Straight Lines
Exercise 9.2 Solutions
Key Equations:
Point-Slope: $y – y_1 = m(x – x_1)$
Two-Point: $y – y_1 = \frac{y_2 – y_1}{x_2 – x_1}(x – x_1)$
Slope-Intercept: $y = mx + c$
Intercept Form: $\frac{x}{a} + \frac{y}{b} = 1$
Point-Slope: $y – y_1 = m(x – x_1)$
Two-Point: $y – y_1 = \frac{y_2 – y_1}{x_2 – x_1}(x – x_1)$
Slope-Intercept: $y = mx + c$
Intercept Form: $\frac{x}{a} + \frac{y}{b} = 1$
Q1
1. Write the equations for the x-and y-axes.
x-axis
Every point on the x-axis has y-coordinate 0. Equation: $y = 0$.
y-axis
Every point on the y-axis has x-coordinate 0. Equation: $x = 0$.
$y = 0$ and $x = 0$
Q2
2. Passing through $(-4, 3)$ with slope $1/2$.
Formula
$y – y_1 = m(x – x_1)$
Substitute
$y – 3 = \frac{1}{2}(x – (-4)) \Rightarrow 2(y – 3) = x + 4$
Simplify
$2y – 6 = x + 4 \Rightarrow x – 2y + 10 = 0$
$x – 2y + 10 = 0$
Q3
3. Passing through $(0, 0)$ with slope m.
Formula
$y – 0 = m(x – 0)$
$y = mx$
Q4
4. Passing through $(2, 2\sqrt{3})$ and inclined with x-axis at $75^\circ$.
Slope m
$\tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{1 + 1/\sqrt{3}}{1 – 1/\sqrt{3}} = 2 + \sqrt{3}$.
Equation
$y – 2\sqrt{3} = (2+\sqrt{3})(x – 2)$
Simplify
$y – 2\sqrt{3} = (2+\sqrt{3})x – 4 – 2\sqrt{3}$
Result
$(2+\sqrt{3})x – y – 4 = 0$
$(2+\sqrt{3})x – y – 4 = 0$
Q5
5. Intersecting x-axis at 3 units to the left ($d = -3$) with slope -2.
Formula
$y = m(x – d)$
Substitute
$y = -2(x – (-3)) \Rightarrow y = -2(x + 3)$
Simplify
$y = -2x – 6 \Rightarrow 2x + y + 6 = 0$
$2x + y + 6 = 0$
Q6
6. Intersecting y-axis at 2 units above origin ($c = 2$), angle $30^\circ$.
Slope m
$\tan 30^\circ = \frac{1}{\sqrt{3}}$.
Formula
$y = mx + c \Rightarrow y = \frac{1}{\sqrt{3}}x + 2$
Simplify
$\sqrt{3}y = x + 2\sqrt{3} \Rightarrow x – \sqrt{3}y + 2\sqrt{3} = 0$
$x – \sqrt{3}y + 2\sqrt{3} = 0$
Q7
7. Passing through points $(-1, 1)$ and $(2, -4)$.
Slope
$m = \frac{-4 – 1}{2 – (-1)} = \frac{-5}{3}$
Equation
$y – 1 = -\frac{5}{3}(x + 1)$
Simplify
$3(y – 1) = -5(x + 1) \Rightarrow 3y – 3 = -5x – 5$
$5x + 3y + 2 = 0$
Q8
8. Find equation of median through R for $\Delta PQR$ with $P(2,1), Q(-2,3), R(4,5)$.
Midpoint M
Midpoint of PQ = $(\frac{2-2}{2}, \frac{1+3}{2}) = (0, 2)$.
Line RM
Passes through $R(4, 5)$ and $M(0, 2)$.
Slope
$m = \frac{2 – 5}{0 – 4} = \frac{-3}{-4} = \frac{3}{4}$
Equation
$y – 2 = \frac{3}{4}(x – 0) \Rightarrow 4y – 8 = 3x$
$3x – 4y + 8 = 0$
Q9
9. Line through $(-3, 5)$ and perpendicular to line joining $(2, 5)$ and $(-3, 6)$.
Slope 1
Slope of line joining points: $m_1 = \frac{6-5}{-3-2} = \frac{1}{-5}$.
Slope 2
Perpendicular slope $m_2 = \frac{-1}{-1/5} = 5$.
Equation
$y – 5 = 5(x – (-3)) \Rightarrow y – 5 = 5(x + 3)$
Simplify
$y – 5 = 5x + 15 \Rightarrow 5x – y + 20 = 0$
$5x – y + 20 = 0$
Q10
10. Line perp to segment joining $(1, 0)$ and $(2, 3)$ divides it in 1:n. Find eq.
Point P
Section Formula (1:n): $x = \frac{1(2)+n(1)}{1+n}, y = \frac{1(3)+n(0)}{1+n}$.
$P(\frac{n+2}{n+1}, \frac{3}{n+1})$.
$P(\frac{n+2}{n+1}, \frac{3}{n+1})$.
Slope
Slope of segment $m_1 = \frac{3-0}{2-1} = 3$. Perp slope $m_2 = -1/3$.
Equation
$y – \frac{3}{n+1} = -\frac{1}{3}(x – \frac{n+2}{n+1})$
Simplify
Multiply by $3(n+1)$: $3((n+1)y – 3) = -1((n+1)x – (n+2))$
Rearrange
$3(n+1)y – 9 = -(n+1)x + n + 2$
$(n+1)x + 3(n+1)y = n + 11$
Q11
11. Line passing through $(2, 3)$ cutting equal intercepts on axes.
Intercepts
Equal intercepts $\Rightarrow a = b$. Form: $\frac{x}{a} + \frac{y}{a} = 1 \Rightarrow x + y = a$.
Substitute
Passes through $(2, 3)$. $2 + 3 = a \Rightarrow a = 5$.
$x + y – 5 = 0$
Q12
12. Line through $(2, 2)$, sum of intercepts is 9.
Conditions
$\frac{x}{a} + \frac{y}{b} = 1$ and $a + b = 9 \Rightarrow b = 9 – a$.
Substitute
$\frac{2}{a} + \frac{2}{9-a} = 1 \Rightarrow 2(9-a) + 2a = a(9-a)$.
Solve
$18 – 2a + 2a = 9a – a^2 \Rightarrow a^2 – 9a + 18 = 0$.
Factors
$(a-6)(a-3) = 0$. Cases: $a=6, b=3$ OR $a=3, b=6$.
$x + 2y – 6 = 0$ OR $2x + y – 6 = 0$
Q13
13. Line through $(0, 2)$ with angle $2\pi/3$. Find parallel line crossing y-axis at -2.
Slope
$m = \tan(120^\circ) = -\sqrt{3}$.
Line 1
$y – 2 = -\sqrt{3}(x – 0) \Rightarrow \sqrt{3}x + y – 2 = 0$.
Line 2
Parallel ($m = -\sqrt{3}$) and passes through $(0, -2)$.
Eq 2
$y = -\sqrt{3}x – 2 \Rightarrow \sqrt{3}x + y + 2 = 0$.
Line 1: $\sqrt{3}x + y – 2 = 0$, Line 2: $\sqrt{3}x + y + 2 = 0$
Q14
14. Perpendicular from origin meets line at $(-2, 9)$. Find equation.
Slope OP
Slope of normal (Origin to P) is $\frac{9-0}{-2-0} = -\frac{9}{2}$.
Line Slope
Line is perpendicular to OP. $m = \frac{-1}{-9/2} = \frac{2}{9}$.
Equation
$y – 9 = \frac{2}{9}(x + 2)$.
Simplify
$9y – 81 = 2x + 4 \Rightarrow 2x – 9y + 85 = 0$.
$2x – 9y + 85 = 0$
Q15
15. Linear relation L vs C. L=124.942 at C=20; L=125.134 at C=110.
Points
$(20, 124.942)$ and $(110, 125.134)$.
Slope
$m = \frac{125.134 – 124.942}{110 – 20} = \frac{0.192}{90}$.
Equation
$L – 124.942 = \frac{0.192}{90}(C – 20)$.
$L = \frac{0.192}{90}(C – 20) + 124.942$
Q16
16. Sell 980L at Rs 14, 1220L at Rs 16. Linear relation. Find litres at Rs 17.
Points
$(14, 980)$ and $(16, 1220)$. Let Price P be x, Litres L be y.
Slope
$m = \frac{1220 – 980}{16 – 14} = \frac{240}{2} = 120$.
Equation
$L – 980 = 120(P – 14)$.
Calculate
At $P=17$: $L – 980 = 120(17 – 14) = 120(3) = 360$.
$L = 980 + 360 = 1340$.
$L = 980 + 360 = 1340$.
1340 Litres
Q17
17. P(a, b) is mid-point of segment between axes. Show $\frac{x}{a} + \frac{y}{b} = 2$.
Intercepts
Let line meet axes at $A(h, 0)$ and $B(0, k)$.
Midpoint
Midpoint of AB is $(\frac{h}{2}, \frac{k}{2})$. Given as $(a, b)$.
Relation
$a = h/2 \Rightarrow h = 2a$. $b = k/2 \Rightarrow k = 2b$.
Equation
Intercept form: $\frac{x}{h} + \frac{y}{k} = 1 \Rightarrow \frac{x}{2a} + \frac{y}{2b} = 1$.
$\frac{x}{a} + \frac{y}{b} = 2$. Proved.
Q18
18. Point R(h, k) divides segment between axes in ratio 1:2. Find eq.
Points
Let points on axes be $A(a, 0)$ and $B(0, b)$.
Section
Ratio 1:2. $h = \frac{1(0) + 2(a)}{3} \Rightarrow 3h = 2a \Rightarrow a = \frac{3h}{2}$.
Section
$k = \frac{1(b) + 2(0)}{3} \Rightarrow 3k = b \Rightarrow b = 3k$.
Equation
$\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow \frac{x}{3h/2} + \frac{y}{3k} = 1 \Rightarrow \frac{2x}{3h} + \frac{y}{3k} = 1$.
$2kx + hy = 3hk$
Q19
19. Prove (3, 0), (-2, -2), (8, 2) are collinear using equation of line.
Slope 1
Slope between (3,0) and (-2,-2): $m = \frac{-2-0}{-2-3} = \frac{-2}{-5} = \frac{2}{5}$.
Line Eq
$y – 0 = \frac{2}{5}(x – 3) \Rightarrow 5y = 2x – 6 \Rightarrow 2x – 5y – 6 = 0$.
Check Pt 3
Put $(8, 2)$ in eq: $2(8) – 5(2) – 6 = 16 – 10 – 6 = 0$.
Point satisfies equation. Hence collinear.