In $\triangle ABC : A(-2,7)$ and $C(7,-5)$. The length of the altitude to $AC$ is $5$ and the length of the altitude to $BC$ is $\sqrt45$. I need to find the coordinates of $B$, given that this point is below $AC$.
I have started by finding the equation of $AC$ to be: $y=\frac{-4}{3}x + \frac{13}{3}$. Then I moved on and using the formula for distance between a line and a point I found that the $y$ coordinate of $B$ equals $\frac{-4}{3}$ times the $x$ coordinates of $B - 4$ .... (In other words, $Y_B = \frac{-4}{3}X_B -4$).
Then I got stuck. Since I know the final answer, I could plot this, but it did not help me solve the problem. I am attaching the plot in case my explanation wasn't clear enough.
I have a problem both with the way of solving and the algebra here. I will need any appropriate help. Thank you in advance !
If $B=(x,y)$, you can build 2 equations of the variables $x,y$ and calculate them.
Using the distance equation between point and line, the distance between $B$ and $AC$ is $\frac{|4x+3y-13|}{5}$. and the distance between $A$ and $BC$ is $\frac{|12x+9y-39|}{\sqrt{(x-7)^2+(y-5)^2}}$.
using the details in the question you need to solve the following equations:
$\frac{|4x+3y-13|}{5}=5$
$\frac{|12x+9y-39|}{\sqrt{(x-7)^2+(y-5)^2}}=\sqrt{45}$
Solving the equation you get $(x,y)=(-3,0)$