If the vertices of ΔABC be A(1, -3) B(4, p) and C(-9, 7) and its area is 15 square units, find the values of p
If the vertices of a triangle are (1, −3), (4, p) and (−9, 7) and its area is 15 sq. units, find the value(s) of p?
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Solution
`Let A(x_1,y_1)= A (1,-3),B(x_2,y_2)=B(4,P) and C (x_3,y_3)= C(-9,7) Now`
`"Area" (ΔABC) = 1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`
`⇒15=1/2 [1(p-7)+4(7+3)-9(-3-p)]`
`⇒15=1/2[10p+60]`
`⇒ |10p +60|=30`
Therefore
⇒ 10p + 60 = -30 or 30
⇒ 10p = -90 or -30
⇒ p = -9 or -3
Hence , p= -9 or p= -3.
Concept: Coordinate Geometry
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