Answers

2015-05-03T19:30:46+08:00

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Please check the picture attached to understand it more:

The Pythagorean Theorem states that for a right triangle 
a^2+b^2=c^2

In terms of points in the Cartesian plane a is equal to x_a-x_b=a
And b=y_a-y_b

Therefore, 
(x_a-x_b)^{2} + (y_a-y_b)^{2} =c^2

Here since 3\ \textgreater \ -3,  (3,y)=(x_a,y_a) \\ (-3,-1)=(x_b, y_b)

We already know the value of c
(3-(-3))^{2} +(y-(-1))^{2} = \sqrt{45} ^2 \\ 6^2+(y+1)^2=45 \\ 36+(y+1)^2=45 \\ (y+1)^2=9

There are 2 cases 

1st Case :
y+1 is equal to the positive squareroot of 9
(y+1)^2=9 \\ y+1= \sqrt{9}  \\ y+1=3 \\ y=2

2nd case:
y+1 is equal to the negative squareroot of 9
(y+1)^2=- \sqrt{9}  \\ y+1=-3 \\ y=-4

Therefore y={-4, 2}
Either of the two are possible

0