InterviewSolution
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InterviewSolution
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Find the value of k so that the area of the triangle with vertices A(k+1, 1), B(4, -3) and C(7, -k) is 6 square units. |
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Answer» Correct Answer - k = 3 Let `A(x_(1) = k + 1, y_(1) =1), B(x_(2) = 4, y_(2) = -3) " and "C(x_(3) = 7, y_(3) = -k)` be the vertices of `Delta ABC`. Then, `ar (Delta ABC) = (1)/(2) |x_(1) (y_(2) -y_(3)) + x_(2) (y_(3) -y_(1)) + x_(3)(y_(1) -y_(2))|` ` = (1)/(2)|(k+1)(-3+k) + 4(-k-1) + 7(1+3)|` ` = (1)/(2)|(k^(2) -6k + 21)| = (1)/(2)|(k^(2) -6k + 9) +12|` ` = (1)/(2) |(k-3)^(2) + 12| = (1)/(2)[(k-3)^(2) + 12].` `therefore (1)/(2)[(k-3)^(2) +12] = 6 rArr (k-3)^(2) + 12 = 12 rArr (k-3)^(2) = 0` Hence, k = 3. |
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