Question

P(1,1), Q(-2,8) and R(3-3)show the points are collinear​

Answer 1

Correct Question:

P(1 , 1), Q(- 2 , 7) and R(3 , - 3)bshow the points are collinear.

Answer:-

We have to prove that P(1 , 1) , Q (- 2 , 7) & R (3 , - 3).

We know that,

If three points are collinear then the area of the triangle formed by them will be 0 (zero).

→ Area of ∆PQR = 0 sq.units.

We know,

Area of a triangle with vertices $\sf (x_1 , y_1) ; (x_2 , y_2) \:\: \& \:\: (x_3,y_3)$ is :

$\sf \large\dfrac{1}{2} \begin{vmatrix} \sf \: x_1 - x_2 & \: \sf \: x_1 - x_3 \\ \\ \sf \: y_1 - y_2 & \: \sf \: y_1 - y_3 \end{vmatrix} \quad$

Let,

• $\sf x_1$ = 1

• $\sf x_2$ = - 2

• $\sf x_3$ = 3

• $\sf y_1$ = 1

• $\sf y_2$ = 7

• $\sf y_3$ = - 3

Hence,

$\sf \dfrac {1}{2} \begin{vmatrix} \sf 1 - (-2) &\sf 1-3\\\\\sf 1 - 7 & \sf1 - (-3) \end{vmatrix} = 0 \\\\\implies \begin{vmatrix} \sf 1 +2 &\sf -2\\\\\sf -6& \sf 1 + 3 \end{vmatrix} = 0$

$\implies | \sf(3)(4) - ( - 6)( - 2)| = 0 \\ \\ | \sf \: 12 - 12| = 0 \\ \\ \sf \implies \large \: 0 = 0$

Hence, Proved.