# Find the equation of the side of a triangle having B(-4, - 5) as a vertex and 5x+3y-4 =0, 3x+8y+13 =0 as the equation of two of its altitudes

Arun
25750 Points
5 years ago
Since the equations of altitudes both not satisfy B(-4,-5), these are equations of perpendiculars from A and C to the opposite sides.

We will assume 5x+3y-4=0 is of altitude AD and
……………….3x+8y+13=0 is of altitude CF.

Side BC has slope 3/5…..eqn is 5y= 3x+k

Side AB has slope 8/3….eqn is 3y= 8x+ k’

Intersection B is given by
15y= 9x+3k
15y= 40x+5k’
31x= 3k-5k’

Also
40y=24x+8k
9y=24x+3k’
31y= 8k-3k’

Given B is x= -4 and y= -5
3k-5k’= -124
8k-3k’=-155
Solving
9k-40k= -372+775
k=- 403/31= -13
k’= 1/3(-104+155)= 51/3=17

BC is 5y=3x-13
AB is 3y= 8x+17

For CA , we must find intersection orthocenter ( from eqn of 2 altitudes) and find equation of altitude from B which is the line joining them. On above lines, we should arrive at that equation also

DIRECT AND EASIER SOLUTION

After slopes of BA and BC are known, since B (-4,-5) is a point on each of them, equations can be written right away as below,
for BC: y+5= 3/5*(x+4)
5y=3x-13
for BA: y+5= 8/3*(x+4)
3y=8x+17
for AC, we need data on point A or C coordinates, as the line having given slope will be many parallels