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  • Please solve RD Sharma class 12 chapter Determinants exercise 5.3 question 13 subquestion (i) maths textbook solution

Please solve RD Sharma class 12 chapter Determinants exercise 5.3 question 13 subquestion (i) maths textbook solution

Answers (1)

Answer: k=0

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

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

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

\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} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array}\right|=4

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

     = \frac{1}{2}[k(0-2)-0+1(8-0)]=4

     \Rightarrow -2k+8=8

     \Rightarrow -2k=0

     \Rightarrow k=0

Posted by

Gurleen Kaur

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