Line joining the points (3,-4) and (-2,6) is perpendicular to the line joining the points (-3,6) and (9,-18).

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(3,-4)

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Concept understanding — Perpendicular Slopes Condition

Perpendicular Slopes Condition

Imagine two roads crossing at a right angle — that's perpendicular lines. The question is: how do their slopes relate?

The Intuition

Take a line with slope $2$ . That means for every $1$ unit you move right, you go up $2$ units — a fairly steep climb. Now picture a line perpendicular to it. If the first line is climbing steeply, the perpendicular line must be falling gently, or climbing very shallowly in the opposite direction.

Why? Because a right angle means the two lines "flip" the rise and run. One line's steepness becomes the other's shallowness, but in the opposite sign.

Try this: a line with slope $2$ (rise $2$ , run $1$ ). A perpendicular line should have rise $1$ and run $-2$ — that gives slope $-\frac{1}{2}$ . Notice: $2 \times (-\frac{1}{2}) = -1$ .

That's the pattern: the slopes are negative reciprocals of each other.

The Precise Statement

m_1 \cdot m_2 = -1

Two non-vertical lines are perpendicular if and only if the product of their slopes is $-1$ .

Equivalently: $m_2 = -\frac{1}{m_1}$ (provided $m_1 \neq 0$ ).

What About Vertical and Horizontal Lines?

A vertical line has undefined slope. A horizontal line has slope $0$ . Their product? Undefined — not $-1$ . Yet they are clearly perpendicular.

Watch out

The formula $m_1 \cdot m_2 = -1$ only works when both slopes are defined (neither line is vertical). For a vertical line ( $x = c$ ) and a horizontal line ( $y = d$ ), they are perpendicular by definition — no slope calculation needed.

Quick Check

Are $y = 3x + 2$ and $y = -\frac{1}{3}x - 5$ perpendicular?

$3 \times (-\frac{1}{3}) = -1$ . Yes.

Are $y = 4x$ and $y = 4x + 1$ perpendicular?

$4 \times 4 = 16 \neq -1$ . No — they're parallel.

Why It Works (A Short Proof)

›Proof

Two lines with slopes $m_1$ and $m_2$ make angles $\theta_1$ and $\theta_2$ with the positive $x$ -axis, where $\tan \theta_1 = m_1$ and $\tan \theta_2 = m_2$ . …

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The key idea is that two lines are perpendicular if the product of their slopes is $-1$ . After computing the slopes, we find the product is $(-2) \times (-2) = 4 \neq -1$ , so the lines are not perpendicular.

The problem asks whether the line through $(3,-4)$ and $(-2,6)$ is perpendicular to the line through $(-3,6)$ and $(9,-18)$ . To decide, we need the slopes of both lines and the perpendicular condition.

Why the Perpendicular Slopes Condition Works

Two lines are perpendicular (at right angles) if and only if the product of their slopes equals $-1$ , provided neither line is vertical. This comes from geometry: if one line makes an angle $\theta$ with the horizontal, its slope is $\tan \theta$ . A perpendicular line makes angle $\theta + 90^\circ$ , and $\tan(\theta + 90^\circ) = -\cot \theta = -\frac{1}{\tan \theta}$ . Multiplying gives $\tan \theta \times (-\frac{1}{\tan \theta}) = -1$ .

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Watch out

A common mistake is to check if slopes are negative reciprocals without actually multiplying. Always compute the product — it's faster and avoids sign errors.

Step-by-Step Solution

1. Find the slope of the first line.

The slope formula for points $(x_1, y_1)$ and $(x_2, y_2)$ is:

m = \frac{y_2 - y_1}{x_2 - x_1}

For $(3,-4)$ and $(-2,6)$ :

m_1 = \frac{6 - (-4)}{-2 - 3} = \frac{6 + 4}{-5} = \frac{10}{-5} = -2

So the first line has slope $-2$ .

2. Find the slope of the second line.

For $(-3,6)$ and $(9,-18)$ : …

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Q.Line joining the points (3,−4)(3,-4)(3,−4) and (−2,6)(-2,6)(−2,6) is perpendicular to the line joining the points (−3,6)(-3,6)(−3,6) and (9,−18)(9,-18)(9,−18).

Why the Perpendicular Slopes Condition Works

Step-by-Step Solution

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Q.Line joining the points $(3,-4)$ and $(-2,6)$ is perpendicular to the line joining the points $(-3,6)$ and $(9,-18)$ .

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