NCERT Class 11 Mathematics — Straight Lines, Exercise 9.2. All 19 questions solved.

Five standard forms, each suited to the data you happen to be given:

Given	Form
A point and the slope	$y – y_1 = m(x – x_1)$
Two points	$y – y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x – x_1)$
Slope and $y$-intercept	$y = mx + c$
Both intercepts	$\dfrac xa + \dfrac yb = 1$
Anything	$\mathrm Ax + \mathrm By + \mathrm C = 0$

Key insight. Choosing the right form is most of the work. When the question talks about intercepts — questions 11, 12, 17 and 18 — use $\tfrac xa + \tfrac yb = 1$ and treat $a$ and $b$ as the unknowns; the given conditions then become two ordinary equations in $a$ and $b$. Forcing such a question into $y = mx + c$ is possible but three times longer.

In questions 1 to 8, find the equation of the line which satisfies the given conditions.

Question 1

Write the equations for the $x$- and $y$-axes.

Solution. Every point on the $x$-axis has ordinate zero, and every point with ordinate zero lies on it. So the $x$-axis is

$$y = 0$$

Similarly the $y$-axis is the set of points with abscissa zero:

$$x = 0$$

Question 2

Passing through the point $(-4, 3)$ with slope $\dfrac12$.

Solution. Point-slope form with $(x_1, y_1) = (-4, 3)$ and $m = \tfrac12$:

$$y – 3 = \frac12\left(x – (-4)\right) = \frac12(x + 4)$$

Multiplying by $2$:

$$2y – 6 = x + 4 \quad\Longrightarrow\quad x – 2y + 10 = 0$$

Question 3

Passing through $(0, 0)$ with slope $m$.

Solution. With $(x_1, y_1) = (0,0)$ the point-slope form collapses:

$$y – 0 = m(x – 0) \quad\Longrightarrow\quad y = mx$$

Every line through the origin has this shape — except the $y$-axis, whose slope is undefined.

Question 4

Passing through $\left(2,\ 2\sqrt3\right)$ and inclined with the $x$-axis at an angle of $75^\circ$.

Solution. First find the slope:

$$m = \tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{1 + \frac{1}{\sqrt3}}{1 – \frac{1}{\sqrt3}} = \frac{\sqrt3 + 1}{\sqrt3 – 1}$$

Rationalising gives $m = 2 + \sqrt3$, but the unrationalised form is more convenient here. Point-slope form:

$$y – 2\sqrt3 = \frac{\sqrt3+1}{\sqrt3-1}(x – 2)$$

Multiplying through by $\left(\sqrt3 – 1\right)$:

$$\left(\sqrt3-1\right)y – 2\sqrt3\left(\sqrt3-1\right) = \left(\sqrt3+1\right)(x-2)$$

$$\left(\sqrt3+1\right)x – \left(\sqrt3-1\right)y = 2\left(\sqrt3+1\right) – 2\sqrt3\left(\sqrt3-1\right)$$

The right side is $2\sqrt3 + 2 – 6 + 2\sqrt3 = 4\sqrt3 – 4$, so

$$\left(\sqrt3+1\right)x – \left(\sqrt3-1\right)y = 4\left(\sqrt3-1\right)$$

Question 5

Intersecting the $x$-axis at a distance of $3$ units to the left of the origin, with slope $-2$.

Solution. “Three units to the left of the origin” on the $x$-axis is the point $(-3, 0)$. With $m = -2$:

$$y – 0 = -2\left(x + 3\right) \quad\Longrightarrow\quad y = -2x – 6 \quad\Longrightarrow\quad 2x + y + 6 = 0$$

Question 6

Intersecting the $y$-axis at a distance of $2$ units above the origin and making an angle of $30^\circ$ with the positive direction of the $x$-axis.

Solution. The $y$-intercept is $c = 2$ and the slope is $m = \tan 30^\circ = \dfrac{1}{\sqrt3}$. Slope-intercept form is the natural choice:

$$y = \frac{1}{\sqrt3}x + 2$$

Multiplying by $\sqrt3$:

$$\sqrt3\,y = x + 2\sqrt3 \quad\Longrightarrow\quad x – \sqrt3\,y + 2\sqrt3 = 0$$

Question 7

Passing through the points $(-1, 1)$ and $(2, -4)$.

Solution. The slope is

$$m = \frac{-4 – 1}{2 – (-1)} = \frac{-5}{3}$$

Using the point $(-1, 1)$:

$$y – 1 = -\frac53(x + 1) \quad\Longrightarrow\quad 3y – 3 = -5x – 5 \quad\Longrightarrow\quad 5x + 3y + 2 = 0$$

Question 8

The vertices of $\triangle \mathrm{PQR}$ are $\mathrm P(2,1)$, $\mathrm Q(-2,3)$ and $\mathrm R(4,5)$. Find the equation of the median through the vertex $\mathrm R$.

Solution. A median joins a vertex to the midpoint of the opposite side, so find the midpoint of $\mathrm{PQ}$ first:

$$\mathrm M = \left(\frac{2 + (-2)}{2},\ \frac{1 + 3}{2}\right) = (0, 2)$$

Now the line through $\mathrm R(4,5)$ and $\mathrm M(0,2)$:

$$m = \frac{5 – 2}{4 – 0} = \frac34$$

$$y – 2 = \frac34(x – 0) \quad\Longrightarrow\quad 4y – 8 = 3x \quad\Longrightarrow\quad 3x – 4y + 8 = 0$$

Question 9

Find the equation of the line passing through $(-3, 5)$ and perpendicular to the line through the points $(2, 5)$ and $(-3, 6)$.

Solution. The slope of the given line is

$$m_1 = \frac{6 – 5}{-3 – 2} = -\frac15$$

Perpendicularity requires $m_1m_2 = -1$, so

$$m_2 = 5$$

Through $(-3, 5)$:

$$y – 5 = 5(x + 3) \quad\Longrightarrow\quad y = 5x + 20 \quad\Longrightarrow\quad 5x – y + 20 = 0$$

Question 10

A line perpendicular to the line segment joining the points $(1, 0)$ and $(2, 3)$ divides it in the ratio $1 : n$. Find the equation of the line.

Solution. Two pieces are needed: the point the line passes through, and its slope.

Point. By the section formula, the point dividing $(1,0)$ and $(2,3)$ in the ratio $1:n$ is

$$\left(\frac{1 \cdot 2 + n \cdot 1}{1+n},\ \frac{1 \cdot 3 + n \cdot 0}{1+n}\right) = \left(\frac{n+2}{n+1},\ \frac{3}{n+1}\right)$$

Slope. The segment has slope $\dfrac{3-0}{2-1} = 3$, so the perpendicular has slope $-\dfrac13$.

Point-slope form:

$$y – \frac{3}{n+1} = -\frac13\left(x – \frac{n+2}{n+1}\right)$$

Multiplying through by $3(n+1)$:

$$3(n+1)y – 9 = -(n+1)x + (n+2)$$

$$(n+1)x + 3(n+1)y = n + 11$$

Question 11

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point $(2, 3)$.

Solution. Equal intercepts means $a = b$, so the intercept form becomes

$$\frac xa + \frac ya = 1 \quad\Longrightarrow\quad x + y = a$$

Substituting $(2,3)$:

$$2 + 3 = a \quad\Longrightarrow\quad a = 5$$

$$x + y = 5$$

Question 12

Find the equation of the line passing through the point $(2, 2)$ and cutting off intercepts on the axes whose sum is $9$.

Solution. Use the intercept form with unknowns $a$ and $b$:

$$\frac xa + \frac yb = 1$$

Condition 1 — the sum of intercepts:

$$a + b = 9$$

Condition 2 — the line passes through $(2,2)$:

$$\frac2a + \frac2b = 1 \quad\Longrightarrow\quad 2(a + b) = ab \quad\Longrightarrow\quad ab = 18$$

So $a$ and $b$ are the roots of $t^2 – 9t + 18 = 0$, that is $t = 3$ or $t = 6$. Both assignments give a valid line:

$$\frac x3 + \frac y6 = 1 \quad\Longrightarrow\quad 2x + y – 6 = 0$$

$$\frac x6 + \frac y3 = 1 \quad\Longrightarrow\quad x + 2y – 6 = 0$$

Question 13

Find the equation of the line through the point $(0, 2)$ making an angle $\dfrac{2\pi}{3}$ with the positive $x$-axis. Also find the equation of the line parallel to it and crossing the $y$-axis at a distance of $2$ units below the origin.

Solution. $\dfrac{2\pi}{3} = 120^\circ$, so

$$m = \tan 120^\circ = -\sqrt3$$

First line, with $y$-intercept $2$:

$$y = -\sqrt3\,x + 2 \quad\Longrightarrow\quad \sqrt3\,x + y – 2 = 0$$

Second line, parallel (so the same slope) with $y$-intercept $-2$:

$$y = -\sqrt3\,x – 2 \quad\Longrightarrow\quad \sqrt3\,x + y + 2 = 0$$

Question 14

The perpendicular from the origin to a line meets it at the point $(-2, 9)$. Find the equation of the line.

Solution. The segment from the origin to $(-2, 9)$ is perpendicular to the line, so start by finding its slope:

$$m_{\mathrm{OP}} = \frac{9 – 0}{-2 – 0} = -\frac92$$

The line therefore has slope

$$m = \frac{2}{9}$$

and passes through $(-2, 9)$:

$$y – 9 = \frac29(x + 2) \quad\Longrightarrow\quad 9y – 81 = 2x + 4 \quad\Longrightarrow\quad 2x – 9y + 85 = 0$$

Question 15

The length $\mathrm L$ (in centimetres) of a copper rod is a linear function of its Celsius temperature $\mathrm C$. In an experiment, $\mathrm L = 124.942$ when $\mathrm C = 20$ and $\mathrm L = 125.134$ when $\mathrm C = 110$. Express $\mathrm L$ in terms of $\mathrm C$.

Solution. “Linear function” means the graph of $\mathrm L$ against $\mathrm C$ is a straight line through the two data points $(20,\ 124.942)$ and $(110,\ 125.134)$.

The slope is

$$m = \frac{125.134 – 124.942}{110 – 20} = \frac{0.192}{90}$$

Using the point $(20,\ 124.942)$ in point-slope form:

$$\mathrm L – 124.942 = \frac{0.192}{90}\left(\mathrm C – 20\right)$$

$$\mathrm L = \frac{0.192}{90}\left(\mathrm C – 20\right) + 124.942$$

(The slope $\tfrac{0.192}{90} \approx 0.00213$ cm per degree is the rod’s thermal expansion, so the physical meaning is visible in the answer.)

Question 16

The owner of a milk store finds that he can sell $980$ litres of milk each week at Rs $14$/litre and $1220$ litres of milk each week at Rs $16$/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs $17$/litre?

Solution. Take the price as $x$ and the demand as $y$, giving the points $(14, 980)$ and $(16, 1220)$.

$$m = \frac{1220 – 980}{16 – 14} = \frac{240}{2} = 120$$

$$y – 980 = 120(x – 14) \quad\Longrightarrow\quad y = 120x – 1680 + 980 = 120x – 700$$

At $x = 17$:

$$y = 120(17) – 700 = 2040 – 700 = 1340$$

Note the demand rises with price here, which is unusual economically but is exactly what the two data points state — the mathematics simply follows them.

Question 17

$\mathrm P(a, b)$ is the mid-point of a line segment between the axes. Show that the equation of the line is $\dfrac xa + \dfrac yb = 2$.

Solution. Let the line meet the $x$-axis at $\mathrm A(p, 0)$ and the $y$-axis at $\mathrm B(0, q)$. The midpoint of $\mathrm{AB}$ is

$$\left(\frac{p + 0}{2},\ \frac{0 + q}{2}\right) = \left(\frac p2,\ \frac q2\right)$$

This is given to be $(a, b)$, so

$$\frac p2 = a \ \Longrightarrow\ p = 2a, \qquad \frac q2 = b \ \Longrightarrow\ q = 2b$$

The intercept form of the line is therefore

$$\frac{x}{2a} + \frac{y}{2b} = 1$$

Multiplying by $2$:

$$\frac xa + \frac yb = 2 \qquad \blacksquare$$

Question 18

Point $\mathrm R(h, k)$ divides a line segment between the axes in the ratio $1 : 2$. Find the equation of the line.

Solution. Let the line cut the axes at $\mathrm A(p, 0)$ and $\mathrm B(0, q)$, with $\mathrm R$ dividing $\mathrm{AB}$ in the ratio $1 : 2$ (measuring from $\mathrm A$). By the section formula:

$$\mathrm R = \left(\frac{1 \cdot 0 + 2 \cdot p}{1+2},\ \frac{1 \cdot q + 2 \cdot 0}{1+2}\right) = \left(\frac{2p}{3},\ \frac q3\right)$$

Equating to $(h, k)$:

$$\frac{2p}{3} = h \ \Longrightarrow\ p = \frac{3h}{2}, \qquad \frac q3 = k \ \Longrightarrow\ q = 3k$$

The intercept form gives

$$\frac{x}{3h/2} + \frac{y}{3k} = 1 \quad\Longrightarrow\quad \frac{2x}{3h} + \frac{y}{3k} = 1$$

Multiplying by $3hk$:

$$2kx + hy = 3hk$$

Question 19

By using the concept of the equation of a line, prove that the three points $(3, 0)$, $(-2, -2)$ and $(8, 2)$ are collinear.

Solution. Find the equation of the line through two of the points and check that the third satisfies it.

Through $(3,0)$ and $(-2,-2)$:

$$m = \frac{-2 – 0}{-2 – 3} = \frac{-2}{-5} = \frac25$$

$$y – 0 = \frac25(x – 3) \quad\Longrightarrow\quad 5y = 2x – 6 \quad\Longrightarrow\quad 2x – 5y – 6 = 0$$

Now substitute the third point $(8, 2)$:

$$2(8) – 5(2) – 6 = 16 – 10 – 6 = 0 \quad ✓$$

Since $(8,2)$ satisfies the equation of the line through the other two, all three lie on one line — they are collinear. $\blacksquare$

Common mistakes

• Question 1, writing the $x$-axis as $x = 0$. The $x$-axis is the set of points whose $y$-coordinate is zero, so its equation is $y = 0$. Testing a point such as $(5, 0)$ settles it instantly.
• Question 5, reading “3 units to the left” as $(3, 0)$. Left of the origin means a negative abscissa, so the point is $(-3, 0)$.
• Question 8, joining $\mathrm R$ to a vertex. A median goes to the midpoint of the opposite side, not to another vertex — that would be a side of the triangle.
• Question 9, using the given slope directly. The line through $(2,5)$ and $(-3,6)$ has slope $-\tfrac15$; the required line is perpendicular, so its slope is $+5$.
• Question 10, applying the section formula in the wrong order. For ratio $m:n$ from $\mathrm A$ to $\mathrm B$, the point is $\left(\tfrac{mx_2 + nx_1}{m+n}, \tfrac{my_2 + ny_1}{m+n}\right)$ — the far point’s coordinates carry $m$.
• Question 12, stopping at one solution. The quadratic $t^2 – 9t + 18 = 0$ gives two intercepts, and swapping which is $a$ and which is $b$ produces two genuinely different lines.
• Question 14, taking $-\tfrac92$ as the line’s own slope. That is the slope of the perpendicular from the origin; the line’s slope is its negative reciprocal, $\tfrac29$.
• Question 16, treating the price as the dependent variable. The question asks how much he can sell at a given price, so demand is the function of price, not the other way round.

Practise next

• Exercise 9.3 — the general form $\mathrm Ax + \mathrm By + \mathrm C = 0$, plus the distance of a point from a line and the distance between parallel lines.
• Exercise 9.1 — worth revising if the slope calculations in questions 7 to 10 felt slow.