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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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