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

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.