NCERT Class 11 Mathematics — Straight Lines, Exercise 9.3. All 17 questions solved.

The general form $\mathrm Ax + \mathrm By + \mathrm C = 0$ can be rearranged into any of the earlier forms, and two distance formulae depend on it.

Distance of a point $(x_1, y_1)$ from a line:

$$d = \frac{\left|\mathrm Ax_1 + \mathrm By_1 + \mathrm C\right|}{\sqrt{\mathrm A^2 + \mathrm B^2}}$$

Distance between the parallel lines $\mathrm Ax + \mathrm By + \mathrm C_1 = 0$ and $\mathrm Ax + \mathrm By + \mathrm C_2 = 0$:

$$d = \frac{\left|\mathrm C_1 – \mathrm C_2\right|}{\sqrt{\mathrm A^2 + \mathrm B^2}}$$

Key insight. The parallel-lines formula only works when the coefficients of $x$ and $y$ are identical in both equations, not merely proportional. Before subtracting the constants, scale one equation so that $\mathrm A$ and $\mathrm B$ match — otherwise the answer is out by the scale factor. Question 5(ii) has this built in, with $l$ appearing in both.

Question 1

Reduce the following equations into slope-intercept form and find their slopes and $y$-intercepts.

(i) $x + 7y = 0$    (ii) $6x + 3y – 5 = 0$    (iii) $y = 0$

Solution. Slope-intercept form is $y = mx + c$, so make $y$ the subject.

(i)

$$7y = -x \quad\Longrightarrow\quad y = -\frac17 x + 0$$

Slope $= -\dfrac17$, $y$-intercept $= 0$.

(ii)

$$3y = -6x + 5 \quad\Longrightarrow\quad y = -2x + \frac53$$

Slope $= -2$, $y$-intercept $= \dfrac53$.

(iii) Already solved for $y$:

$$y = 0\cdot x + 0$$

Slope $= 0$, $y$-intercept $= 0$. (This is the $x$-axis itself.)

Question 2

Reduce the following equations into intercept form and find their intercepts on the axes.

(i) $3x + 2y – 12 = 0$    (ii) $4x – 3y = 6$    (iii) $3y + 2 = 0$

Solution. Intercept form is $\dfrac xa + \dfrac yb = 1$, so move the constant to the right and divide through by it.

(i)

$$3x + 2y = 12 \quad\Longrightarrow\quad \frac{3x}{12} + \frac{2y}{12} = 1 \quad\Longrightarrow\quad \frac x4 + \frac y6 = 1$$

Intercepts: $4$ on the $x$-axis, $6$ on the $y$-axis.

(ii)

$$\frac{4x}{6} – \frac{3y}{6} = 1 \quad\Longrightarrow\quad \frac{x}{3/2} + \frac{y}{-2} = 1$$

Intercepts: $\dfrac32$ and $-2$.

(iii)

$$3y = -2 \quad\Longrightarrow\quad y = -\frac23$$

This is a horizontal line. It meets the $y$-axis at $-\tfrac23$ and is parallel to the $x$-axis, so it has no $x$-intercept and cannot be written in intercept form at all.

Question 3

Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y – 2)$.

Solution. Put the line in general form first:

$$12x + 72 = 5y – 10 \quad\Longrightarrow\quad 12x – 5y + 82 = 0$$

So $\mathrm A = 12$, $\mathrm B = -5$, $\mathrm C = 82$, and

$$d = \frac{\left|12(-1) – 5(1) + 82\right|}{\sqrt{12^2 + (-5)^2}} = \frac{\left|-12 – 5 + 82\right|}{\sqrt{169}} = \frac{65}{13} = 5$$

Question 4

Find the points on the $x$-axis whose distances from the line $\dfrac x3 + \dfrac y4 = 1$ are $4$ units.

Solution. Clear the fractions to get the general form. Multiplying by $12$:

$$4x + 3y = 12 \quad\Longrightarrow\quad 4x + 3y – 12 = 0$$

A point on the $x$-axis is $(a, 0)$, and its distance is

$$\frac{\left|4a + 0 – 12\right|}{\sqrt{16 + 9}} = \frac{\left|4a – 12\right|}{5} = 4$$

so $\left|4a – 12\right| = 20$, giving two cases:

$$4a – 12 = 20 \ \Longrightarrow\ a = 8 \qquad\text{or}\qquad 4a – 12 = -20 \ \Longrightarrow\ a = -2$$

The modulus is what produces two points, one on each side of the line.

Question 5

Find the distance between the parallel lines

(i) $15x + 8y – 34 = 0$ and $15x + 8y + 31 = 0$ (ii) $l(x + y) + p = 0$ and $l(x + y) – r = 0$

Solution.

(i) The coefficients already match, so

$$d = \frac{\left|-34 – 31\right|}{\sqrt{15^2 + 8^2}} = \frac{65}{\sqrt{289}} = \frac{65}{17}$$

(ii) Expanding, the lines are $lx + ly + p = 0$ and $lx + ly – r = 0$, so $\mathrm A = \mathrm B = l$ and the constants are $p$ and $-r$:

$$d = \frac{\left|p – (-r)\right|}{\sqrt{l^2 + l^2}} = \frac{\left|p + r\right|}{l\sqrt2} = \frac{1}{\sqrt2}\left|\frac{p+r}{l}\right|$$

Question 6

Find the equation of the line parallel to the line $3x – 4y + 2 = 0$ and passing through the point $(-2, 3)$.

Solution. A parallel line has the same $x$ and $y$ coefficients, differing only in the constant:

$$3x – 4y + k = 0$$

Substituting $(-2, 3)$ to find $k$:

$$3(-2) – 4(3) + k = 0 \quad\Longrightarrow\quad -6 – 12 + k = 0 \quad\Longrightarrow\quad k = 18$$

$$3x – 4y + 18 = 0$$

Question 7

Find the equation of the line perpendicular to the line $x – 7y + 5 = 0$ and having $x$-intercept $3$.

Solution. The given line has slope $\tfrac17$ (from $y = \tfrac17 x + \tfrac57$), so the perpendicular has slope $-7$.

An $x$-intercept of $3$ means the line passes through $(3, 0)$:

$$y – 0 = -7(x – 3) \quad\Longrightarrow\quad y = -7x + 21 \quad\Longrightarrow\quad 7x + y = 21$$

Question 8

Find the angles between the lines $\sqrt3\,x + y = 1$ and $x + \sqrt3\,y = 1$.

Solution. Read off the slopes:

$$m_1 = -\sqrt3, \qquad m_2 = -\frac{1}{\sqrt3}$$

$$\tan\theta = \left|\frac{m_2 – m_1}{1 + m_1m_2}\right| = \left|\frac{-\frac{1}{\sqrt3} + \sqrt3}{1 + \sqrt3 \cdot \frac{1}{\sqrt3}}\right| = \left|\frac{\frac{-1 + 3}{\sqrt3}}{2}\right| = \frac{2}{2\sqrt3} = \frac{1}{\sqrt3}$$

So $\theta = 30^\circ$. Two lines always make two angles, supplementary to each other, so the other is

$$180^\circ – 30^\circ = 150^\circ$$

Question 9

The line through the points $(h, 3)$ and $(4, 1)$ intersects the line $7x – 9y – 19 = 0$ at right angles. Find the value of $h$.

Solution. The slope of the given line is $\tfrac79$ (from $9y = 7x – 19$).

The slope of the line through $(h,3)$ and $(4,1)$ is

$$m = \frac{1 – 3}{4 – h} = \frac{-2}{4 – h}$$

Perpendicularity requires the product to be $-1$:

$$\frac{-2}{4-h} \times \frac79 = -1 \quad\Longrightarrow\quad \frac{-14}{9(4-h)} = -1 \quad\Longrightarrow\quad 14 = 9(4 – h)$$

$$14 = 36 – 9h \quad\Longrightarrow\quad 9h = 22 \quad\Longrightarrow\quad h = \frac{22}{9}$$

Question 10

Prove that the line through the point $(x_1, y_1)$ and parallel to the line $\mathrm Ax + \mathrm By + \mathrm C = 0$ is

$$\mathrm A(x – x_1) + \mathrm B(y – y_1) = 0$$

Solution. The given line has slope $-\dfrac{\mathrm A}{\mathrm B}$ (assuming $\mathrm B \ne 0$), and a parallel line has the same slope. By the point-slope form, the line through $(x_1, y_1)$ is

$$y – y_1 = -\frac{\mathrm A}{\mathrm B}\left(x – x_1\right)$$

Multiplying both sides by $\mathrm B$:

$$\mathrm B(y – y_1) = -\mathrm A(x – x_1)$$

$$\mathrm A(x – x_1) + \mathrm B(y – y_1) = 0 \qquad \blacksquare$$

(The final form is valid even when $\mathrm B = 0$, where the slope argument breaks down: it then reads $\mathrm A(x – x_1) = 0$, that is $x = x_1$, correctly a vertical line parallel to $\mathrm Ax + \mathrm C = 0$.)

Question 11

Two lines passing through the point $(2, 3)$ intersect each other at an angle of $60^\circ$. If the slope of one line is $2$, find the equation of the other line.

Solution. Let the unknown slope be $m$. The angle formula gives

$$\tan 60^\circ = \left|\frac{m – 2}{1 + 2m}\right| = \sqrt3$$

The modulus produces two cases.

Case 1: $\dfrac{m-2}{1+2m} = \sqrt3$.

$$m – 2 = \sqrt3 + 2\sqrt3\,m \quad\Longrightarrow\quad m\left(1 – 2\sqrt3\right) = 2 + \sqrt3 \quad\Longrightarrow\quad m = \frac{2+\sqrt3}{1 – 2\sqrt3}$$

Rationalising (multiply above and below by $1 + 2\sqrt3$):

$$m = \frac{(2+\sqrt3)(1+2\sqrt3)}{1 – 12} = \frac{8 + 5\sqrt3}{-11} = -\frac{8+5\sqrt3}{11}$$

The line through $(2,3)$ with this slope, cleared of fractions, is

$$\left(\sqrt3 + 2\right)x + \left(2\sqrt3 – 1\right)y = 8\sqrt3 + 1$$

Case 2: $\dfrac{m-2}{1+2m} = -\sqrt3$.

$$m – 2 = -\sqrt3 – 2\sqrt3\,m \quad\Longrightarrow\quad m\left(1 + 2\sqrt3\right) = 2 – \sqrt3 \quad\Longrightarrow\quad m = \frac{2 – \sqrt3}{1 + 2\sqrt3}$$

which gives the line

$$\left(\sqrt3 – 2\right)x + \left(1 + 2\sqrt3\right)y = -1 + 8\sqrt3$$

Both are genuine — the second line can be rotated $60^\circ$ either way from the first.

Question 12

Find the equation of the right bisector of the line segment joining the points $(3, 4)$ and $(-1, 2)$.

Solution. The right bisector (perpendicular bisector) passes through the midpoint of the segment and is perpendicular to it.

Midpoint:

$$\left(\frac{3 + (-1)}{2},\ \frac{4 + 2}{2}\right) = (1, 3)$$

Slope of the segment:

$$\frac{2 – 4}{-1 – 3} = \frac{-2}{-4} = \frac12$$

so the bisector has slope $-2$. Through $(1,3)$:

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

Question 13

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

Solution. The foot is the intersection of the given line with the perpendicular dropped from $(-1, 3)$.

The given line has slope $\tfrac34$, so the perpendicular has slope $-\tfrac43$:

$$y – 3 = -\frac43(x + 1) \quad\Longrightarrow\quad 3y – 9 = -4x – 4 \quad\Longrightarrow\quad 4x + 3y = 5$$

Now solve simultaneously with $3x – 4y = 16$. Multiplying the first by $4$ and the second by $3$:

$$16x + 12y = 20, \qquad 9x – 12y = 48$$

Adding: $25x = 68$, so $x = \dfrac{68}{25}$. Substituting back into $4x + 3y = 5$:

$$3y = 5 – \frac{272}{25} = \frac{125 – 272}{25} = -\frac{147}{25} \quad\Longrightarrow\quad y = -\frac{49}{25}$$

Question 14

The perpendicular from the origin to the line $y = mx + c$ meets it at the point $(-1, 2)$. Find the values of $m$ and $c$.

Solution. Two conditions are available.

The point lies on the line:

$$2 = m(-1) + c \quad\Longrightarrow\quad c – m = 2$$

The perpendicularity: the segment from $(0,0)$ to $(-1,2)$ has slope $\dfrac{2-0}{-1-0} = -2$, so

$$m \times (-2) = -1 \quad\Longrightarrow\quad m = \frac12$$

Substituting into the first equation:

$$c = 2 + \frac12 = \frac52$$

Question 15

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the lines $x\cos\theta – y\sin\theta = k\cos 2\theta$ and $x\sec\theta + y\operatorname{cosec}\theta = k$ respectively, prove that $p^2 + 4q^2 = k^2$.

Solution. Apply the distance formula from the origin, where the numerator reduces to the constant term.

First line: $x\cos\theta – y\sin\theta – k\cos2\theta = 0$, so

$$p = \frac{\left|-k\cos2\theta\right|}{\sqrt{\cos^2\theta + \sin^2\theta}} = \left|k\cos 2\theta\right|$$

using $\cos^2\theta + \sin^2\theta = 1$. Hence

$$p^2 = k^2\cos^2 2\theta$$

Second line: $x\sec\theta + y\operatorname{cosec}\theta – k = 0$, so

$$q = \frac{k}{\sqrt{\sec^2\theta + \operatorname{cosec}^2\theta}}$$

Simplify the radicand:

$$\sec^2\theta + \operatorname{cosec}^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}$$

Therefore

$$q = k\,\left|\sin\theta\cos\theta\right| = \frac{k}{2}\left|\sin 2\theta\right| \quad\Longrightarrow\quad 4q^2 = k^2\sin^2 2\theta$$

Adding:

$$p^2 + 4q^2 = k^2\left(\cos^2 2\theta + \sin^2 2\theta\right) = k^2 \qquad \blacksquare$$

The factor $4$ in the statement is exactly what turns $\sin\theta\cos\theta$ into $\tfrac12\sin2\theta$ — the identity is what the question is built around.

Question 16

In the triangle $\mathrm{ABC}$ with vertices $\mathrm A(2, 3)$, $\mathrm B(4, -1)$ and $\mathrm C(1, 2)$, find the equation and length of the altitude from the vertex $\mathrm A$.

Solution. The altitude from $\mathrm A$ is perpendicular to $\mathrm{BC}$ and passes through $\mathrm A$.

Slope of BC:

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

so the altitude has slope $1$. Through $\mathrm A(2,3)$:

$$y – 3 = 1(x – 2) \quad\Longrightarrow\quad y – x = 1$$

Length of the altitude is the distance from $\mathrm A$ to the line $\mathrm{BC}$. First find that line: through $\mathrm B(4,-1)$ with slope $-1$,

$$y + 1 = -(x – 4) \quad\Longrightarrow\quad x + y – 3 = 0$$

$$d = \frac{\left|2 + 3 – 3\right|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt2} = \sqrt2$$

Question 17

If $p$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$, then show that

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

Solution. The line in intercept form is

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

Clear the fractions by multiplying through by $ab$, then move everything to one side:

$$bx + ay – ab = 0$$

The perpendicular distance from the origin $(0,0)$ is

$$p = \frac{\left|b(0) + a(0) – ab\right|}{\sqrt{a^2 + b^2}} = \frac{\left|ab\right|}{\sqrt{a^2+b^2}}$$

Squaring and inverting:

$$p^2 = \frac{a^2b^2}{a^2 + b^2} \quad\Longrightarrow\quad \frac{1}{p^2} = \frac{a^2 + b^2}{a^2b^2} = \frac{a^2}{a^2b^2} + \frac{b^2}{a^2b^2} = \frac{1}{b^2} + \frac{1}{a^2} \qquad \blacksquare$$

Common mistakes

• Question 2(iii), forcing an intercept form. $y = -\tfrac23$ is parallel to the $x$-axis and never meets it, so there is no $x$-intercept and no intercept form. Saying so is the answer.
• Question 3, using the equation before rearranging. The distance formula needs $\mathrm Ax + \mathrm By + \mathrm C = 0$; applying it to $12(x+6) = 5(y-2)$ as written gives nonsense.
• Question 4, dropping the modulus. $\left|4a – 12\right| = 20$ has two solutions, and the question asks for points — plural.
• Question 5, subtracting constants without matching coefficients. The formula $\dfrac{|\mathrm C_1 – \mathrm C_2|}{\sqrt{\mathrm A^2+\mathrm B^2}}$ requires identical $\mathrm A$ and $\mathrm B$. If one line is $6x + 8y = 5$ and the other $3x + 4y = 1$, halve the first before using it.
• Question 8, giving only one angle. Two intersecting lines form two pairs of vertically opposite angles, $30^\circ$ and $150^\circ$ here. Both are correct answers to “find the angles”.
• Question 11, taking only one sign. The modulus in the angle formula yields two lines, one on each side.
• Question 13, stopping at the perpendicular’s equation. The foot is the intersection of the two lines, so the simultaneous equations must still be solved.
• Question 16, confusing the altitude with the median. An altitude is perpendicular to the opposite side; a median goes to its midpoint. The two coincide only in an isosceles configuration.

Practise next

• Miscellaneous Exercise on Chapter 9 — concurrency, angle bisectors, images in a line and shifting the origin, all resting on the distance formula here.
• Exercise 9.2 — worth revising if converting between forms in questions 1 and 2 felt slow.