find coordinate on line at given distance from given coordinate I got two coordinates of a straight line $(-2,-4)$ and $(3,4)$. How can i find a coordinate that lies on this line and is $5$ units away from the $(-2,-4)$ coordinate?

The equation of the line passing through the points $(-2, -4)$ & $(3, 4)$ is given as 
$$y-(-4)=\frac{4-(-4)}{3-(-2)}(x-(-2))$$ $$\implies y+4=\frac{8}{5}(x
+2)$$ $$\implies 5y+20=8x+16$$ $$\implies 8x-5y-4=0$$
Let the coordinates be $(p, q)$ on the line: $8x-5y-4=0$ Then this point will satisfy the equation of the line as follows $$8p-5q-4=0 \implies q=\frac{8p-4}{5}$$ Now, the distance of the point $(p, q)$ from $(-2, -4)$ is $5$ hence we have
$$\sqrt{(p-(-2))^2+(q-(-4))^2}=5$$ $$(p+2)^2+\left(\frac{8p-4}{5}+4\right)^2=25$$
$$(p+2)^2+\frac{64}{25}(p+2)^2=25 $$$$ 89(p+2)^2=25\times 25$$ $$ p+2=\pm\frac{25}{\sqrt{89}}$$ $$\implies p=\pm\frac{25}{\sqrt{89}}-2$$
$$\text{if}\quad \color{blue}{p=\frac{25}{\sqrt{89}}-2} \implies \color{blue}{q=\frac{40}{\sqrt{89}}-4}$$
$$\text{if}\quad \color{blue}{p=-\frac{25}{\sqrt{89}}-2} \implies \color{blue}{q=-\frac{40}{\sqrt{89}}-4}$$
