(B) The slope of the given line $y = x$ is $m_1 = 1$.Since the required line is perpendicular to $y = x$,its slope $m_2$ must satisfy $m_1 \times m_2

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(B) The slope of the given line $y = x$ is $m_1 = 1$.Since the required line is perpendicular to $y = x$,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.Thus,$1 \times m_2 = -1$,which gives $m_2 = -1$.The equation of a line with slope $m = -1$ passing through $(x_1, y_1) = (3, 2)$ is given by $y - y_1 = m(x - x_1)$.Substituting the values,we get $y - 2 = -1(x - 3)$.$y - 2 = -x + 3$.Rearranging the terms,we get $x + y = 5$.

Class 11 Mathematics Straight Lines Slope of line, Equation of line in different forms

Easy

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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Straight Lines

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Slope of line, Equation of line in different forms

Find the equation of a straight line which passes through the point $(-1, -1)$ and makes an angle $150^{\circ}$ with the positive direction of the $X$-axis.

EasyAP EAMCET 2021

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The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$ is:

DifficultTS EAMCET 2003

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The line passing through $(-1, \pi/2)$ and perpendicular to $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$ is

Difficult

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Find the equation of a line that makes an intercept of $-3$ on the $y$-axis and makes an angle of ${\tan ^{ - 1}}\frac{3}{5}$ with the $x$-axis.

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The equation of the line passing through the point $(1, 2)$ and perpendicular to the line $x + y + 1 = 0$ is

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

Similar Questions

Find the equation of a straight line which passes through the point $(-1, -1)$ and makes an angle $150^{\circ}$ with the positive direction of the $X$-axis.

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