Question

verify whether the following points are collinear. (i)(1,-1) (4,1), (-2,-3)
(ii) (1,-1), (2,3),(2,0)
(iii) (1,-6), (3,-4), (4,-3)​

Answer 1

We know that ,

If the area of a Δ is 0 units , then its vertices will be collinear . So ,

$\sf \fbox{ \fbox { \: Area \: of \: Δ = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{1}(y_{3} - y_{1}) + x_{1}(y_{1} - y_{2})| \: \: \: \: }}$

$\sf \underline{ \red{For \: (i) \: \: }}$

The area of triangle formed by the vertices A(1,-1) , B(4,1) , and C(-2,-3) , by using the above formula , is given by

$\sf \implies \frac{1}{2} |1 \bigg(1 - ( - 3) \bigg) + 4 \bigg( (- 3 )- ( - 1) \bigg)+ ( - 2) \bigg( (- 1) - 1 \bigg)| \\ \\ \sf \implies \frac{1}{2} | (4 + 4(2) - 2( - 2) | \\ \\ \sf \implies \frac{1}{2} |4 + 4 + 8| \\ \\ \sf \implies \frac{1}{2} |20| \\ \\ \sf \implies \frac{20}{2} \\ \\ \sf \implies 10 \: \: units$

Therefore , the given points are not collinear

$\sf \underline{ \red{For \: (ii) \: \: }}$

The area of triangle formed by the vertices D(1,-1), E(2,3), and F(2,0) , by using the above formula , is given by

$\sf \implies \frac{1}{2} | 1 \bigg(3 - 0 \bigg) + 2 \bigg(0 - ( - 1) \bigg) + 2 \bigg(( - 1) - 3 \bigg)| \\ \\ \sf \implies \frac{1}{2} |3 + 2(1) + 2( - 4)| \\ \\\sf \implies \frac{1}{2} |3 + 2 - 8| \\ \\ \sf \implies \frac{1}{2} |5 - 8| \\ \\\sf \implies \frac{1}{2} | - 3| \\ \\ \sf \implies \frac{3}{2}$

Therefore , the given points are not collinear

$\sf \underline{ \red{For \: (iii) \: \: }}$

The area of triangle formed by the vertices G(1,-6), H(3,-4), I(4,-3) , by using the above formula , is given by

$\sf \implies \frac{1}{2} |1 \bigg( - 4 - ( - 3) \bigg) + 3 \bigg( (- 3 )- ( - 6) \bigg) + 4 \bigg( (- 6) - ( - 4) \bigg)| \\ \\ \sf \implies \frac{1}{2} ( - 1 + 3(3) + 4( - 2)) \\ \\ \sf \implies \frac{1}{2} | - 1 + 9 - 8| \\ \\\sf \implies \frac{1}{2} |8 - 8| \\ \\\sf \implies \frac{0}{2} \\ \\\sf \implies 0$

Therefore , the given points are collinear

Answer 2

Hope this is helpful!!!