Question

The distance between the points (-2,y) and (3, -7) is 13 units.What are the possible values of y?

Answer 1

To calculate hte possible values of y you have to apply the Pythagoras theorem:

`a^2+b^2=c^2`

Where

c will be the distance between the given points, and the hypothenuse of a right triangle

a will be the base of a theoretical triangle below the hypothenuse, you calculate it as (x2-x1)

b= will be the heigth of said triangle, you calculate it using the y-coordinates (y2-y1)

So:

`\begin{gathered} c^2=a^2+b^2 \\ c^2=(x_2-x_1)^2+(y_2-y_1)^2 \\ c=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}`

Replace the expression with the given measurements to calculate the y-coordinate of the first point:

`\begin{gathered} c=\sqrt[]{(3_{}-(-2))^2+(-7-y)^2} \\ 13=(3+2)+(-7-y) \\ 13=5-7-y \\ 13=-2-y \\ 13+2=-y \\ -15=y \end{gathered}`

The possible values for y, since its