Question

if p(0,y) is equidistant from A(2,5) and B(3,7). find the value of y​

Answer 1

Given:-

✯ Vertices of P(0, y)

✯ Vertices of A(2, 5)

✯ Vertices of B(3, 7)

✯ P is equidistant from A and B i.e. P is midpoint of line AB.

To Find:-

✯ Value of y

Solution:-

$\sf A(x_{1} = 2, y_{1} = 5)$

$\sf B(x_{2} = 3, y_{2} = 7)$

$\sf P(x = 0, y = ?)$

As, P is midpoint of A and B

Therefore, we have

$\sf y = \frac{y_{1} + y_{2}}{2}$

$\sf y = \frac{5 + 7}{2}$

$\sf y = \frac{\cancel{12}}{\cancel{2}}$

$\sf y = 6$

Hence, value of y = 6.

Answer 2

Step-by-step explanation:

Given:-

✯ Vertices of P(0, y)

✯ Vertices of A(2, 5)

✯ Vertices of B(3, 7)

✯ P is equidistant from A and B i.e. P is midpoint of line AB.

To Find:-

✯ Value of y

Solution:-

\sf A(x_{1} = 2, y_{1} = 5)A(x

1

=2,y

1

=5)

\sf B(x_{2} = 3, y_{2} = 7)B(x

2

=3,y

2

=7)

\sf P(x = 0, y = ?)P(x=0,y=?)

As, P is midpoint of A and B

Therefore, we have

\sf y = \frac{y_{1} + y_{2}}{2}y=

2

y

1

+y

2

\sf y = \frac{5 + 7}{2}y=

2

5+7

\sf y = \frac{\cancel{12}}{\cancel{2}}y=

2

12

\sf y = 6y=6

Hence, value of y = 6.