Question

The distance between P(p,3) and Q(-1,2p) is 5 units. Find the distance PQ in terms of p

Answer 1

Use pythagorean distance formula with x1 - x2 and y1 - y2 as the legs. Distance will be expressed in terms of "p".

Answer 2

Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

Answer 3

would that be p^+1+9-2p^ ?

Answer 4

You have to plug in the coordinates for x1, x2, y1, y2 from the points.

Answer 5

x1 is p. y1 is 3. x2 is -1. y2 is 2p. Just plug into formula.

Answer 6

yep thats what i did and i think it simplifies to my above answer. How do i find out what p is?

Answer 7

First, you won't know what p is. p is p, it's left as a variable. Second, your intermediate answer is still off, so write it out before simplifying so I can tell you where you went wrong. I can't give out a final answer. Have to help the asker to get it himself.

Answer 8

ok i did ((p--1)^ + (3-2p)^)

Answer 9

What is this #^ ? You are leaving out a power. Perhaps you mean #^2 ? Fill in your exponents, they are not implied.

Answer 10

If you put a 2 after your ^ it looks like you are on the right track.

Answer 11

If you put in the correct corresponding points, you will come down to distance equaling the positive (since distances are positive, you take the positive sqrt) square root of a particular quadratic in p. Try it again now. Try to get that quadratic in p.

Answer 12

After you set distance = 5, you can square both sides and solve for p.

Answer 13

oh yes sorry ^2

Answer 14

The quadratic in p can be solved by factoring and you should get 2 values for p, so there will be 2 SETS of 2 points each for your solution.

Answer 15

Are you still there?

Answer 16

sorry I had to rush off but thank you very much for your help!