The equation of the straight line passing through the point (3,2) and perpendicular to the line y=x is (A) x-y

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(3,2)

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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 $\tan \theta_2 = m_2$ . …

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The key idea is that perpendicular lines have slopes whose product is $-1$ . The given line $y=x$ has a slope of $1$ , so the perpendicular line has a slope of $-1$ . Using the point-slope form with the given point $(3,2)$ , the equation is $x+y=5$ .

To find the equation of a straight line, we typically need two pieces of information: either two points it passes through, or one point and its slope. In this problem, we are given a point and a condition (perpendicularity to another line) that allows us to determine the slope.

The core concept here is the relationship between the slopes of perpendicular lines.

Identify the given point:

The straight line we need to find passes through the point $(3,2)$ . Let's denote this as $(x_1, y_1) = (3,2)$ .
Determine the slope of the given line:

The given line is $y=x$ . This equation is in the slope-intercept form, $y = mx + c$ , where $m$ is the slope and $c$ is the y-intercept.

Comparing $y=x$ with $y=mx+c$ , we see that $m=1$ and $c=0$ .

So, the slope of the line $y=x$ is $m_1 = 1$ .
Calculate the slope of the required line:

The required line is perpendicular to the line $y=x$ . For two non-vertical lines to be perpendicular, the product of their slopes must be $-1$ .
$m_1$
Let $m_2$ be the slope of the required line. Using the perpendicularity condition:

$m_1 \cdot m_2 = -1$

$1 \cdot m_2 = -1$

$m_2 = -1$

So, the slope of the required line is $-1$ . …

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Q.The equation of the straight line passing through the point (3,2)(3,2)(3,2) and perpendicular to the line y=xy=xy=x is (A) x−y=5x-y=5x−y=5 (B) x+y=5x+y=5x+y=5 (C) x+y=1x+y=1x+y=1 (D) x−y=1x-y=1x−y=1

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Q.The equation of the straight line passing through the point $(3,2)$ and perpendicular to the line $y=x$ is
(A) $x-y=5$
(B) $x+y=5$
(C) $x+y=1$
(D) $x-y=1$

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