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

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

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.

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)