(A) The distance $d$ of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$ is given by the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

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(A) The distance $d$ of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$ is given by the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Given the point $(-2, 3)$ and the line $x - y - 5 = 0$,we have $A = 1, B = -1, C = -5, x_1 = -2, y_1 = 3$.Substituting these values into the formula:$d = \frac{|(1)(-2) + (-1)(3) - 5|}{\sqrt{1^2 + (-1)^2}}$$d = \frac{|-2 - 3 - 5|}{\sqrt{1 + 1}}$$d = \frac{|-10|}{\sqrt{2}}$$d = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.

Class 11 Mathematics Straight Line Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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The distance of point $(-2, 3)$ from the line $x - y - 5 = 0$ is

A
$5\sqrt{2}$
B
$2\sqrt{5}$
C
$3\sqrt{5}$
D
$5\sqrt{3}$

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Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

The base of an equilateral triangle is along the line given by $3x + 4y = 9$. If a vertex of the triangle is $(1, 2)$,then the length of a side of the triangle is

DifficultJEE MAIN 2014

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Find the distance of the point $(2, 3)$ from the line $2x - 3y + 9 = 0$ measured along the direction of the line $x - y + 1 = 0$.

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Let $R$ be the interior region between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin. The set of all values of $a$,for which the points $(a^2, a+1)$ lie in $R$,is :

DifficultJEE MAIN 2024

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Two points $A$ and $B$ have coordinates $(1, 1)$ and $(3, -2)$ respectively. The coordinates of a point $P$ at a distance of $\sqrt{85}$ from $B$ on the line through $B$ perpendicular to $AB$ are:

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The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:

Medium

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The distance of point $(-2, 3)$ from the line $x - y - 5 = 0$ is

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The base of an equilateral triangle is along the line given by $3x + 4y = 9$. If a vertex of the triangle is $(1, 2)$,then the length of a side of the triangle is

Find the distance of the point $(2, 3)$ from the line $2x - 3y + 9 = 0$ measured along the direction of the line $x - y + 1 = 0$.

Let $R$ be the interior region between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin. The set of all values of $a$,for which the points $(a^2, a+1)$ lie in $R$,is :

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