Find the equation of the line passing through the point (5,2) and perpendicular to the line joining the points

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(5,2)

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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 $\tan \theta_2 = m_2$ .

Perpendicular means $\theta_2 = \theta_1 + 90^\circ$ .

Using $\tan(\theta + 90^\circ) = -\cot \theta = -\frac{1}{\tan \theta}$ , we get:

$m_2 = \tan(\theta_1 + 90^\circ) = -\frac{1}{\tan \theta_1} = -\frac{1}{m_1}$

Hence $m_1 m_2 = -1$ .

The One Thing to Remember

Important

Perpendicular slopes are negative reciprocals.

If one slope is $m$ , the perpendicular slope is $-\frac{1}{m}$ (unless $m = 0$ , then the perpendicular is vertical).

The Perpendicular Slopes Condition is a key result from the NCERT Class 11 Mathematics chapter on Straight Lines, and it's exactly what students are looking for when they search "perpendicular lines slope formula" or "straight lines important questions class 11 maths". This negative-reciprocal rule is also a quick, frequently tested check in JEE Main and CET coordinate geometry problems.

The key idea is that perpendicular lines have slopes that are negative reciprocals. The slope of the given line is $-4$ , so the perpendicular slope is $\frac{1}{4}$ . Using the point $(5,2)$ , the equation is $x - 4y + 3 = 0$ .

Concept and Intuition

When two lines are perpendicular, their slopes multiply to $-1$ (provided neither is vertical). This is the Perpendicular Slopes Condition: if $m_1$ and $m_2$ are the slopes of two perpendicular lines, then $m_1 \cdot m_2 = -1$ .

Why does this work? Think of slope as "rise over run." A line that goes steeply upward (large positive slope) is perpendicular to a line that goes gently downward (small negative slope). The negative reciprocal relationship captures this perfectly.

For this problem, we first find the slope of the line through $(2,3)$ and $(3,-1)$ . Then we take its negative reciprocal to get the slope of the perpendicular line. Finally, we use the given point $(5,2)$ to write the equation.

Step-by-Step Solution

1. Find the slope of the line joining $(2,3)$ and $(3,-1)$ .

The slope formula is:

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

Let $(x_1, y_1) = (2,3)$ and $(x_2, y_2) = (3,-1)$ . Then:

m_1 = \frac{-1 - 3}{3 - 2} = \frac{-4}{1} = -4

So the given line has slope $-4$ .

Watch out

A common mistake is to subtract in the wrong order. Always keep the coordinates consistent: $y_2 - y_1$ over $x_2 - x_1$ . Swapping them gives the same magnitude but the wrong sign.

2. Determine the slope of the perpendicular line.

If two lines are perpendicular, the product of their slopes is $-1$ :

m_1 \cdot m_2 = -1

Here $m_1 = -4$ , so:

-4 \cdot m_2 = -1

m_2 = \frac{-1}{-4} = \frac{1}{4}

The perpendicular slope is $\frac{1}{4}$ .

Tip

To get the perpendicular slope quickly: flip the fraction and change the sign. For $-4$ (which is $-\frac{4}{1}$ ), flipping gives $-\frac{1}{4}$ , then changing the sign gives $+\frac{1}{4}$ .

3. Write the equation of the line with slope $\frac{1}{4}$ passing through $(5,2)$ .

Use the point-slope form:

y - y_1 = m(x - x_1)

Substitute $m = \frac{1}{4}$ , $x_1 = 5$ , $y_1 = 2$ :

y - 2 = \frac{1}{4}(x - 5)

4. Simplify to the required form.

Multiply both sides by $4$ to eliminate the fraction:

4(y - 2) = x - 5

4y - 8 = x - 5

Bring all terms to one side:

0 = x - 5 - 4y + 8

0 = x - 4y + 3

Or equivalently:

x - 4y + 3 = 0

This is the equation in standard form.

Note

You could also write it as $x - 4y = -3$ or $y = \frac{1}{4}x + \frac{3}{4}$ , but the standard form $x - 4y + 3 = 0$ is most common in exam contexts.

✓Final answer

\boxed{x - 4y + 3 = 0}

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Q.Find the equation of the line passing through the point (5,2)(5,2)(5,2) and perpendicular to the line joining the points (2,3)(2,3)(2,3) and (3,−1)(3,-1)(3,−1).

Concept and Intuition

Step-by-Step Solution

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Q.Find the equation of the line passing through the point $(5,2)$ and perpendicular to the line joining the points $(2,3)$ and $(3,-1)$ .

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