NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3

Q1

1. Reduce to slope-intercept form ($y=mx+c$), find slope and y-intercept:
(i) $x + 7y = 0$, (ii) $6x + 3y – 5 = 0$, (iii) $y = 0$.

(i) $7y = -x \Rightarrow y = -\frac{1}{7}x + 0$.
Slope $m = -1/7$, y-int $c = 0$.

(ii) $3y = -6x + 5 \Rightarrow y = -2x + \frac{5}{3}$.
Slope $m = -2$, y-int $c = 5/3$.

(iii) $y = 0x + 0$.
Slope $m = 0$, y-int $c = 0$.

Q2

2. Reduce to intercept form ($\frac{x}{a} + \frac{y}{b} = 1$), find intercepts:
(i) $3x + 2y – 12 = 0$, (ii) $4x – 3y = 6$, (iii) $3y + 2 = 0$.

(i) $3x + 2y = 12 \Rightarrow \frac{3x}{12} + \frac{2y}{12} = 1 \Rightarrow \frac{x}{4} + \frac{y}{6} = 1$.
$a=4, b=6$.

(ii) $\frac{4x}{6} – \frac{3y}{6} = 1 \Rightarrow \frac{x}{1.5} + \frac{y}{-2} = 1$.
$a=1.5, b=-2$.

(iii) $3y = -2 \Rightarrow y = -2/3$. No x-term, so parallel to x-axis.
x-int: undefined, y-int: $-2/3$.

Q3

3. Find distance of point $(-1, 1)$ from line $12(x + 6) = 5(y – 2)$. [Image of perpendicular distance from a point to a line]

Equation $12x + 72 = 5y – 10 \Rightarrow 12x – 5y + 82 = 0$.

Formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} = \frac{|12(-1) – 5(1) + 82|}{\sqrt{12^2 + (-5)^2}}$.

Calc $\frac{|-12 – 5 + 82|}{\sqrt{144 + 25}} = \frac{|65|}{\sqrt{169}} = \frac{65}{13} = 5$.

Distance = 5 units

Q4

4. Find points on x-axis whose distance from $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.

Line Eq $4x + 3y = 12 \Rightarrow 4x + 3y – 12 = 0$.

Point Point on x-axis is $(h, 0)$.

Condition $\frac{|4(h) + 3(0) – 12|}{\sqrt{4^2 + 3^2}} = 4 \Rightarrow \frac{|4h – 12|}{5} = 4$.

Solve $|4h – 12| = 20$.
Case 1: $4h – 12 = 20 \Rightarrow 4h = 32 \Rightarrow h = 8$.
Case 2: $4h – 12 = -20 \Rightarrow 4h = -8 \Rightarrow h = -2$.

Points: $(8, 0)$ and $(-2, 0)$

Q5

5. Distance between parallel lines:
(i) $15x + 8y – 34 = 0$ and $15x + 8y + 31 = 0$
(ii) $l(x+y) + p = 0$ and $l(x+y) – r = 0$

(i)

Formula $d = \frac{|C_1 – C_2|}{\sqrt{A^2 + B^2}} = \frac{|-34 – 31|}{\sqrt{15^2 + 8^2}} = \frac{|-65|}{\sqrt{225 + 64}}$.

Result $\frac{65}{\sqrt{289}} = \frac{65}{17}$ units.

(ii)

Rewrite $lx + ly + p = 0$ and $lx + ly – r = 0$. ($A=l, B=l$).

Result $d = \frac{|p – (-r)|}{\sqrt{l^2 + l^2}} = \frac{|p+r|}{\sqrt{2l^2}} = \frac{|p+r|}{l\sqrt{2}}$.

Q6

6. Equation of line parallel to $3x – 4y + 2 = 0$ passing through $(-2, 3)$.

Slope $4y = 3x + 2 \Rightarrow m = 3/4$. Parallel lines have same slope.

Equation $y – 3 = \frac{3}{4}(x – (-2))$.

Simplify $4(y-3) = 3(x+2) \Rightarrow 4y – 12 = 3x + 6 \Rightarrow 3x – 4y + 18 = 0$.

$3x – 4y + 18 = 0$

Q7

7. Equation of line perpendicular to $x – 7y + 5 = 0$ with x-intercept 3.

Given Slope $7y = x + 5 \Rightarrow m_1 = 1/7$.

Perp Slope $m_2 = -1/m_1 = -7$.

Point x-intercept 3 means line passes through $(3, 0)$.

Equation $y – 0 = -7(x – 3) \Rightarrow y = -7x + 21 \Rightarrow 7x + y – 21 = 0$.

$7x + y – 21 = 0$

Q8

8. Find angles between lines $y – \sqrt{3}x – 5 = 0$ and $\sqrt{3}y – x – 6 = 0$.

Slopes $L_1: y = \sqrt{3}x + 5 \Rightarrow m_1 = \sqrt{3}$.
$L_2: y = \frac{1}{\sqrt{3}}x + \frac{6}{\sqrt{3}} \Rightarrow m_2 = \frac{1}{\sqrt{3}}$.

Formula $\tan \theta = |\frac{m_2 – m_1}{1 + m_1 m_2}| = |\frac{1/\sqrt{3} – \sqrt{3}}{1 + (\sqrt{3})(1/\sqrt{3})}|$.

Calc $\tan \theta = |\frac{(1-3)/\sqrt{3}}{1+1}| = |\frac{-2/\sqrt{3}}{2}| = |-1/\sqrt{3}| = \frac{1}{\sqrt{3}}$.

Angle $\theta = 30^\circ$.

Angles: $30^\circ$ and $150^\circ$

Q9

9. Line through $(h, 3)$ and $(4, 1)$ intersects $7x – 9y – 19 = 0$ at right angle. Find h.

Slope 1 $m_1 = \frac{1 – 3}{4 – h} = \frac{-2}{4 – h}$.

Slope 2 $9y = 7x – 19 \Rightarrow m_2 = 7/9$.

Perp Cond $m_1 m_2 = -1 \Rightarrow (\frac{-2}{4-h})(\frac{7}{9}) = -1$.

Solve $\frac{-14}{9(4-h)} = -1 \Rightarrow 14 = 36 – 9h \Rightarrow 9h = 22 \Rightarrow h = 22/9$.

h = 22/9

Q10

10. Prove line through $(x_1, y_1)$ parallel to $Ax + By + C = 0$ is $A(x – x_1) + B(y – y_1) = 0$.

Given Slope Slope of $Ax + By + C = 0$ is $m = -A/B$.

Parallel Parallel line has same slope $m = -A/B$.

Equation Point-slope form: $y – y_1 = -\frac{A}{B}(x – x_1)$.

Rearrange $B(y – y_1) = -A(x – x_1) \Rightarrow A(x – x_1) + B(y – y_1) = 0$.

Proved.

Q11

11. Two lines through $(2, 3)$ intersect at $60^\circ$. Slope of one is 2. Find equation of other.

Given $m_1 = 2, \quad \theta = 60^\circ$.

Formula $\tan 60^\circ = |\frac{m_2 – 2}{1 + 2m_2}| \Rightarrow \sqrt{3} = |\frac{m – 2}{1 + 2m}|$.

Case 1 $\frac{m-2}{1+2m} = \sqrt{3} \Rightarrow m-2 = \sqrt{3} + 2\sqrt{3}m \Rightarrow m(1-2\sqrt{3}) = 2+\sqrt{3} \Rightarrow m = \frac{2+\sqrt{3}}{1-2\sqrt{3}}$.

Case 2 $\frac{m-2}{1+2m} = -\sqrt{3} \Rightarrow m-2 = -\sqrt{3} – 2\sqrt{3}m \Rightarrow m(1+2\sqrt{3}) = 2-\sqrt{3} \Rightarrow m = \frac{2-\sqrt{3}}{1+2\sqrt{3}}$.

Eq: $y – 3 = m(x – 2)$ using slopes found.

Q12

12. Find equation of right bisector of segment joining $(3, 4)$ and $(-1, 2)$. [Image of a line segment and its perpendicular bisector]

Midpoint $M = (\frac{3-1}{2}, \frac{4+2}{2}) = (1, 3)$.

Slope AB $m_{AB} = \frac{2-4}{-1-3} = \frac{-2}{-4} = \frac{1}{2}$.

Perp Slope $m_{\perp} = -2$.

Equation $y – 3 = -2(x – 1) \Rightarrow y – 3 = -2x + 2$.

$2x + y – 5 = 0$

Q13

13. Find coordinates of foot of perpendicular from $(-1, 3)$ to $3x – 4y – 16 = 0$.

Slope Line $4y = 3x – 16 \Rightarrow m = 3/4$.

Perp Slope $m’ = -4/3$. Line through $(-1, 3)$.

Perp Eq $y – 3 = -\frac{4}{3}(x + 1) \Rightarrow 3y – 9 = -4x – 4 \Rightarrow 4x + 3y – 5 = 0$.

Intersect Solve $3x – 4y – 16 = 0$ and $4x + 3y – 5 = 0$.

Solve Multiply eq1 by 3, eq2 by 4… $x = 68/25, y = -49/25$.

Foot: $(\frac{68}{25}, -\frac{49}{25})$

Q14

14. Perpendicular from origin to $y = mx + c$ meets at $(-1, 2)$. Find m and c.

Slope Normal Line connecting $(0,0)$ and $(-1, 2)$. $m_N = \frac{2-0}{-1-0} = -2$.

Slope m Line is perpendicular to normal. $m = -1/(-2) = 1/2$.

Find c Point $(-1, 2)$ lies on line. $2 = (1/2)(-1) + c \Rightarrow 2 = -0.5 + c \Rightarrow c = 2.5$.

$m = 1/2, \quad c = 5/2$

Q15

15. If p and q are perp lengths from origin to $x \cos \theta – y \sin \theta = k \cos 2\theta$ and $x \sec \theta + y \csc \theta = k$, prove $p^2 + 4q^2 = k^2$.

Dist p $p = \frac{|-k \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = |k \cos 2\theta|$. (Since denom is 1).

Dist q $q = \frac{|-k|}{\sqrt{\sec^2 \theta + \csc^2 \theta}} = \frac{k}{\sqrt{\frac{1}{\cos^2} + \frac{1}{\sin^2}}} = \frac{k}{\sqrt{\frac{1}{\sin^2 \cos^2}}} = k \sin \theta \cos \theta$.

Calc $p^2 + 4q^2 = k^2 \cos^2 2\theta + 4k^2 \sin^2 \theta \cos^2 \theta$.

Simplify $4 \sin^2 \theta \cos^2 \theta = (2 \sin \theta \cos \theta)^2 = \sin^2 2\theta$.
$k^2 (\cos^2 2\theta + \sin^2 2\theta) = k^2$.

Proved: $p^2 + 4q^2 = k^2$

Q16

16. $\Delta ABC$ with A(2,3), B(4,-1), C(1,2). Find equation and length of altitude from A.

Slope BC $m_{BC} = \frac{2 – (-1)}{1 – 4} = \frac{3}{-3} = -1$.

Alt Slope Altitude is perp to BC. $m_{alt} = 1$.

Eq Alt Through A(2,3): $y – 3 = 1(x – 2) \Rightarrow y – x – 1 = 0$.

Length Perp dist from A to line BC ($x + y – 3 = 0$).
Line BC eq: $y – 2 = -1(x – 1) \Rightarrow x + y – 3 = 0$.

Distance $d = \frac{|2 + 3 – 3|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Eq: $x – y + 1 = 0$, Length: $\sqrt{2}$

Q17

17. p is length of perpendicular from origin to line with intercepts a, b. Show $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.

Line Eq $\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow bx + ay – ab = 0$.

Dist p $p = \frac{|-ab|}{\sqrt{b^2 + a^2}} = \frac{ab}{\sqrt{a^2 + b^2}}$.

Square $p^2 = \frac{a^2 b^2}{a^2 + b^2} \Rightarrow \frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2}$.

Split $\frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}$.

Proved.

Straight Lines