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`        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 `
4 months 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
```
4 months ago
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