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Grade: 12th pass
        
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 
2 months ago

Answers : (1)

Arun
22825 Points
							
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
2 months ago
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