SSC CGL Maths · Coordinate Geometry

Parallel and Perpendicular Lines

Compare the slopes of two lines to tell parallel from perpendicular, build the perpendicular bisector of a segment, and find the shortest distance from a point to a line.

All Coordinate Geometry notes

Why this matters

Six questions in the bank, and four of the chapter's six MODERATE rows are here. Each joins two earlier steps (a midpoint and a slope, or a slope and a squared unknown), so the page is where the chapter's speed is won or lost. Two are a single negative-reciprocal step and take seconds.

Concept 1 of 3: Slopes of parallel and perpendicular lines

Parallel lines climb at the same rate, so their slopes are equal. Turn a line through a right angle and its slope flips over and changes sign, so the two slopes multiply to minus 1.

Definition

Parallel: $m_1 = m_2$ . For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ this is $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ . If $\frac{c_1}{c_2}$ is the same ratio too, the two equations are one line, not two parallel lines.
Perpendicular: $m_1 \times m_2 = -1$ , so the new slope is the negative reciprocal: slope 2 becomes $-\frac{1}{2}$ . In coefficient form, $a_1a_2 + b_1b_2 = 0$ .
Fast route: a line parallel to $ax + by + c = 0$ is $ax + by + k = 0$ (keep the coefficients). A line perpendicular to it is $bx - ay + k = 0$ (swap them and change one sign). Then fix $k$ with the given point.
A horizontal line and a vertical line are perpendicular, though the product rule cannot be used: a vertical line has no slope.

Parallel and perpendicular conditions

m_1 = m_2 \qquad m_1 m_2 = -1 \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} \qquad a_1a_2 + b_1b_2 = 0

$m_1, m_2$ the slopes of the two lines
$a, b$ the coefficients of x and y in ax + by + c = 0

Worked example

For the line

3x - 4y + 2 = 0

, find the slope of a parallel line and of a perpendicular line, and write the perpendicular line through the origin.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 3 · Q73Easy

Example 1 · Coordinate Geometry · Parallel and Perpendicular Lines

What is the slope of line perpendicular to

y = -3x + 7

Taking only the reciprocal, or only the negative

The perpendicular slope to 6 is

-\frac{1}{6}

. The options print

\frac{1}{6}

-6

and 6 as well; check that the two slopes multiply to exactly

-1

Keeping both roots when the unknown is squared

A parallel condition such as

\frac{2}{k + 1} = \frac{k + 1}{8}

gives

k + 1 = \pm 4

, two values of

k

. Keep only the one the question allows (for example

k > 0

), and check it does not make the two equations the same line.

Drill 2 more on slopes of parallel and perpendicular lines

Concept 2 of 3: Perpendicular bisector of a segment

The perpendicular bisector cuts a segment in half at a right angle. Every point on it is the same distance from the two ends, which gives both a way to build it and a fast way to check an option.

Definition

Build it in three steps: find the midpoint of the segment, find the segment's slope, then write the line through the midpoint with the negative reciprocal slope.
Equal-distance check: any point on the bisector is equally far from both ends. Take the y-intercept of an option and compare its squared distances from the two ends; the right option makes them equal.
A horizontal segment has a vertical bisector $x =$ (midpoint x), and a vertical segment has a horizontal one.

Worked example

Find the perpendicular bisector of the segment joining

A(-1, 1)

and

B(7, 5)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 17 Sep 2025, Shift 3 · Q69Moderate

Example 2 · Coordinate Geometry · Parallel and Perpendicular Lines

What is the equation of the perpendicular bisector of the line segment connecting the points (2, 8) and (6, 4)?

Drawing the bisector through an end instead of the midpoint

The perpendicular bisector passes through the midpoint of the segment. A line through one end with the perpendicular slope is perpendicular, but it is not the bisector.

Giving the bisector the segment's own slope

The bisector crosses the segment at a right angle, so its slope is the negative reciprocal of the segment's slope. A line with the segment's own slope through the midpoint is the segment's line itself.

Drill 1 more on perpendicular bisector of a segment

Concept 3 of 3: Shortest distance from a point to a line

The shortest path from a point to a line meets the line at a right angle. The formula measures that perpendicular directly, so you never need to find where it lands.

Definition

Write the line as $ax + by + c = 0$ . Put the point into the left side, take the absolute value, and divide by $\sqrt{a^2 + b^2}$ .
From the origin the top is just $|c|$ .
Two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ are $\frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$ apart. Make the x and y coefficients identical first.
Fast route: a line with intercepts $p$ and $q$ makes a right triangle with the axes, and the distance from the origin is the altitude to its hypotenuse: $\frac{pq}{\sqrt{p^2 + q^2}}$ .
A segment, not a whole line: the perpendicular distance counts only if its foot lands between the two ends. If it lands outside, the nearest point is the nearer end.

Distance from a point to a line

d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \qquad d_{\text{parallel}} = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}

$(x_0, y_0)$ the point

Worked example

Find the distance of the point

(1, -2)

from the line

6x + 8y + 5 = 0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 2 · Q69Moderate

Example 3 · Coordinate Geometry · Parallel and Perpendicular Lines

What is the minimum distance from the origin (0,0) to a point on the line segment connecting the points A(3, 0) and B(0, 4)?

Leaving out the square root on the bottom

The bottom is

\sqrt{a^2 + b^2}

, not

a^2 + b^2

and not

a + b

. For

12x + 5y = 39

it is 13, and the distance from the origin is

\frac{39}{13} = 3

Using unequal coefficients for parallel lines

2x + 3y = 1

and

4x + 6y = 9

are parallel, but subtract the constants only after rewriting the first as

4x + 6y = 2

. The distance is then

\frac{7}{\sqrt{52}}

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Slopes of parallel and perpendicular lines
Parallel and perpendicular conditions
$m_1 = m_2 \qquad m_1 m_2 = -1 \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} \qquad a_1a_2 + b_1b_2 = 0$
Shortest distance from a point to a line
Distance from a point to a line
$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \qquad d_{\text{parallel}} = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$

Watch out for (6)

Taking only the reciprocal, or only the negative
→ Slopes of parallel and perpendicular lines
Keeping both roots when the unknown is squared
→ Slopes of parallel and perpendicular lines
Drawing the bisector through an end instead of the midpoint
→ Perpendicular bisector of a segment
Giving the bisector the segment's own slope
→ Perpendicular bisector of a segment
Leaving out the square root on the bottom
→ Shortest distance from a point to a line
Using unequal coefficients for parallel lines
→ Shortest distance from a point to a line

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