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Explain solution RD Sharma class 12 chapter Determinants exercise 5.3 question 9 maths

Answers (1)

$\boldsymbol{Answer\! :} k=-1,\frac{1}{2}$

Hints: If points are collinear then determinant will be zero.

$\boldsymbol{Given\! :} (k,2-2k), (-k+1, 2k), (-4-k, 6-2k)$

$\boldsymbol{Explanation\! :} V\! ertices\; are\; (k,2-2k), (-k+1, 2k), (-4-k, 6-2k)$

$\text { Determinant}=\left|\begin{array}{lll} \; \; \; \; \; \; k & 2-2k & 1 \\ -k+1 & \; \; \; 2k & 1 \\ -4-k & 6-2k & 1 \end{array}\right|=0$

Expanding along row1.

$\Rightarrow k(2k-(6-2k))-(2-2k)(-k+1-(-4-k))+1[(-k+1)(6-2k)-2k(-4-k)]$

$\Rightarrow [2k^{2}-6k+2k^{2}]-(2-2k)(-k+1+4+k)+[-6k+6+2k^{2}-2k]+8k+2k^{2}$

$\Rightarrow [2k^{2}-6k+2k^{2}]-[(2-2k)(5)+(6+4k^{2})]=0$

$\Rightarrow [4k^{2}-6k-10+10k+6+4k^{2}]=0$

$\Rightarrow [8k^{2}-4+4k]=0$

$\Rightarrow 2k^{2}+k-1=0$

$\Rightarrow 2k^{2}+2k-k-1=0$

$\Rightarrow 2k(k+1)-1(k+1)=0$

$\Rightarrow (2k-1)(k+1)=0$

$\begin{bmatrix} k+1 &=0 \\ k &=-1 \end{bmatrix}$ $\begin{bmatrix} 2k-1&=0 \\ k&=\frac{1}{2} \end{bmatrix}$

$k=-1,\frac{1}{2}$

Posted by

Gurleen Kaur

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