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

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(C) The perpendicular distance $d$ of a line $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Here,the line is $3x + 4y - 10 = 0$ and the origin is $(0, 0)$.Substituting the values,we get $d = \frac{|3(0) + 4(0) - 10|}{\sqrt{3^2 + 4^2}}$.$d = \frac{|-10|}{\sqrt{9 + 16}} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2$.Thus,the perpendicular distance is $2$.

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

Easy

What is the perpendicular distance of the line $3x + 4y = 10$ from the origin?

A
$5$
B
$1$
C
$2$
D
$4$

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

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$A$ variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$,where $O$ is the origin,is:

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The distance of the line $2x - 3y = 4$ from the point $(1, 1)$ measured parallel to the line $x + y = 1$ is

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What is the perpendicular distance of the line $3x + 4y = 10$ from the origin?

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