Question

In a triangle ABC, A=(4,6), mid point of AB is (12,10) and midpoint of BC is (18,4). The length of side BC is

Answer 1

Given:

A=(4,6), mid point of AB is (12,10) and midpoint of BC is (18,4)

To find:

The length of the BC

Solution:

As it is given in the question that the mid-point of AB is (12,10)

Let the coordinate of B is (a,b) and C is (c,d)

Then coordinate of B is given by

$\frac{x1+x2}{2} ,\frac{y1+y2}{2} = (x,y)$

$\frac{4+a}{2} ,\frac{6+b}{2} = (12,10)$

on comparing we get

• $\frac{4+a}{2} = 12 ,\frac{6+b}{2} = 10$
• ${4+a} = 24 ,{6+b}= 20$
• $a = 20 , b = 14$
• B(20,14)

Again it is given in the question that the mid-point of BC is (18,4)

Then coordinate of C is given by

$\frac{x1+x2}{2} ,\frac{y1+y2}{2} = (x,y)$

$\frac{20+c}{2} ,\frac{14+d}{2} = (18,4)$

on comparing we get

• $\frac{20+c}{2} = 18 ,\frac{14+b}{2} = 4$
• $20+c} = 36 ,{14+d}= 8$
• $c = 16 , b = -6$
• C(16,-6)

We will find the distance of BC by the distance formula

• $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$
• $\sqrt{(20-16)^{2}+(14-(-6))^{2}}$
• $\sqrt{(20-16)^{2}+(14+6)^{2}}$
• $\sqrt{4^{2}+20^{2}}$
• $\sqrt{16+400}$
• $\sqrt{416}$
• 4√26

So the length of the BC is 4√26 units