Question

Find the distance of the point P(-2.3) from the line AB which
is x-y=5.​

Answer 1

Answer:

$\sf 2 \ units$

Step-by-step explanation:

Given :

x - y = 5

x + ( - y ) + ( - 5 ) = 0

Point P ( - 2 , 3 )

We know :

$\sf d=\left| \dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|$

Putting values here we get :

$\sf d=\left| \dfrac{1\times(-2)+(-1)\times3+(-5)}{\sqrt{(-2)^2+(3)^2}}\right|$

$\sf d=\left| \dfrac{-2-3-5}{5}\right|$

$\sf d=\left| \dfrac{-10}{5} \right|$

$\sf d=2 \ units$

Therefore , we get required distance.