1.

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.

Answer»

Let A(1, −3), B(4, p) and C(−9, 7) be the vertices of the ∆ABC. 

Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3

\(\frac{1}2\) |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| 

Given that area of ∆ABC = 15 

∴ 15 = \(\frac{1}2\) |1(p – 7) +4 (7 – (-3)) - 9(-3 - p)| 

∴ 30 = |p – 7 + 40 + 27 + 9p| 

∴ 30 = | 10p + 60| 

∴ 10p + 60 = ± 30 

Taking positive sign, 

10p + 60 = 30 

p = -3 

Taking negative sign, 

10p + 60 = - 30 

p = -9 

Hence, 

the value of p are -3 and -9



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