Question

Point on Y-axis which is equidistant from P(7,-8) , Q(1,4)

Answer 1

Ⓐ︎Ⓝ︎Ⓢ︎Ⓦ︎Ⓔ︎Ⓡ︎

Let A(0,y) be be the point on Y axis which is equidistant from P(7,-8) and Q(1,4)

d(PA) = d(QA)

By distance formula,

√(0 - 7)² + [ y - (-8)]² = √(0 - 1)² + (y - 4)²

√ (-7)² + (y + 8)² = √ (-1)² + (y - 4)²

squaring both the side,

49 + (y + 8)² = 1 + (y - 4)²

49 + y² + 16y + 64 = 1 + y² - 8y + 16

16y + 113 = -8y + 17

16y + 8y = 17 - 113

24y = -96

$y = \frac{ - 96}{24} = - 4$

co-ordinate of point A(0, -4)

Answer 2

Step-by-step explanation:

P (7,-8)

R (0,y)

Q (1,4)

PR = RQ. (Equidistant Given)

By using distance formula.

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

$PR = \sqrt{ {(7_{} - 0_{}})^{2} + {( - 8_{} - y_{}})^{2}}$

$PR = \sqrt{49 + 64 + {y}^{2} + 16y}$

$PR = \sqrt{113 + {y}^{2} + 16y}$

$RQ = \sqrt{( {0 - 1)}^{2} + {(y - 4)}^{2} }$

$RQ = \sqrt{1 + {y}^{2} + 16 - 8y }$

$RQ = \sqrt{17 + {y}^{2} - 8y }$

PR = RQ

Compare

$\sqrt{113 + {y}^{2} + 16y} = \sqrt{17 + {y}^{2} - 8y }$

$113 + {y}^{2} + 16y = 17 + {y}^{2} - 8y$

${y}^{2} - {y}^{2} + 16y + 8y = 17 - 113$

$24y = - 96$

$y = - \frac{96}{24}$

$y = - 4$

So

required Cordinate

(0,y) =(0,-4) Ans

I hope its help

And make sure hit like and follow and brilliant marks