनाएको कोण \(\theta\) छ । यदि \(\theta=\alpha\) भए बन्ने साङ्ङ्किक क्षेत्रलाई के भनिन्छ ?)
- a) Parabola (पाराबोला)
- b) Ellipse (दीर्घवृत्त)
- c) Circle (वृत्त)
- d) Hyperbola (हाइपरबोला)
Correct Answer: a) Parabola (पाराबोला)
Solution:
When \(\theta=\alpha\), the plane is parallel to one generator, and the curve formed is a parabola.
(\(\theta=\alpha\) हुँदा पाराबोला बन्छ ।)
Question 1(G). If the semi-vertical angle of a cone is \(\alpha\), the angle made by the cutting plane with the axis is \(\theta\) and \(\theta<\alpha\), which conic section is formed? (सोलीको अर्धशीर्षकोण \(\alpha\) र अक्षसँग समतल सतहले बनाएको कोण \(\theta\) छ । यदि \(\theta<\alpha\) भए बन्ने साङ्ङ्किक क्षेत्रलाई के भनिन्छ ?)
- a) Parabola (पाराबोला)
- b) Ellipse (दीर्घवृत्त)
- c) Circle (वृत्त)
- d) Hyperbola (हाइपरबोला)
Correct Answer: d) Hyperbola (हाइपरबोला)
Solution:
When \(\theta<\alpha\), the plane cuts both the upper and lower nappes of the double cone, giving a two-piece curve, a hyperbola.
(\(\theta<\alpha\) हुँदा हाइपरबोला बन्छ ।)
Question 1(H). Which of the following is not a conic section? (दिइएका मध्ये कुन साङ्ङ्किक क्षेत्र होइन ?)
- a) Parabola (पाराबोला)
- b) Ellipse (दीर्घवृत्त)
- c) Circle (वृत्त)
- d) Straight line (सिधा रेखा)
Correct Answer: d) Straight line (सिधा रेखा)
Solution:
The circle, ellipse, parabola and hyperbola are the conic sections. A straight line appears only in the degenerate case when the plane passes through the vertex, so it is not counted as a conic section.
(वृत्त, दीर्घवृत्त, पाराबोला र हाइपरबोला साङ्ङ्किक क्षेत्र हुन्; सिधा रेखा होइन ।)
Exercise 8.3: Written Questions (लेख्ने प्रश्न)
From the figure of the right circular cone:
- Point \(A\) at the top is the vertex.
- Angle \(\alpha\) between the generator and the axis is the semi-vertical angle.
- The slant line \(AC\) from the vertex to the edge of the base is the generator (slant height \(l\)).
- The line \(AB\) through the centre of the base is the axis of the cone (height \(h\)).
- The segment \(BC\) from the centre to the edge of the base is the base radius (\(r\)).
- (a) A round closed curve whose points are all at the same distance from one centre: circle.
- (b) An elongated oval closed curve: ellipse.
- (c) A single open U-shaped symmetrical curve: parabola.
- (d) Two open symmetrical curves facing opposite ways: hyperbola.
A circle is the conic section formed when a plane cuts a right circular cone perpendicular to its axis, that is, \(\theta=90^\circ\). It is the set of points in a plane at a constant distance (radius) from a fixed point (centre).
An ellipse is the conic section formed when a plane cuts one nappe of a right circular cone so that the angle \(\theta\) with the axis is greater than the semi-vertical angle \(\alpha\) and less than \(90^\circ\). It is a closed oval curve.
A parabola is the conic section formed when a plane cuts a cone parallel to one of its generators, so that \(\theta=\alpha\). It is an open U-shaped curve.
A hyperbola is the conic section formed when a plane cuts both nappes (upper and lower halves) of a double right circular cone, so that \(\theta<\alpha\). It has two separate open symmetrical curves.
Let \(\alpha\) be the semi-vertical angle of the cone and \(\theta\) the angle between the cutting plane and the axis.
A circle is formed when the cutting plane is perpendicular to the axis of the cone.
Let \(\alpha\) be the semi-vertical angle of the cone and \(\theta\) the angle between the cutting plane and the axis.
An ellipse is formed when the plane cuts all the generators on one side of the vertex, so \(\theta\) is greater than \(\alpha\) and less than \(90^\circ\).
Let \(\alpha\) be the semi-vertical angle of the cone and \(\theta\) the angle between the cutting plane and the axis.
A parabola is formed when the plane is parallel to a generator of the cone.
Let \(\alpha\) be the semi-vertical angle of the cone and \(\theta\) the angle between the cutting plane and the axis.
A hyperbola is formed when the plane cuts both the upper and lower nappes of the double cone.
1. A. c B. c C. d D. a E. b F. a G. d H. d
3. a. Circle b. Ellipse c. Parabola d. Hyperbola
5. a. \(\theta=90^\circ\) b. \(\alpha<\theta<90^\circ\) c. \(\theta=\alpha\) d. \(\theta<\alpha\)
5. Exercise 8.4 Equation of a Circle: Class 10 Optional Mathematics Coordinate Geometry
Key Formulas for Exercise 8.4
Exercise 8.4: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)
Question 1(A). Which of the following equations represents a circle with centre \((0,0)\) and radius \(10\) units? (दिइएका समीकरणहरूमध्ये केन्द्रबिन्दु \((0,0)\) र अर्धव्यास \(10\) एकाइ कुन समीकरणले जनाउँछ ?)
- a) \(x^2-y^2=10\)
- b) \(x^2+y^2=10\)
- c) \(x^2-y^2=100\)
- d) \(x^2+y^2=100\)
Correct Answer: d) \(x^2+y^2=100\)
Solution:
The circle with centre \((h,k)\) and radius \(r\) is \((x-h)^2+(y-k)^2=r^2\). Put \(h=0,\ k=0,\ r=10\):
\[x^2+y^2=10^2=100\](\(h=k=0\) र \(r=10\) राख्दा \(x^2+y^2=100\) हुन्छ ।)
Question 1(B). Which of the following points lies on the circle \((x-5)^2+(y-4)^2=18\)? (दिइएका कुन बिन्दु वृत्त \((x-5)^2+(y-4)^2=18\) मा पर्छ ?)
- a) \((0,0)\)
- b) \((1,0)\)
- c) \((2,1)\)
- d) \((0,2)\)
Correct Answer: c) \((2,1)\)
Solution:
Test \((2,1)\) in the equation:
\[(2-5)^2+(1-4)^2=9+9=18=\text{RHS}\]So \((2,1)\) satisfies the equation.
(\((2,1)\) राख्दा बायाँ पक्ष = दायाँ पक्ष = \(18\) हुन्छ ।)
Question 1(C). A circle with centre \((0,0)\) passes through the point \((-8,-15)\). What is its radius? (केन्द्रबिन्दु \((0,0)\) भएको वृत्त बिन्दु \((-8,-15)\) भएर जान्छ भने अर्धव्यास कति हुन्छ ?)
- a) 8 units
- b) 15 units
- c) 17 units
- d) 21 units
Correct Answer: c) 17 units
Solution:
The radius is the distance from the centre to the point:
\[r=\sqrt{(-8)^2+(-15)^2}=\sqrt{64+225}=\sqrt{289}=17\](दूरी सूत्रबाट \(r=\sqrt{289}=17\) एकाइ ।)
Question 1(D). For the circle \(x^2+y^2-4x-8y-45=0\), which of the following is its centre and radius? (समीकरण \(x^2+y^2-4x-8y-45=0\) भएको वृत्तको केन्द्रबिन्दु र अर्धव्यास कुन हुन्छ ?)
- a) Centre \((2,4)\), radius \(5\) units
- b) Centre \((4,8)\), radius \(65\) units
- c) Centre \((4,8)\), radius \(\sqrt{65}\) units
- d) Centre \((2,4)\), radius \(\sqrt{65}\) units
Correct Answer: d) Centre \((2,4)\), radius \(\sqrt{65}\) units
Solution:
Compare with \(x^2+y^2+2gx+2fy+c=0\): \(2g=-4,\ 2f=-8,\ c=-45\), so \(g=-2,\ f=-4\).
\[\text{Centre}=(-g,-f)=(2,4)\]\[r=\sqrt{g^2+f^2-c}=\sqrt{4+16+45}=\sqrt{65}\](केन्द्र \((-g,-f)=(2,4)\) र अर्धव्यास \(\sqrt{65}\) एकाइ ।)
Question 1(E). The equation of the circle with centre \((1,-2)\) and radius \(4\) units is: (केन्द्रबिन्दु \((1,-2)\) र अर्धव्यास \(4\) एकाइ भएको वृत्तको समीकरण ...... हो ।)
- a) \((x+1)^2+(y-2)^2=4^2\)
- b) \((x+1)^2+(y+2)^2=4^2\)
- c) \((x-1)^2+(y-2)^2=4^2\)
- d) \((x-1)^2+(y+2)^2=4^2\)
Correct Answer: d) \((x-1)^2+(y+2)^2=4^2\)
Solution:
Put \(h=1,\ k=-2,\ r=4\) in \((x-h)^2+(y-k)^2=r^2\):
\[(x-1)^2+(y+2)^2=4^2\](\(h=1,\ k=-2\) राख्दा \((x-1)^2+(y+2)^2=4^2\) हुन्छ ।)
Exercise 8.4: Written Questions (लेख्ने प्रश्न)
Use \((x-h)^2+(y-k)^2=r^2\).
\[(x+4)^2+(y-1)^2=16\]\[x^2+8x+16+y^2-2y+1-16=0\]Use \((x-h)^2+(y-k)^2=r^2\).
\[(x+3)^2+(y+4)^2=49\]\[x^2+6x+9+y^2+8y+16-49=0\]Use \((x-h)^2+(y-k)^2=r^2\).
\[(x-2)^2+(y+1)^2=9\]\[x^2-4x+4+y^2+2y+1-9=0\]Use \((x-h)^2+(y-k)^2=r^2\).
\[x^2+(y-10)^2=25\]\[x^2+y^2-20y+100-25=0\]Use \((x-h)^2+(y-k)^2=r^2\).
\[(x+a)^2+(y+a)^2=(a\sqrt2)^2=2a^2\]\[x^2+2ax+a^2+y^2+2ay+a^2-2a^2=0\]Use \((x-h)^2+(y-k)^2=r^2\).
\[(x-a)^2+(y-b)^2=b^2\]\[x^2-2ax+a^2+y^2-2by+b^2-b^2=0\]Compare with \((x-h)^2+(y-k)^2=r^2\): \(h=0,\ k=0,\ r^2=25\implies r=5\).
Compare with \((x-h)^2+(y-k)^2=r^2\): \(h=0,\ k=0,\ r^2=36\implies r=6\).
Compare with \((x-h)^2+(y-k)^2=r^2\): \(h=0,\ k=-2,\ r^2=36\implies r=6\).
Write it as \(x^2+y^2-2y-24=0\).
Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=0,\ f=-1,\ c=-24\).
\[\text{Centre}=(-g,-f)=(0,1)\]\[r=\sqrt{0+1+24}=\sqrt{25}=5\]Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=-5,\ f=2,\ c=13\).
\[\text{Centre}=(5,-2)\]\[r=\sqrt{25+4-13}=\sqrt{16}=4\]Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=3,\ f=-4,\ c=-11\).
\[\text{Centre}=(-3,4)\]\[r=\sqrt{9+16+11}=\sqrt{36}=6\]Divide by 2:
\[x^2+y^2-4x-6y+\frac12=0\]Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=-2,\ f=-3,\ c=\frac12\).
\[\text{Centre}=(2,3)\]\[r=\sqrt{4+9-\frac12}=\sqrt{\frac{25}2}=\frac5{\sqrt2}\]Divide by 9:
\[x^2+y^2-4x+\frac23y-\frac{107}9=0\]Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=-2,\ f=\frac13,\ c=-\frac{107}9\).
\[\text{Centre}=\left(2,-\frac13\right)\]\[r=\sqrt{4+\frac19+\frac{107}9}=\sqrt{\frac{144}9}=4\]Use the diameter form: \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\).
\[(x-a)(x+a)+(y+1)(y-2)=0\]\[x^2-a^2+y^2-y-2=0\]Use the diameter form: \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\).
\[(x-2)(x-3)+(y+6)(y+7)=0\]\[x^2-5x+6+y^2+13y+42=0\]Use the diameter form: \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\).
\[x(x+3)+(y+7)(y-2)=0\]\[x^2+3x+y^2+5y-14=0\]Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=-2,\ f=3,\ c=-12\).
\[\text{Centre}=(-g,-f)=(2,-3)\]From (a), the centre is \(C(2,-3)\). It is the midpoint of the diameter with end points \(A(5,3)\) and \(B(x_2,y_2)\).
\[\frac{5+x_2}2=2\implies x_2=-1\]\[\frac{3+y_2}2=-3\implies y_2=-9\]Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=-3,\ f=-4,\ c=-11\).
\[\text{Centre}=(3,4)\]From (a), \(g=-3,\ f=-4,\ c=-11\).
\[r=\sqrt{g^2+f^2-c}=\sqrt{9+16+11}=\sqrt{36}=6\]The centre \(C(3,4)\) is the midpoint of \(A(-2,-3)\) and \(B(x_2,y_2)\).
\[\frac{-2+x_2}2=3\implies x_2=8\]\[\frac{-3+y_2}2=4\implies y_2=11\]Divide by 3:
\[x^2+y^2+2x-y+1=0\]Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=1,\ f=-\frac12,\ c=1\).
\[\text{Centre}=(-g,-f)=\left(-1,\frac12\right)\]From (a), \(g=1,\ f=-\dfrac12,\ c=1\).
\[r=\sqrt{1+\frac14-1}=\sqrt{\frac14}=\frac12\]The centre \(C\left(-1,\dfrac12\right)\) is the midpoint of \(A(5,3)\) and \(B(x_2,y_2)\).
\[\frac{5+x_2}2=-1\implies x_2=-7\]\[\frac{3+y_2}2=\frac12\implies y_2=-2\]A circle in the third quadrant touching both axes has centre \((-r,-r)=(-5,-5)\).
\[(x+5)^2+(y+5)^2=25\]\[x^2+10x+25+y^2+10y+25-25=0\]A circle in the fourth quadrant touching both axes has centre \((r,-r)=(6,-6)\).
\[(x-6)^2+(y+6)^2=36\]\[x^2-12x+36+y^2+12y+36-36=0\]The circle passes through the origin and cuts the axes at \((4,0)\) and \((0,6)\). The angle at the origin is \(90^\circ\), so these two points are the end points of a diameter.
\[(x-4)(x-0)+(y-0)(y-6)=0\]\[x^2-4x+y^2-6y=0\]The circle passes through the origin and cuts the axes at \((5,0)\) and \((0,10)\). The angle at the origin is \(90^\circ\), so these two points are the end points of a diameter.
\[(x-5)(x-0)+(y-0)(y-10)=0\]\[x^2-5x+y^2-10y=0\]The circle passes through the origin and cuts the axes at \((12,0)\) and \((0,8)\). The angle at the origin is \(90^\circ\), so these two points are the end points of a diameter.
\[(x-12)(x-0)+(y-0)(y-8)=0\]\[x^2-12x+y^2-8y=0\]Step 1: \(C\) lies on the line, so \(2h-3k+1=0\ \cdots(1)\).
Step 2: \(CM=CN\) (both are radii), so \(CM^2=CN^2\):
\[(h+2)^2+k^2=h^2+(k+2)^2\]\[4h=4k\implies h=k\]Step 3: Put \(h=k\) in (1): \(2k-3k+1=0\implies k=1\). So the centre is \(C(1,1)\).
Step 4: Radius squared:
\[r^2=(1+2)^2+(1-0)^2=10\]Step 5: Equation:
\[(x-1)^2+(y-1)^2=10\]\[x^2+y^2-2x-2y-8=0\]Step 1: \(C\) lies on the line, so \(4h+k=16\implies k=16-4h\ \cdots(1)\).
Step 2: \(CP^2=CQ^2\):
\[(h-4)^2+(k-1)^2=(h-6)^2+(k-5)^2\]\[-8h-2k+17=-12h-10k+61\]\[4h+8k=44\implies h+2k=11\]Step 3: Put \(k=16-4h\):
\[h+32-8h=11\implies h=3,\quad k=4\]The centre is \(C(3,4)\).
Step 4: Radius squared:
\[r^2=(3-4)^2+(4-1)^2=10\]Step 5: Equation:
\[(x-3)^2+(y-4)^2=10\]\[x^2+y^2-6x-8y+15=0\]The angle at the origin is \(90^\circ\), so \((4,0)\) and \((0,2)\) are the end points of a diameter.
\[(x-4)(x-0)+(y-0)(y-2)=0\]\[x^2-4x+y^2-2y=0\]The angle at the origin is \(90^\circ\), so \((5,0)\) and \((0,5)\) are the end points of a diameter.
\[(x-5)(x-0)+(y-0)(y-5)=0\]\[x^2-5x+y^2-5y=0\]Step 1: For \(x^2+y^2-8x+4=0\), \(g=-4,\ f=0\), so the centre is \((4,0)\).
Step 2: A concentric circle has the same centre \((4,0)\). It passes through \((4,2)\):
\[r^2=(4-4)^2+(2-0)^2=4\]Step 3: Equation:
\[(x-4)^2+y^2=4\]\[x^2-8x+16+y^2-4=0\]Step 1: For \(x^2+y^2-6x-10y-2=0\), \(g=-3,\ f=-5\), so its centre is \((3,5)\).
Step 2: The required circle has centre \((-1,2)\) and passes through \((3,5)\):
\[r^2=(3+1)^2+(5-2)^2=16+9=25\]Step 3: Equation:
\[(x+1)^2+(y-2)^2=25\]\[x^2+2x+1+y^2-4y+4-25=0\]Distance from the centre \(C(2,-1)\) to the student \(P(8,-1)\):
\[d=\sqrt{(8-2)^2+(-1+1)^2}=\sqrt{36}=6\text{ m}\]Since \(d=6<7=r\), the student is inside the flower bed.
1. A. d B. c C. c D. d E. d
2. a. \(x^2+y^2+8x-2y+1=0\) b. \(x^2+y^2+6x+8y-24=0\) c. \(x^2+y^2-4x+2y-4=0\) d. \(x^2+y^2-20y+75=0\) e. \(x^2+y^2+2ax+2ay=0\) f. \(x^2+y^2-2ax-2by+a^2=0\)
3. a. \((0,0),5\) b. \((0,0),6\) c. \((0,-2),6\) d. \((0,1),5\) e. \((5,-2),4\) f. \((-3,4),6\) g. \((2,3),\frac5{\sqrt2}\) h. \(\left(2,-\frac13\right),4\)
4. a. \(x^2+y^2-y-a^2-2=0\) b. \(x^2+y^2-5x+13y+48=0\) c. \(x^2+y^2+3x+5y-14=0\)
5. a. \((2,-3)\) b. \((-1,-9)\) 6. a. \((3,4)\) b. \(6\) c. \((8,11)\) 7. a. \(\left(-1,\frac12\right)\) b. \(\frac12\) c. \((-7,-2)\)
8. a. \(x^2+y^2+10x+10y+25=0\) b. \(x^2+y^2-12x+12y+36=0\)
9. a. \(x^2+y^2-4x-6y=0\) b. \(x^2+y^2-5x-10y=0\) c. \(x^2+y^2-12x-8y=0\)
10. a. \(x^2+y^2-2x-2y-8=0\) b. \(x^2+y^2-6x-8y+15=0\) 11. a. \(x^2+y^2-4x-2y=0\) b. \(x^2+y^2-5x-5y=0\)
12. \(x^2+y^2-8x+12=0\) 13. \(x^2+y^2+2x-4y-20=0\) 14. a. \(x^2+y^2-4x+2y-44=0\) b. Inside
6. Frequently Asked Questions on Class 10 Optional Mathematics Coordinate Geometry
What is covered in Class 10 Optional Mathematics Coordinate Geometry Exercise 8.1? (कक्षा १० ऐच्छिक गणित निर्देशाङ्क ज्यामिति अभ्यास ८.१ मा के के समावेश छ ?)
Exercise 8.1 covers the equation of a straight line in two-point form, finding the equation from a graph, collinear points, the midpoint of a segment, the length and equation of a median, and the distance between two points. This page solves all 37 questions. Exercise 8.2 covers the angle between two lines, parallel and perpendicular lines, perpendicular bisectors and lines at a given angle, with another 37 solved questions.
What is the two-point form of a straight line? (सिधा रेखाको दुई बिन्दु रूप भनेको के हो ?)
The line through \(P_1(x_1,y_1)\) and \(P_2(x_2,y_2)\) has the equation \(y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\).
How do I find the slope of a line joining two points? (दुई बिन्दु जोड्ने रेखाको ढाल कसरी निकाल्ने ?)
\(m=\dfrac{y_2-y_1}{x_2-x_1}\). The slope is not defined when \(x_1=x_2\), because the line is vertical.
How do I write the answer in the general form \(Ax+By+C=0\)? (उत्तरलाई \(Ax+By+C=0\) रूपमा कसरी लेख्ने ?)
Cross-multiply to remove the fraction, expand the brackets and move every term to one side. Keep the coefficient of \(x\) positive. Example: Exercise 8.1, Question 4(b) gives \(x+6y-22=0\).
What is the equation of the line through \((a,0)\) and \((0,b)\)? (\((a,0)\) र \((0,b)\) भएर जाने रेखाको समीकरण के हो ?)
It is \(\dfrac xa+\dfrac yb=1\), the intercept form, with \(x\)-intercept \(a\) and \(y\)-intercept \(b\) (Exercise 8.1, Question 1C).
How do I prove that three points are collinear? (तीन बिन्दु समरेखीय छन् भनी कसरी प्रमाणित गर्ने ?)
Find the slopes of \(AB\) and \(BC\). If \(m_{AB}=m_{BC}\), the points lie on one line because \(B\) is common to both segments (Exercise 8.1, Question 5).
What is the midpoint formula? (मध्यबिन्दुको सूत्र के हो ?)
The midpoint of \((x_1,y_1)\) and \((x_2,y_2)\) is \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
What is the distance formula? (दुरीको सूत्र के हो ?)
The distance between \((x_1,y_1)\) and \((x_2,y_2)\) is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) (Exercise 8.1, Question 11c gives \(4\sqrt2\)).
How do I find the equation of a median of a triangle? (त्रिभुजको मध्यिकाको समीकरण कसरी निकाल्ने ?)
First find the midpoint of the opposite side, then use the two-point form with the vertex and that midpoint (Exercise 8.1, Questions 9 and 10).
What if the two points have the same x-coordinate? (दुवै बिन्दुको x-निर्देशाङ्क बराबर भए के हुन्छ ?)
The line is vertical and its equation is \(x=x_1\). In Exercise 8.1, Question 8(c), \(P(1,0)\) and \(M(1,2)\) give \(x=1\).
Are these questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू नेपालमा SEE को तयारीका लागि उपयोगी छन् ?)
Yes, for students who study Optional Mathematics. This set covers every question type of Exercise 8.1: equations of lines, collinearity, midpoints, distances and medians. Practising them builds the skills needed for SEE-style questions.
Are the solutions available in Nepali? (के समाधान नेपाली भाषामा पनि उपलब्ध छन् ?)
Every question is shown in English with the Nepali question in brackets. Multiple choice solutions also have Nepali in brackets. All other solutions are in English only, because the steps and symbols of mathematics are the same in both languages.
What is the formula for the angle between two lines in Exercise 8.2? (दुई रेखाबिचको कोणको सूत्र के हो ?)
If the slopes are \(m_1\) and \(m_2\), then \(\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|\). This gives the acute angle; the obtuse angle is \(180^\circ-\theta\) (Exercise 8.2, Questions 1(A) and 3).
What are the conditions for parallel and perpendicular lines? (समानान्तर र लम्ब रेखाका सर्तहरू के के हन् ?)
Parallel lines have equal slopes, \(m_1=m_2\). Perpendicular lines satisfy \(m_1m_2=-1\) (Exercise 8.2, Questions 1(B) and 1(C)).
How do I find the equation of a perpendicular bisector? (लम्बार्धकको समीकरण कसरी पत्ता लगाउने ?)
Find the midpoint of the segment, find the slope of the segment, take its negative reciprocal, then use the point-slope form through the midpoint. Example: Exercise 8.2, Question 13(a) gives \(3x+2y-13=0\).
How do I find lines that make a given angle with a line? (दिइएको कोण बनाउने रेखाहरू कसरी पत्ता लगाउने ?)
Put \(\tan\theta=\pm\dfrac{m_1-m}{1+m_1m}\) with the given slope \(m_1\), solve both signs for \(m\), then use the given point. Each sign gives one line (Exercise 8.2, Question 14).
How is each conic section formed from a cone? (सोलीबाट प्रत्येक साङ्ङ्किक क्षेत्र कसरी बन्छ ?)
With semi-vertical angle \(\alpha\) and plane angle \(\theta\): circle when \(\theta=90^\circ\), ellipse when \(\alpha<\theta<90^\circ\), parabola when \(\theta=\alpha\), hyperbola when \(\theta<\alpha\) (Exercise 8.3, Question 5).
How do I find the centre and radius from \(x^2+y^2+2gx+2fy+c=0\)? (\(x^2+y^2+2gx+2fy+c=0\) बाट केन्द्र र अर्धव्यास कसरी निकाल्ने ?)
The centre is \((-g,-f)\) and the radius is \(\sqrt{g^2+f^2-c}\). Divide by the coefficient of \(x^2\) first if it is not \(1\) (Exercise 8.4, Questions 3(g) and 3(h).)
How do I find the equation of a circle from the end points of a diameter? (व्यासका दुई छेउका बिन्दुबाट वृत्तको समीकरण कसरी निकाल्ने ?)
Use \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\). Example: Exercise 8.4, Question 4(c) gives \(x^2+y^2+3x+5y-14=0\).
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