# Q : 5     Find the points on the x-axis, whose distances from the line  $\frac{x}{3}+\frac{y}{4}=1$  are  $4$  units.

G Gautam harsolia

Given equation of line is
$\frac{x}{3}+\frac{y}{4}=1$
we can rewrite it as
$4x+3y-12=0$
Now, we know that
$d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
In this problem A = 4 , B = 3 C = -12 and d = 4
point is on x-axis therefore  $(x_1,y_1)$ = (x ,0)
Now,
$4= \frac{|4.x+3.0-12|}{\sqrt{4^2+3^2}}= \frac{|4x-12|}{\sqrt{16+9}}= \frac{|4x-12|}{\sqrt{25}}= \frac{|4x-12|}{5}$
$20=|4x-12|\\ 4|x-3|=20\\ |x-3|=5$
Now if x > 3
Then,
$|x-3|=x-3\\ x-3=5\\ x = 8$
Therefore, point is (8,0)
and if x < 3
Then,
$|x-3|=-(x-3)\\ -x+3=5\\ x = -2$
Therefore, point is (-2,0)
Therefore, the points on the x-axis, whose distances from the line  $\frac{x}{3}+\frac{y}{4}=1$  are  $4$  units are  (8 , 0) and (-2 , 0)

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