Given a vertex and 2 triangle height equations, get the equations, on which the triangle's sides lay

Question

One vertex of the triangle: $(3; -4)$

2 height equations:

$$ 7x-2y-1=0 $$ $$ 2x-7y-6=0 $$

The task is to find the straight line equations, on which the triangle sides lay.

So far, I haven't figured out how to proceed with this.

Also as I now made a comment, I did figure one point out: I do have a point, which isn't on any of the 2 given altitude equations, means that the vertex is on the 3rd altitude equation. The 2 altitude equations intersect eachother, which means I can make the third altitude equation by using 2 points and finding the intersection between the 2 equations.

Answer 1

Hints:

(1) Check the point isn't on any of those two lines (and thus those heights are to the sides of which the point is part and from the other two triangle's vertices);

(2) The two lines through the given vertex and perpendicular to the given lines are the one on which the triangle's vertices are. For example, one of them is:

$$y+4=-\frac27(x-3)$$

(3) The other two vertices are on the intersection points of the lines one which the sides of the triangle lay and the given heights (why?)