Question

Find the values of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units.

Answer 1

√[(10-2)²+(y+3)²]=10

64+(y+3)^2=100

(y+3)²=36

y+3=6                 or           y+3=-6

y=3                       or          y=-9

Answer 2

Answer:

y=3 or -9

Step-by-step explanation:

If the coordinates of two points on the coordinate plane are known to be (x1, y1) and  (x2, y2), then the distance between them is given by

$\sqrt{(x1-x2)^{2}+(y1-y2)^{2}  }$

Now, Coordinates of P(2,-3) and Q(10,y) are given. And the distance PQ is given as 10 units. The value of y is to be determined.

So, $\sqrt{(2-10)^{2}+(-3-y)^{2}  }$=10

{Squaring both sides}

⇒ 64+ (3+y)² =100

⇒ (3+y)² =36

⇒ (3+y) = ±6

⇒ y=3 or -9. (Answer)