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  • Need solution for RD Sharma maths class 12 chapter Determinants exercise 5.3 question 4

Need solution for RD Sharma maths class 12 chapter Determinants exercise 5.3 question 4

Answers (1)

Answer: Yes, points are collinear.

Hints: Use the values of vertices and find the determinant. If determinant is zero, points will be collinear.

\mathbf{Given\! :} V\! ertices\; are \left ( a,b \right )\! ,\left ( a',b' \right ) and \left ( a-a',b-b' \right ) and\; ab' = a'b

Explanation:

Let,

x_{1}=a, x_{2}=a', x_{3}=a-a'

y_{1}=a, y_{2}=a', y_{3}=b-b'

Determinant=\begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}

                            =\begin{vmatrix} a &b &1 \\ a' &b' &1 \\ a-a' &b-b' &1 \end{vmatrix}

                            =a\! \begin{vmatrix} b' &1 \\ b-b' &1 \end{vmatrix}-b\! \begin{vmatrix} a' &1 \\ a-a' &1 \end{vmatrix}+1\! \begin{vmatrix} a' &b' \\ a-a' &b-b' \end{vmatrix}

                            =a\! \left ( b'-\! \left ( b-b' \right ) \right )-b\! \left ( a'-\! \left ( a-a' \right ) \right )+1\! \left [ a'\! \left ( b-b' \right )-b'\! \left ( a-a' \right ) \right ]

                            =a\left(b^{\prime}-b+b^{\prime}\right)-b\left(a^{\prime}-a+a^{\prime}\right)+\left(a^{\prime} b-\not a^{\prime} \phi^{\prime}-b^{\prime} a^{\prime}+\not b^{\prime} \alpha^{\prime}\right)

                            =a\left(2 b^{\prime}-b\right)-b\left(2 a^{\prime}-a\right)+1\left(a^{\prime} b-b^{\prime} a\right)

                            =2 a b^{\prime}-a b-2 a^{\prime} b+a b+a^{\prime} b-b^{\prime} a

                                        P u t\; a b^{\prime}=a^{\prime} b

                            \Rightarrow 2 a^{\prime} b-a b-2 a^{\prime} b+a b+a^{\prime} b-a^{\prime} b

                            \Rightarrow 0

Hence, points are collinear.

Posted by

Gurleen Kaur

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