If the area of the triangle formed by the vertices (-2, 4),(3,-1) and (5, k) is 10 square units, find the value of 'k'.
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The values of k are 1,-7
Given vertices of triangle are
A(x_{1},y_{1})=(-2,4)A(x1,y1)=(−2,4)
B(x_{2},y_{2})=(3,-1)B(x2,y2)=(3,−1)
C(x_{3},y_{3})=(5,k)C(x3,y3)=(5,k)
Area of triangle ABC is given by
\frac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|=1021∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣=10
\frac{1}{2}|-2(-1-k)+3(k-4)+5(4+1)|=1021∣−2(−1−k)+3(k−4)+5(4+1)∣=10
\frac{1}{2}|2+2k+3k-12+25|=1021∣2+2k+3k−12+25∣=10
5k + 15 = ±20
Case (i):
5k + 15 = 20
5k = 20 - 15
5k = 5
k = 1
Case (ii):
5k + 15 = -20
5k = -20 - 15
5k = -35
k = -7
∴ The values of k are 1,-7.