InterviewSolution
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InterviewSolution
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Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and(-k, -5) is 24 sq. units. |
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Answer» Let A(1, -1), B(-4, 2k) and C(-k, -5) be the vertices of the given `Delta ABC.` Then, `A(x_(1) = 1, y_(1) = -1), B(x_(2) =-4, y_(2) = 2k), " and " C(x_(3) = -k, y_(3) = -5)`. `therefore 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)|*(2k+5) -4 * (-5+1) -k(-1) -2k)|` `=(1)/(2)|2k + 5 +16 +k + 2k^(2)|` `= (1)/(2)|2k^(2) + 3k +21|` sq units. But, it is given that `ar(Delta ABC)` = 24 sq units. `therefore (1)/(2)|2k^(2) + 3k + 21| =24 rArr |k^(2) + (3)/(2) k + (21)/(2)| = 24.` But `{k^(2) + (3)/(2)k + (21)/(2)} = (k^(2) + (3)/(2)k + (9)/(16)) + ((21)/(2) - (9)/(16)) = {(k + (3)/(2))^(2) + (159)/(16)} gt 0.` So, we may write (i) as `k^(2) + (3)/(2)k + (21)/(2) = 24 rArr 2k^(2) + 3k+ 21 = 48` `rArr 2k^(2) +3k -27 = 0 rArr 2k^(2) +9k-6k -27 = 0` `rArr k(2k+9)-3(2k +9) = 0rArr (2k +9)(k-3) = 0` `rArr k -3 = 0 "or" 2k +9 =9 rArr k = 3 "or" k = -(9)/(2)` Hence, k = 3 or`k = -(9)/(2).` |
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