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:
[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Distance from R to Q:
Here [tex]x_1=3, y_1=3, x_2=3, y_2=6[/tex]
Substitute these in the given formula, we get
Distance = [tex]\sqrt{(3-3)^2+(6-3)^2}[/tex]
[tex]=\sqrt{0+9}[/tex]
= 3 units
Distance from R to Q is 3 units.
Distance from P to Q:
Here [tex]x_1=-3, y_1=6, x_2=3, y_2=6[/tex]
Substitute these in the given formula, we get
Distance = [tex]\sqrt{(3-(-3))^2+(6-6)^2}[/tex]
[tex]=\sqrt{36+0}[/tex]
= 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.
