Question

What is the sum of the distances of R to Q and P to Q?

Answer 1

The sum of the distances of R to Q and P to Q is 9 units.

Solution:

Given points are P(–3, 6), Q(3, 6) and R(3, 3).

Distance between two points formula:

   $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance from R to Q:

Here $x_1=3, y_1=3, x_2=3, y_2=6$

Substitute these in the given formula, we get

Distance = $\sqrt{(3-3)^2+(6-3)^2}$  

               $=\sqrt{0+9}$

               = 3 units

Distance from R to Q is 3 units.

Distance from P to Q:

Here $x_1=-3, y_1=6, x_2=3, y_2=6$

Substitute these in the given formula, we get

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

               $=\sqrt{36+0}$

               = 6 units

Distance from P to Q is 6 units.

Sum of the distances = 3 + 6 = 9 units

Hence the sum of the distances of R to Q and P to Q is 9 units.