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?
 A: The line passing through two points $(-2,-4),(3,4)$ can be expressed as
$$y-4=\frac{4-(-4)}{3-(-2)}(x-3),$$
i.e.
$$y=\frac{8}{5}x-\frac 45.$$
So, every point on this line can be expressed as $(t,\frac{8}{5}t-\frac 45)$ for some $t\in\mathbb R$.
Now, solve the following equation for $t$ :
$$5=\sqrt{(t-(-2))^2+\left(\frac{8}{5}t-\frac 45-(-4)\right)^2}$$
A: We define vector from first point to second: 
$ \vec{w} = [3,4]-[-2,-4] = [5,8] $
Module of this vector is:
$|w| = \sqrt{5^2+8^2} \approx 9,434  $
Hence a the unit vector in this direction has value:
$ \vec{u} := \vec{w} / |\vec{w}| = [5/|\vec{w}|, 8/|\vec{w}| ] = [0.53,  0.85] $
And distance from $(-2,-4)$ is:
$(-2,-4) + 5\vec{u} = (-2,-4) + [2.65, 4.25] = (0.65, 0.25) $
or in the reverse direction:
$(-2,-4) - 5\vec{u} = (-2,-4) - [2.65, 4.25] = ( -4.65, -8.25) $.
A: If $(h,k)$ be the required coordinates, 
the slope :$\dfrac{k-4}{h-3}=\dfrac{4+4}{3+2}$
$\iff\dfrac{k-4}8=\dfrac{h-3}5=t$(say)
$\implies k=4+8t,h=3+5t$
Now using the distance: $\{h-(-2)\}^2+\{k-(-4)\}^2=5^2$
Put the values of $h,k$
Can you take it from here?
A: 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
$$\sqrt{(p-(-2))^2+(q-(-4))^2}=5$$ $$(p+2)^2+\left(\frac{8p-4}{5}+4\right)^2=25$$
$$(p+2)^2+\frac{64}{25}(p+2)^2=25 $$$$ 89(p+2)^2=25\times 25$$ $$ p+2=\pm\frac{25}{\sqrt{89}}$$ $$\implies p=\pm\frac{25}{\sqrt{89}}-2$$
$$\text{if}\quad \color{blue}{p=\frac{25}{\sqrt{89}}-2} \implies \color{blue}{q=\frac{40}{\sqrt{89}}-4}$$
$$\text{if}\quad \color{blue}{p=-\frac{25}{\sqrt{89}}-2} \implies \color{blue}{q=-\frac{40}{\sqrt{89}}-4}$$
