Question

Find the co-ordinates of a point on the line $x+y-4=0$ whose distance from the line $4x+3y=10$ is $1$.

Answer 1

We know distance of a $pt(x_1,y_1)$

from line $a_{1}x+b_{1}y+c_1=0$ is

$\frac{|a_1{x}_1+b_1{y}_1+c_1|}{\sqrt{a_1^2+b_1^2}}$

Putting $a_1=4,b_1=3,c_1=-10$

& considering $(x_1,y_1)$ the required $pt$,

$1=\frac{|4x_1+3y_1-10|}{\sqrt{16+9}}$

$\therefore 5=|4x_1+3y_1-10|$

$\therefore 4x_1+3y_1-10=5$ or $4x_1+3y_1-10=-5$

$\therefore 4x_1+3y_1=15$ or $4x_1+3y_1=5$

$(x_1,y_1)$ is on the line $x+y-4=0$. i.e $x_1+y_1=4$.

$\therefore$ We can substitute $x_1=4-y_1$

$\therefore 4(4-y_1)+3y_1=15$ or $4(4-y_1)+3y_1=5$

$\therefore 16-y_1=15$ or $16-y_1=5$

$y_1=1$ or $y_1=11$

$\therefore x_1=4-1=3$ mor $x_1=4-11=-7$

$\therefore$ The $pt$ can be $(3,1)$ or $(-7,11)$.