Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 9 Exercise 9.2 This exercise requires finding the equations of straight lines using various given conditions, such as point-slope form, two-point form, intercept form, and normal form.
Finding the Equation of a Line (Exercises 1–8)
1. Equations for the $x$- and $y$-axes.
- $x$-axis: Every point on the $x$-axis has a $y$-coordinate of 0.$$\mathbf{y = 0}$$
- $y$-axis: Every point on the $y$-axis has an $x$-coordinate of 0.$$\mathbf{x = 0}$$
2. Passing through the point $(-4, 3)$ with slope $\frac{1}{2}$.
We use the point-slope form: $y – y_1 = m(x – x_1)$.
Given: $(x_1, y_1) = (-4, 3)$ and $m = \frac{1}{2}$.
$$y – 3 = \frac{1}{2}(x – (-4))$$
$$y – 3 = \frac{1}{2}(x + 4)$$
Multiply by 2 to clear the fraction:
$$2(y – 3) = x + 4$$
$$2y – 6 = x + 4$$
Rearrange into the general form:
$$\mathbf{x – 2y + 10 = 0}$$
3. Passing through $(0, 0)$ with slope $m$.
We use the slope-intercept form ($y = mx + c$), where the $y$-intercept $c$ is 0 since it passes through the origin.
$$y = mx + 0$$
$$\mathbf{y = mx}$$
(Alternatively, using point-slope form with $(0, 0)$: $y – 0 = m(x – 0) \implies y = mx$)
4. Passing through $(\sqrt{3}, 2)$ and inclined with the $x$-axis at an angle of $75^\circ$.
First, find the slope $m$:
$$m = \tan(75^\circ)$$
Recall $\tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B}$. Use $75^\circ = 45^\circ + 30^\circ$.
$$m = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 – \tan 45^\circ \tan 30^\circ}$$
$$m = \frac{1 + 1/\sqrt{3}}{1 – 1/\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} – 1}$$
Rationalize the denominator:
$$m = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} – 1)(\sqrt{3} + 1)} = \frac{3 + 1 + 2\sqrt{3}}{3 – 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$
Now use the point-slope form $y – y_1 = m(x – x_1)$ with $(x_1, y_1) = (\sqrt{3}, 2)$ and $m = 2 + \sqrt{3}$.
$$y – 2 = (2 + \sqrt{3})(x – \sqrt{3})$$
$$y – 2 = (2 + \sqrt{3})x – \sqrt{3}(2 + \sqrt{3})$$
$$y – 2 = (2 + \sqrt{3})x – 2\sqrt{3} – 3$$
Rearrange:
$$\mathbf{(2 + \sqrt{3})x – y + 5 – 2\sqrt{3} = 0}$$
5. Intersecting the $x$-axis at a distance of 3 units to the left of origin with slope $-2$.
- Intersecting $x$-axis at 3 units to the left means the $x$-intercept is $a = -3$. The line passes through $(-3, 0)$.
- Slope $m = -2$.
Using the point-slope form $y – y_1 = m(x – x_1)$ with $(-3, 0)$:
$$y – 0 = -2(x – (-3))$$
$$y = -2(x + 3)$$
$$y = -2x – 6$$
Rearrange:
$$\mathbf{2x + y + 6 = 0}$$
6. Intersecting the $y$-axis at a distance of 2 units above the origin and making an angle of $30^\circ$ with positive direction of the $x$-axis.
- Intersecting $y$-axis at 2 units above the origin means the $y$-intercept is $c = 2$.
- Angle with positive $x$-axis is $\theta = 30^\circ$, so slope $m = \tan(30^\circ) = \frac{1}{\sqrt{3}}$.
We use the slope-intercept form: $y = mx + c$.
$$y = \frac{1}{\sqrt{3}}x + 2$$
Multiply by $\sqrt{3}$:
$$\sqrt{3}y = x + 2\sqrt{3}$$
Rearrange:
$$\mathbf{x – \sqrt{3}y + 2\sqrt{3} = 0}$$
7. Passing through the points $(-1, 1)$ and $(2, -4)$.
We use the two-point form: $y – y_1 = \frac{y_2 – y_1}{x_2 – x_1} (x – x_1)$.
First, find the slope $m$:
$$m = \frac{-4 – 1}{2 – (-1)} = \frac{-5}{3}$$
Now use point-slope form with $(-1, 1)$:
$$y – 1 = -\frac{5}{3}(x – (-1))$$
$$y – 1 = -\frac{5}{3}(x + 1)$$
Multiply by 3:
$$3(y – 1) = -5(x + 1)$$
$$3y – 3 = -5x – 5$$
Rearrange:
$$\mathbf{5x + 3y + 2 = 0}$$
8. Vertices of $\triangle PQR$ are $P(2, 1)$, $Q(-2, 3)$ and $R(4, 5)$. Find equation of the median through the vertex $R$.
A median from a vertex connects that vertex to the mid-point of the opposite side. The median from $R(4, 5)$ goes to the mid-point $M$ of side $PQ$.
1. Find the mid-point $M$ of $PQ$: $P(2, 1), Q(-2, 3)$.
$$M = \left(\frac{2 + (-2)}{2}, \frac{1 + 3}{2}\right) = \left(\frac{0}{2}, \frac{4}{2}\right) = (0, 2)$$
2. Find the equation of the line passing through $R(4, 5)$ and $M(0, 2)$.
Use the two-point form. First, find the slope $m$:
$$m = \frac{2 – 5}{0 – 4} = \frac{-3}{-4} = \frac{3}{4}$$
Now use the slope-intercept form ($y = mx + c$) since $M(0, 2)$ gives the $y$-intercept $c=2$:
$$y = \frac{3}{4}x + 2$$
Multiply by 4:
$$4y = 3x + 8$$
Rearrange:
$$\mathbf{3x – 4y + 8 = 0}$$
Perpendicular and Parallel Lines (Exercises 9–14)
9. Find the equation of the line passing through $(-3, 5)$ and perpendicular to the line through $(2, 5)$ and $(-3, 6)$.
1. Find the slope $m_L$ of the line $L$ joining $(2, 5)$ and $(-3, 6)$.
$$m_L = \frac{6 – 5}{-3 – 2} = \frac{1}{-5} = -\frac{1}{5}$$
2. Find the slope $m_{req}$ of the required perpendicular line.
For perpendicular lines, $m_{req} \cdot m_L = -1$.
$$m_{req} = -\frac{1}{m_L} = -\frac{1}{-1/5} = 5$$
3. Find the equation of the line with $m_{req} = 5$ passing through $(-3, 5)$.
Use the point-slope form:
$$y – 5 = 5(x – (-3))$$
$$y – 5 = 5x + 15$$
Rearrange:
$$\mathbf{5x – y + 20 = 0}$$
10. A line perpendicular to the line segment joining $(1, 0)$ and $(2, 3)$ divides it in the ratio $1:n$. Find the equation of the line.
The required line passes through a point $P$ and is perpendicular to the segment $AB$.
1. Find the coordinates of the division point $P$.
The point $P(x, y)$ divides $A(1, 0)$ and $B(2, 3)$ in the ratio $1:n$.
$$x = \frac{1(2) + n(1)}{1 + n} = \frac{2 + n}{1 + n}$$
$$y = \frac{1(3) + n(0)}{1 + n} = \frac{3}{1 + n}$$
The required line passes through $P\left(\frac{2 + n}{1 + n}, \frac{3}{1 + n}\right)$.
2. Find the slope $m_{AB}$ of the segment $AB$.
$$m_{AB} = \frac{3 – 0}{2 – 1} = 3$$
3. Find the slope $m_{req}$ of the required perpendicular line.
$$m_{req} = -\frac{1}{m_{AB}} = -\frac{1}{3}$$
4. Find the equation of the line.
Use the point-slope form $y – y_P = m_{req}(x – x_P)$:
$$y – \frac{3}{1 + n} = -\frac{1}{3} \left(x – \frac{2 + n}{1 + n}\right)$$
Multiply by $3(1 + n)$:
$$3(1 + n)y – 3(3) = -(1 + n)x + (2 + n)$$
$$3(1 + n)y – 9 = -(1 + n)x + 2 + n$$
Rearrange:
$$(1 + n)x + 3(1 + n)y = 9 + 2 + n$$
$$\mathbf{(n + 1)x + 3(n + 1)y = n + 11}$$
11. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point $(2, 3)$.
Equal intercepts means $a = b$. We use the intercept form: $\frac{x}{a} + \frac{y}{b} = 1$.
$$\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a$$
The line passes through $(2, 3)$. Substitute $(x, y) = (2, 3)$ into the equation:
$$2 + 3 = a \implies a = 5$$
The equation of the line is $x + y = 5$.
$$\mathbf{x + y – 5 = 0}$$
12. Find equation of the line passing through the point $(2, 2)$ and cutting off intercepts on the axes whose sum is 9.
Let the intercepts be $a$ (on $x$-axis) and $b$ (on $y$-axis).
Given: $a + b = 9 \implies b = 9 – a$.
The equation is $\frac{x}{a} + \frac{y}{b} = 1$.
The line passes through $(2, 2)$. Substitute $(x, y) = (2, 2)$:
$$\frac{2}{a} + \frac{2}{b} = 1$$
Substitute $b = 9 – a$:
$$\frac{2}{a} + \frac{2}{9 – a} = 1$$
Multiply by $a(9 – a)$:
$$2(9 – a) + 2a = a(9 – a)$$
$$18 – 2a + 2a = 9a – a^2$$
$$18 = 9a – a^2$$
$$a^2 – 9a + 18 = 0$$
Factor the quadratic: $(a – 3)(a – 6) = 0$.
$$a = 3 \quad \text{or} \quad a = 6$$
This gives two possible lines:
- Case 1: $a = 3$. Then $b = 9 – 3 = 6$.$$\frac{x}{3} + \frac{y}{6} = 1 \implies 2x + y = 6 \implies \mathbf{2x + y – 6 = 0}$$
- Case 2: $a = 6$. Then $b = 9 – 6 = 3$.$$\frac{x}{6} + \frac{y}{3} = 1 \implies x + 2y = 6 \implies \mathbf{x + 2y – 6 = 0}$$
13. Line through $(0, 2)$ making an angle $\frac{2\pi}{3}$ with the positive $x$-axis. Also, find the equation of a line parallel to it and crossing the $y$-axis at 2 units below the origin.
1. Equation of the first line ($L_1$):
- Passes through $(0, 2)$, so $y$-intercept $c_1 = 2$.
- Angle $\theta = \frac{2\pi}{3} = 120^\circ$.
- Slope $m_1 = \tan(120^\circ) = -\sqrt{3}$.
- Using $y = mx + c$:$$y = -\sqrt{3}x + 2$$$$\mathbf{\sqrt{3}x + y – 2 = 0}$$
2. Equation of the parallel line ($L_2$):
- $L_2$ is parallel to $L_1$, so $m_2 = m_1 = -\sqrt{3}$.
- $L_2$ crosses the $y$-axis 2 units below the origin, so $y$-intercept $c_2 = -2$.
- Using $y = mx + c$:$$y = -\sqrt{3}x – 2$$$$\mathbf{\sqrt{3}x + y + 2 = 0}$$
14. The perpendicular from the origin to a line meets it at the point $(-2, 9)$, find the equation of the line.
The perpendicular line is the segment from the origin $O(0, 0)$ to $P(-2, 9)$. The required line $L$ is perpendicular to $OP$ and passes through $P(-2, 9)$.
1. Find the slope $m_{OP}$ of the perpendicular segment $OP$.
$$m_{OP} = \frac{9 – 0}{-2 – 0} = -\frac{9}{2}$$
2. Find the slope $m_L$ of the required line $L$.
$$m_L = -\frac{1}{m_{OP}} = -\frac{1}{-9/2} = \frac{2}{9}$$
3. Find the equation of the line $L$ with $m_L = \frac{2}{9}$ passing through $P(-2, 9)$.
Use the point-slope form:
$$y – 9 = \frac{2}{9}(x – (-2))$$
$$y – 9 = \frac{2}{9}(x + 2)$$
Multiply by 9:
$$9(y – 9) = 2(x + 2)$$
$$9y – 81 = 2x + 4$$
Rearrange:
$$\mathbf{2x – 9y + 85 = 0}$$
Word Problems and Proofs (Exercises 15–19)
15. Linear relationship between length $L$ and temperature $C$.
Given two points $(C, L)$: $(20, 124.942)$ and $(110, 125.134)$.
The relationship is linear, so we use the two-point form to express $L$ in terms of $C$.
1. Find the slope $m = \frac{L_2 – L_1}{C_2 – C_1}$.
$$m = \frac{125.134 – 124.942}{110 – 20} = \frac{0.192}{90} = \frac{192}{90000} = \frac{8}{3750} = \frac{4}{1875} \approx 0.002133$$
2. Use the point-slope form $L – L_1 = m(C – C_1)$ with $(20, 124.942)$.
$$L – 124.942 = 0.002133 (C – 20)$$
$$L = 0.002133 C – 0.002133(20) + 124.942$$
$$L = 0.002133 C – 0.04266 + 124.942$$
$$L = \mathbf{0.002133 C + 124.89934}$$
(Using fractional slope for accuracy):
$$L – 124.942 = \frac{4}{1875} (C – 20)$$
$$\mathbf{1875L – 4C = 234133.75}$$
(or $L = \frac{4}{1875} C + \frac{234277.5}{1875}$)
16. Linear relationship between selling price ($P$) and demand ($D$).
Given two points $(P, D)$: $(14, 980)$ and $(16, 1220)$.
We need to find the demand $D$ when the price $P = 17$.
1. Find the slope $m = \frac{D_2 – D_1}{P_2 – P_1}$.
$$m = \frac{1220 – 980}{16 – 14} = \frac{240}{2} = 120$$
(The demand increases by 120 litres for every Rs 1 increase in price.)
2. Use the point-slope form $D – D_1 = m(P – P_1)$ with $(14, 980)$.
$$D – 980 = 120(P – 14)$$
3. Substitute $P = 17$ to find $D$.
$$D – 980 = 120(17 – 14)$$
$$D – 980 = 120(3)$$
$$D – 980 = 360$$
$$D = 980 + 360 = \mathbf{1340}$$
He could sell 1340 litres weekly at Rs 17/litre.
17. $P(a, b)$ is the mid-point of a line segment between axes. Show that equation of the line is $\frac{x}{2a} + \frac{y}{2b} = 1$.
Let the line cut the axes at $A(A, 0)$ (x-intercept) and $B(0, B)$ (y-intercept).
The mid-point $P(a, b)$ is given by:
$$a = \frac{A + 0}{2} \implies A = 2a$$
$$b = \frac{0 + B}{2} \implies B = 2b$$
The equation of the line in intercept form is $\frac{x}{A} + \frac{y}{B} = 1$.
Substituting the intercepts $A=2a$ and $B=2b$:
$$\mathbf{\frac{x}{2a} + \frac{y}{2b} = 1}$$
(Proved)
18. Point $R(h, k)$ divides a line segment between the axes in the ratio $1:2$. Find equation of the line.
Let the line cut the axes at $A(A, 0)$ and $B(0, B)$.
The point $R(h, k)$ divides the segment $AB$ in the ratio $1:2$.
Using the section formula:
$$h = \frac{1(0) + 2(A)}{1 + 2} = \frac{2A}{3} \implies A = \frac{3h}{2}$$
$$k = \frac{1(B) + 2(0)}{1 + 2} = \frac{B}{3} \implies B = 3k$$
The equation of the line in intercept form is $\frac{x}{A} + \frac{y}{B} = 1$.
Substituting the intercepts $A = \frac{3h}{2}$ and $B = 3k$:
$$\frac{x}{3h/2} + \frac{y}{3k} = 1$$
$$\mathbf{\frac{2x}{3h} + \frac{y}{3k} = 1} \quad \text{or} \quad \mathbf{2kx + hy = 3hk}$$
19. Prove that the three points $A(3, 0)$, $B(-2, -2)$ and $C(8, 2)$ are collinear.
Points are collinear if they lie on the same line, which means the slope between any pair of points must be equal.
1. Slope of $AB$, $m_{AB}$:
$$m_{AB} = \frac{-2 – 0}{-2 – 3} = \frac{-2}{-5} = \frac{2}{5}$$
2. Slope of $BC$, $m_{BC}$:
$$m_{BC} = \frac{2 – (-2)}{8 – (-2)} = \frac{4}{10} = \frac{2}{5}$$
Since $m_{AB} = m_{BC} = \frac{2}{5}$, the points share the same slope and thus lie on the same line.
The points are collinear.
(Alternatively, find the equation of $AB$ and check if $C$ lies on it. Line $AB$: $y – 0 = \frac{2}{5}(x – 3) \implies 5y = 2x – 6 \implies 2x – 5y – 6 = 0$. Check $C(8, 2)$: $2(8) – 5(2) – 6 = 16 – 10 – 6 = 0$. True.)
