Login/Sign Up
Home » 2. Using the determinants show that the following points are collinear:(iii) , and (iv) , and
2. Using the determinants show that the following points are collinear:(iii) (3,-2), (8,8) and (5,2) (iv) (2,3), (-1,-2) and (5,8)
CBSE | Class 12 | Determinants | Exercise 6.3 | Maths | RD Sharma
2. Using the determinants show that the following points are collinear:(iii) (3,-2), (8,8) and (5,2) (iv) (2,3), (-1,-2) and (5,8)

(iii) As per the question it is given that (3,-2), (8,8) and (5,2)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are \left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\And \left( {{x}_{3}},{{y}_{3}} \right), then the area of the triangle is given by,

\vartriangle =\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=0

Then, put given values in above formula

\vartriangle =\frac{1}{2}\left| \begin{matrix} 3 & -2 & 1 \\ 8 & 8 & 1 \\ 5 & 2 & 1 \\ \end{matrix} \right|=0

\frac{1}{2}\left| \begin{matrix} 3 & -2 & 1 \\ 8 & 8 & 1 \\ 5 & 2 & 1 \\ \end{matrix} \right|

Expanding along {{R}_{1}}

=\frac{1}{2}\left[ 3\left| \begin{matrix} 8 & 1 \\ 2 & 1 \\ \end{matrix} \right|-2\left| \begin{matrix} 8 & 1 \\ 5 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} 8 & 8 \\ 5 & 2 \\ \end{matrix} \right| \right]

=\frac{1}{2}\left[ 3\left( 6 \right)-2\left( 3 \right)+1\left( -24 \right) \right]

=\frac{1}{2}\left[ 0 \right]

=0

Therefore, Area of triangle is zero so, the points are collinear.

(iv) As per the information given in the question (2,3), (-1,-2) and (5,8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are \left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\And \left( {{x}_{3}},{{y}_{3}} \right), then the area of the triangle is given by,

\vartriangle =\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=0

Put the value in above formula

\vartriangle =\frac{1}{2}\left| \begin{matrix} 2 & 3 & 1 \\ -1 & 2 & 1 \\ 5 & 8 & 1 \\ \end{matrix} \right|=0

\frac{1}{2}\left| \begin{matrix} 2 & 3 & 1 \\ -1 & -2 & 1 \\ 5 & 8 & 1 \\ \end{matrix} \right|

Expanding along {{R}_{1}}

=\frac{1}{2}\left[ 2\left| \begin{matrix} -2 & 1 \\ 8 & 1 \\ \end{matrix} \right|-3\left| \begin{matrix} -1 & 1 \\ 5 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} -1 & -2 \\ 5 & 8 \\ \end{matrix} \right| \right]

=\frac{1}{2}\left[ 2\left( -10 \right)-3\left( -1-5 \right)+1\left( -8+10 \right) \right]

=\frac{1}{2}\left[ -20+18+2 \right]

=0

=\frac{1}{2}\left[ 2\left( -10 \right)-3\left( -1-5 \right)+1\left( -8+10 \right) \right]

=\frac{1}{2}\left[ -20+18+2 \right]

=0

i

More Material