Answer:
[tex]\displaystyle d_{RP}=50\sqrt{2}\ units[/tex]
Step-by-step explanation:
Distance between points in [tex]R^2[/tex]
If P(p1,p2) and Q(q1,q2) are points on the plane [tex]R^2[/tex], the distance between them is
[tex]\displaystyle d=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}[/tex]
We have Q(-10,-20) plotted on the coordinate grid. We also know that P is at (40, -20). We can see they have the same y-coordinate, so the distance between them is computed simply by subtracting their x-coordinates
[tex]\displaystyle d_{PQ}=40-(-10)=50[/tex]
We must locate R knowing it's vertically above Q (x-coordinate = -10) and at the same distance from point Q as point P is from point Q. That means that from R to Q there are 50 units. They-coordinate of R will be -20+50=30.
The point R is located at (-10,30)
The distance from R to P is
[tex]\displaystyle d_{RP}=\sqrt{(-10-40)^2+(30+20)^2}[/tex]
[tex]\displaystyle d_{RP}=\sqrt{(-50)^2+50^2}[/tex]
[tex]\displaystyle d_{RP}=\sqrt{2500+2500}[/tex]
[tex]\displaystyle d_{RP}=\sqrt{5000}[/tex]
[tex]\displaystyle d_{RP}=50\sqrt{2}\ units[/tex]