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If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is
  • -12, -2
  • 12
  • 12, -2
  • -2
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If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is
  • -12, -2
  • 12
  • 12, -2
  • -2

Answer

Given vertices of the triangle are (2, -6), (5, 4) and (k, 4).
If the vertices of the triangle are given by (x1, y1), (x2, y2), (x3, y3), therefore
Area of triangle is given by \(\triangle\) = \(\frac{1}{2}\left|\begin{array}{lll} {\mathrm{x}_{1}} & {\mathrm{y}_{1}} & {1} \\ {\mathrm{x}_{2}} & {\mathrm{y}_{2}} & {1} \\ {\mathrm{x}_{3}} & {\mathrm{y}_{3}} & {1} \end{array}\right|\) 
Given, Area of triangle  = 35 sq. units
\(\Rightarrow\) \(\frac{1}{2}\left\|\begin{array}{ccc} {2} & {-6} & {1} \\ {5} & {4} & {1} \\ {k} & {4} & {1} \end{array}\right\|=35\) 
\(\Rightarrow\) \(35=\frac{1}{2} \times[2 \times(4 \times 1-4 \times 1)-(-6) \times(5 \times 1-k \times 1)+1 \times(5 \times 4-k \times 4)]\) 
\(\Rightarrow\) \(35=\frac{1 }{ 2} \times[2 \times(4-4)+6 \times(5-k)+1 \times(20-4 k)]\) 
\(\Rightarrow\) \(35 \times 2=(2 \times 0+30-6 k+20-4 k)\) 
\(\Rightarrow\) 70 = 30 - 6 k + 20 - 4 k
\(\Rightarrow\) 70 = 50 - 10 k
\(\Rightarrow\) 70 - 50 = -10 k 
\(\Rightarrow\) 20 = \(-10k\)
\(\Rightarrow\) k = \(\frac{-20}{10}\) 
\(\Rightarrow\) k = -2 
Therefore the choice is: D
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