A triangle ABC. given B(-4,-5). the altitudes d1,

d1: 5x+3y-4=0, and d2: 5x+8y+13=0. Find the equations for the sides of the triangle, and locate A and C. thanks in advance. Please help if you can. thanks.

The two given altitudes are not from vertex B because the point B(-4, -5) does not satisfy either d1 or d2.
Because the problem didn't state d1 and d2 starting from A or C, we have to set it.

1). Case #1

- If d1 is from vertex A, then d1 is perpendicular to BC

d1: 5x + 3y - 4 = 0
y = 4/3 - (5/3)x

because d1 is perpendicular to BC so slope m1 for line segment BC is 3/5
and BC passes through B(-4, -5)
therefore equation for BC is:

y + 5 = (3/5)(x + 4)
y + 5 = (3/5)x + 12/5

3x - 5y - 13 = 0 <====equation for line BC
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- d2 is from point C so it is perpendicular to AB
d2: 5x + 8y + 13 = 0
y = (-5/8)x - 13/8

so slope of AB is 8/5
and AB passes through point B(-4, -5)

y + 5 = (8/5)(x + 4)
y + 5 = (8/5)x + 32/5
8x - 5y + 7 = 0 <======equation for AB

From equation for AB and d1 you solve this system to find point A
8x - 5y + 7 = 0
5x + 3y - 4 = 0
----------------------solve this system to get the coordinate for point A

From d2 and BC,
5x + 8y + 13 = 0
3x - 5y - 13 = 0
-------------------------solve this system to get the coordinate for point C

Once you get point A and point C, find the equation for the line that passes through those twp points and that is the equation for side AC.
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Case #2
2). If d1 is from point C and d2 is from point A.

Do the same approach as above.
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Good luck.