Question

Using distance formula, show that A(2, 3), B(4, 7) and C(0, -1) are collinear points.

Answer 1

HELLO DEAR,

given vertices are A(2 , 3) , B(4 , 7) and C(0 , -1).

distance formula = $\bold{\sf{\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}}}$

AB = √{(7 - 3)² + (4 - 2)²} = √(16 + 4) = 2√5.
BC = √{(0 - 4)² + (-1 - 7)²} = √(16 + 64) = 4√5.
AC = √{(3 + 1)² + (2 - 0)²} = √(16 + 4) = 2√5.

therefore, consider BC = AB + AC
4√5 = 2√5 + 2√5 = 4√5
HENCE, A,B,C are collinear.

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Dear Student,

Solution:

Distance formula : $\sqrt{(x_{2} - x_{1})^{2}+(y_{2} - y_{1})^{2}}$

A( 2,3 ) B ( 4,7)

Distance between AB = $\sqrt{(4 - 2)^{2}+(7 - 3)^{2}}$

AB =$\sqrt{2^{2}+4^{2}} \\ =\sqrt{4+16} \\ \\ =\sqrt{20} \\ \\ 2\sqrt{5}$

A(2,3 ) C (0,-1)

Distance between AC = $\sqrt{(0 - 2)^{2}+(-1 - 3)^{2}}$

AC =$\sqrt{2^{2}+4^{2}} \\ =\sqrt{4+16} \\ \\ =\sqrt{20} \\ \\ 2\sqrt{5}$

B(4,7) C (0 ,-1)

Distance between BC = $\sqrt{(0 - 4)^{2}+(-1-7 )^{2}}$

BC =$\sqrt{4^{2}+8^{2}} \\ =\sqrt{16+64} \\ \\ =\sqrt{80} \\ \\ 4\sqrt{5}$

So , we can verify that AB+ AC = BC

Thus A ,B and C points are collinear.

Hope it helps you.