Question

Prove that the points A(1,2) B(-3,4) C(7,-1) are collinear and the ratio in which A divides BC

Answer 1

hope it will help youm

Answer 2

A divide BC in 2:3.

Step-by-step explanation:

The given points are A(1,2), B(-3,4), C(7,-1).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula we get

$AB=\sqrt{(-3-1)^2+(4-2)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$

$BC=\sqrt{(7-(-3))^2+(-1-4)^2}=\sqrt{100+25}=\sqrt{125}=5\sqrt{5}$

$AC=\sqrt{(7-1)^2+(-1-2)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}$

Now,

$AB+AC=2\sqrt{5}+3\sqrt{5}$

$AB+AC=5\sqrt{5}$

$AB+AC=BC$

It means points A(1,2) B(-3,4) C(7,-1) are collinear.

$\dfrac{BA}{AC}=\dfrac{AB}{AC}=\dfrac{2\sqrt{5}}{3\sqrt{5}}=\dfrac{2}{3}=2:3$

Therefore, point A divide BC in 2:3.

#Learn more

Distance between the points (-2,3) (2,3).

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