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If the vertices of a triangle are (1, k), (4, -3), (-9, 7) and its area is 15 sq units, find the value(s) of k.. [Ans: -3 ,21/13 ]
Ans: A(1, k) B(4, -3) C(-9, 7)
Area of Δ ABC =1/2 [x1(y2 -y1 )+x2 (y3 -y1 ) + x1 (y1 -y2 )]
= 1/2 [1(-3-7)+4(7-k)+(-9)(k+3)] = 15
-10 + 28 - 4k - 9k - 27 = 30
- 9 - 13k = 30 ⇒ k = -3
| -9 - 13k | = 30
9 + 13k = 30
k = 21/13
k = -3, 21/13
R={(r, ?):1=r= 2cos? ,-p/3= ? =p/3
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