Question

find the value of y for which the distance between the points A(3,-1) and B(11,y) is 10 units​

Answer 1

Given:-

Two points A(3,1) and B(11,y) and AB = 10 units.

To find:-

Value of y

Solution:-

We have, Coordinates of A and B as (3,1) and (11,y) respectively.

Now,

$AB = \sqrt{ {(11 - 3)}^{2} + {(y - 1)}^{2} }$

$AB = \sqrt{ {8}^{2} + {y}^{2} - 2y + 1 }$

$10 = \sqrt{64 + 1 + {y}^{2} + 2y}$

${10}^{2} = {y}^{2} + 2y + 1$

$100 = {y}^{2} - 2y + 1$

${y}^{2} - 2y - 99 = 0$

${y}^{2} - 11y + 2y - 99 = 0$

$y(y - 11) + 2(y - 11) = 0$

$(y - 11)(y + 2) = 0$

y - 11 = 0 or y + 2 = 0

y = 11,-2

Hence, the value of y is 11 or -2