If the points p(-3,9) Q(A,B) and R(4,-5) are collinear and A+B=1 then find the value of Aand B
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if points are collinear then area = 0
so triangle pqr will have area 0
by formula
area = 1/2×(-3 (b-5)×a(-5+9)×4(9-b))
0=-3b+15-4a+36-4b
0=-7b+51-4a
7b-4a=51 (1)
b+a=1 (2)
by equating
7b-4a=51
7b+7a=7
-11a=44
a=-4
b=5
so triangle pqr will have area 0
by formula
area = 1/2×(-3 (b-5)×a(-5+9)×4(9-b))
0=-3b+15-4a+36-4b
0=-7b+51-4a
7b-4a=51 (1)
b+a=1 (2)
by equating
7b-4a=51
7b+7a=7
-11a=44
a=-4
b=5
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