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Provide solution for RD Sharma maths class 12 chapter Determinants exercise 5.3 question 13 subquestion (ii)

Answers (1)

Answer: k=8

Hints: Putting the values of vertices, in the formula of area of triangle.

Given: (-2,0), (0,4) and (0,k)
              area of triangle= 4 sq. units

Explanation: Vertices are (-2,0), (0,4) and (0,k)

\text { Area of triangle is }=\Delta=\frac{1}{2}\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|=4

\Rightarrow \Delta=\frac{1}{2}\left|\begin{array}{lll} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right|=4

\Delta=\frac{1}{2}\left(-2\left|\begin{array}{cc} 4 & 1 \\ k & 1 \end{array}\right|-0\left|\begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array}\right|+1\left|\begin{array}{cc} 0 & 4 \\ 0 & k \end{array}\right|\right)=4

     = \frac{1}{2}[-2(4-k)-0+1(0)]=4

     \Rightarrow -8+2k=8

     \Rightarrow 2k=16

     \Rightarrow k=8

Posted by

Gurleen Kaur

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