I have got a coordinate $(x_1,y_1)$ say, $(10,12)$ and a slope of $3$. Now I need to find a coordinate $(x_2,y_2)$ such that is $4$ units away from $(x_1,y_1)$.
I know the formula $d = \sqrt{(y_2-y_1)^2 + (x_2-x_1)^2}$ but I don't have either $x_2$ or $y_2$ to proceed with it.
On
You are given the point (10, 12) and the straight line through (10, 12) with slope 3. That line can be written as $y= 3(x- 10)+ 12= 3x- 18$. The set of all points at distance 4 from (10, 12) is the circle given by $(x- 10)^2+ (y- 12)^2= 16$. The points (there are two of them) on that line that are 4 units from (10, 12) satisfy both equations. Since $y= 3x- 18$, $(x- 10)^2+ (y- 12)^2= (x- 10)^2+ (3x- 18- 10)^2= (x- 10)^2+ (3x- 28)^2$ $= x^2- 20x+ 100+ 9x^2- 168x+ 784= 10x^2- 188x+ 884= 16$ or $5x^2- 94x+ 434= 0$. Solve that for the two values of x and then use $y= 3x- 18$ to find the corresponding values of y.
Observe that, since the slope is $3$, you have that $$\frac{y_1-y_2}{x_1-x_2}=3\implies x_2=x_1-\frac{y_1-y_2}3$$ Substitute this in your distance formula to obtain $y_2$ and the compute $x_2$.