The length of the perpendicular drawn from th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the line passing through $(x', y')$ and $(x'', y'')$ is given by: $(y - y') = \frac{y'' - y'}{x'' - x'}(x - x')$ $(y - y')(x'' - x

Vedclass · Accepted Answer

$\frac{|x'y'' - y'x''|}{\sqrt{(x'' - x')^2 + (y'' - y')^2}}$

Vedclass · Answer

(B) The equation of the line passing through $(x', y')$ and $(x'', y'')$ is given by: $(y - y') = \frac{y'' - y'}{x'' - x'}(x - x')$ $(y - y')(x'' - x') = (y'' - y')(x - x')$ $x(y'' - y') - y(x'' - x') + y'x'' - y'x' - x'y'' + x'y' = 0$ $x(y'' - y') - y(x'' - x') + (x'y'' - y'x'') = 0$ The length of the perpendicular from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is $\frac{|C|}{\sqrt{A^2 + B^2}}$. Here,$A = (y'' - y')$,$B = -(x'' - x')$,and $C = (x'y'' - y'x'')$. Thus,the length is $\frac{|x'y'' - y'x''|}{\sqrt{(y'' - y')^2 + (-(x'' - x'))^2}} = \frac{|x'y'' - y'x''|}{\sqrt{(x'' - x')^2 + (y'' - y')^2}}$.

The length of the perpendicular drawn from the origin $(0, 0)$ to the line joining the points $(x', y')$ and $(x'', y'')$ is:

Similar Questions

If the points $(1, 2)$ and $(3, 4)$ lie on the same side of the straight line $3x - 5y + a = 0$,then $a$ lies in the set

If a line $l$ passes through $(k, 2k), (3k, 3k)$ and $(3, 1)$,where $k \neq 0$,then the distance from the origin to the line $l$ is

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

What is the length of the perpendicular from the point $(3, 4)$ to the line $3x + 4y + 10 = 0$?

Find the distance between the lines $3x + 4y + 7 = 0$ and $3x + 4y + 22 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module