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`        If (8,2) is right angle vertex, x-3y-2=0 is one side and (7,15) is mid point of hypotenuse. Find equations of sides and vertices of a triangle. `
10 months ago

```							note that x-3y-2=0 passes through (8,2). so we know for sure that it is not the hypotenuse but is in fact a side.now, the side perpendicular to the above line will have the eqn:y – 2= – 3(x – 8) [ since slope of x-3y-2=0 is 1/3 ]or y+3x= 26 is eqn of another sidenow, let us take a point on x-3y-2=0 in parametric form (3k+2, k).and on y+3x= 26 as (p, 26 – 3p). so, we are assuming that the hypotenuse passes through (3k+2, k), (7, 15). (p, 26 – 3p). but given that (7,15) is mid point.so, (3k+2+p)/2= 7 and (k+26 – p)/2= 15solving, we get k= 4, p= 0so, the vertices of triangle are (14, 4) and (0, 26).you can easily find eqn of hypotenuse as y+11x/7= 26kindly approve :)
```
10 months ago
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