Q.Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre lies on the line $x - 3y - 11 = 0$.

Slopes of Perpendicular Lines

Imagine standing at the corner of a rectangular field. One side runs straight ahead of you — that's one line. The other side runs exactly to your right, making a perfect right angle (90°). These two sides are perpendicular.

Now think about walking along those sides. The first side might be flat (horizontal), and the other goes straight up (vertical). But what if the field is on a hill? Then both sides are tilted, yet they still meet at a right angle. The question is: how are their tilts — their slopes — related?

The Intuition: Flip and Reverse

Slope measures how steep a line is: rise over run. A line that goes up gently has a small positive slope; a line that plunges steeply has a large negative slope.

For two lines to be perpendicular, one must be the "opposite" of the other in a specific way. If one line rises 2 units for every 1 unit it runs (slope = 2), the perpendicular line must fall 1 unit for every 2 units it runs (slope = -1/2). The steepness gets flipped upside down, and the direction gets reversed.

The Precise Statement

If two non-vertical lines are perpendicular, the product of their slopes is $-1$.

Let line $L_1$ have slope $m_1$ and line $L_2$ have slope $m_2$. Then:

Equivalently, each slope is the negative reciprocal of the other:

Why It Works

Consider two lines through the origin: $y = m_1 x$ and $y = m_2 x$. They are perpendicular if the angle between them is 90°. Using the tangent formula for the angle between two lines:

For $\theta = 90^\circ$, $\tan 90^\circ$ is undefined, which happens when the denominator is zero:

That's the entire derivation. The denominator must vanish for the lines to be at right angles.

Quick Examples

| Slope of one line ($m_1$) | Slope of perpendicular ($m_2$) | Check: $m_1 \cdot m_2$ | …