Question

Find x so the distance between (3,-3) and (x,2) is 13

Answer 1

The distance between two points is: Square root of { (difference in x coordinates)^2 + (difference in y coordinates)^2 }

Answer 2

Difference in x coordinates is (x-3).
Difference in y coordinates is 2 - (-3) = 5.
Square root of { (x-3)^2 + 5^2 } = 13
You can now solve for x.

Answer 3

I got passed the sqrt of (x-3)^2-5^2=13 but couln't come out with the right answer. can you show the steps in order for me to solve for x?

Answer 4

To get rid of the square root, square BOTH sides.
(x-3)^2 + 5^2 = 13^2
x^2 - 6x + 9 + 25 = 169
x^2 - 6x - 135 = 0
Solve the quadratic equation for x

Answer 5

Ooooh okay, the part I forgot to do was square both sides. Thank you :)

Answer 6

When you solve the quadratic equation you will get two values for x and BOTH are valid solutions to the problem.

Answer 7

What x values are you getting?

Answer 8

I ended up doing a different problem that was similar which was "Find x so the distance between (-9,-4) and (x,-1) is 13" my x values were 21 and -3

Answer 9

For this problem, x = 15 and -9

Answer 10

oops i mean (9,4) and (x,-1)

Answer 11

Yes, for (9,4) and (x,-1) your solutions x = 21 & -3 are correct!

Answer 12

Just remember that the formula for distance between two points comes from the Pythagoras theorem (a^2 + b^2 = c^2) where the hypotenuse of a right triangle is the square root of the sum of squares of the other two sides. If you graph any two points (x1, y1) and (x2, y2) and draw a line between them, the length of that line will be the distance between the two points. The difference in x coordinates gives the length of one side of a right triangle, the difference in y coordinates gives the length of the other side and the distance between the two pints that we are interested in forms the hypotenuse of the right triangle.