Aditya Gupta
Last Activity: 5 Years 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 side
now, 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= 15
solving, we get k= 4, p= 0
so, the vertices of triangle are (14, 4) and (0, 26).
you can easily find eqn of hypotenuse as y+11x/7= 26
kindly approve :)
