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InterviewSolution
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The four vertices of a quadrilateral are (1, 2), (-5, 6), (7, -4) and (k, -2) taken in order. If the area of the quadrilateral is zero, find the value of k. |
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Answer» Let four vertices of quadrilateral be A (1, 2) and B (−5, 6) and C (7, −4) and D (k, −2) Area of □ ABCD = Area of ∆ABC + Area of ∆ACD = 0 sq. unit 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)| Area of ∆ABC = \(\frac{1}2\) |1(6 – (-4)) - 5(-4 -2) + 7(2 – 6)| = \(\frac{1}2\) |10 + 30 -28| = 6 sq. units Area of ∆ACD = \(\frac{1}2\) |1(-2 – (-4)) + k(-4 -2) + 7(2 – (-2))| = \(\frac{1}2\) |2 - 6k + 30| = \(\frac{1}2\) (3k -15) sq. units Area of ∆ABC + Area of ∆ACD = 0 sq. unit ∴ 6 + 3k -15 =0 3k -9 = 0 ∴ k =3 Hence, the value of k is 3 |
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