Question

If A( 1,8) B(2,6) and C(4,2) are three points, show that AC=3AB​

Answer 1

Answer:

Proven below

Step-by-step explanation:

Distance Between Points in the Plane

Given two points A(x,y) and B(w,z), the distance between them is:

$d=\sqrt{(w-x)^2+(z-y)^2}$

Let's calculate the distance AC, where A(1,8) and C(4,2):

$d_{AC}=\sqrt{(4-1)^2+(2-8)^2}$

$d_{AC}=\sqrt{3^2+(-6)^2}$

$d_{AC}=\sqrt{9+36}=\sqrt{45}$

Since 45=9*5:

$d_{AC}=\sqrt{9*5}=3\sqrt{5}$

Calculate the distance AB, where A(1,8) and B(2,6)

$d_{AB}=\sqrt{(2-1)^2+(6-8)^2}$

$d_{AB}=\sqrt{1^2+(-2)^2}$

$d_{AB}=\sqrt{1+4}=\sqrt{5}$

It follows that:

$d_{AC}=3d_{AB}$

Answer 2

Answer:

see explanation

Step-by-step explanation:

Calculate AC and AB using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = A(1,8) and (x₂, y₂ ) = C(4,2)

AC = $\sqrt{(4-1)^2+(2-8)^2}$

      = $\sqrt{3^2+(-6)^2}$

       = $\sqrt{9+36}$

       = $\sqrt{45}$ = $\sqrt{9(5)}$ = 3$\sqrt{5}$

Repeat with

(x₁, y₁ ) = A(1, 8) and (x₂, y₂ ) = B(2, 6)

AB = $\sqrt{(2-1)^2+(6-8)^2}$

      = $\sqrt{1^2+(-2)^2}$

       = $\sqrt{1+4}$

       = $\sqrt{5}$

So AB = $\sqrt{5}$ and AC = 3$\sqrt{5}$

Thus

AC = 3AB