Question

Find the value of y for which the distance between the point A(3,-1) and (11,y) Is 10 units.

Answer 1

Hope this helps u a lottt

Answer 2

Answer:

The value of y are -7 and 5 .

Step-by-step explanation:

Formula

$Distance\ formula = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

As given

Distance between the point A(3,-1) and (11,y) Is 10 units.

Putting the values in the above

$10= \sqrt{(11-3)^{2}+(y-(-1))^{2}}$

$10= \sqrt{(8)^{2}+(y+1)^{2}}$

$10= \sqrt{(8)^{2}+(y+1)^{2}}$

Taking square on the both sides .

( By using this formula (a + b)² = a² + b² + 2ab )

100 = 64 + y² + 1 + 2y

y² + 2y + 65 - 100 = 0

y² + 2y - 35 = 0

y² + 7y - 5y - 35 = 0

y(y + 7) -5 (y + 7 ) = 0

(y + 7)(y - 5) = 0

y = -7 , y = 5

Therefore the value of y are -7 and 5 .