The points on the line that are $5$  units away from A,  if the slope of a line through  $A\left(2,3\right)$  is  $\frac{3}{4}$   are

$r=5units$

$A(x_1,y_1)=(2,3)$

slope=$\tan \theta =\frac{3}{4}$

$\sin \theta =\frac{3}{5}$

$\cos \theta =\frac{4}{5}$

The required points are:

$x=x_1\pm r\cos \theta$  $=2\pm 5\left(\frac{4}{5}\right)$  $=2\pm 4$

$y=y_1\pm r\sin \theta$  $=3\pm 5\left(\frac{3}{5}\right)$  $=3\pm 3$

(or)

$\text{A=}\left(2,3\right)$ , slope $\left(m\right)=\frac{3}{4}$

$B=\left(x,y\right)$

$AB=5units$

$\sqrt{{\left(x-2\right)}^2+{\left(y-3\right)}^2}=5$

$x^2-4x+4+y^2-6y+9=25$

$x^2+y^2-4x-6y-12=0\to \left(1\right)$

Slope of $AB=\frac{3}{4}$

$\frac{y-3}{x-2}=\frac{3}{4}$

$4y-12=3x-6$

$4y=3x+6$

$y=\frac{3x+6}{4}$

From $\left(1\right)$

$x^2+{\left(\frac{3x+6}{4}\right)}^2-4x-6\left(\frac{3x+6}{4}\right)-12=0$

$x^2+\frac{9x^2+36x+36}{16}-4x-\frac{18x+36}{4}-12=0$

$16x^2+9x^2+36x+36-64x-72x-144-192=0$

$25x^2-100x-300=0$

$x^2-4x-12=0$

$x^2-6x+2x-12=0$

$\left(x-6\right)\left(x+2\right)=0$

$x=6,2$

$y=\frac{3\left(6\right)+6}{4}$

$=\frac{24}{6}$

$=6$

$B\left(x,y\right)=\left(6,6\right)$

$y=\frac{3\left(2\right)+6}{4}$ $=3$

$A\left(x,y\right)=\left(2,3\right)$

Simply $C\left(x,y\right)=\left(-2,0\right)$

Option (a) : $\to$  $x=2\pm 4,y=3\pm 3$