# Find the coordinate if slope, distance and one coordinate is known.

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.

• Do both of these points lie on a straight line? – Ak19 May 16 at 11:54
• So, you have a right triangle with hypotenuse $4$ and side lengths $a$ and $b$ with $a/b=3$. You can solve, using Pythagoras for $a$ and $b$. Then you can find the coordinates. – David Mitra May 16 at 11:54
• yes, a straight line. – Shubhank Gupta May 16 at 11:56

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 $$x_2$$.
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.