Question

What point on the y-axis is equidistant from p(0,8) and Q(-4,4).

Answer 1

Answer:0,4

Step-by-step explanation:see the attachment

Answer 2

The point on the y-axis is equidistant from P(0,8) and Q(-4,4) is (0,4)

Explanation:

Distance formula:

Distance between two points (a,b) and (c,d) is given by :-

$\sqrt{(d-b)^2+(c-a)^2}$

Let O(0,y) be the point on the y-axis.

Distance between P(0,8) and O(0,y): $PO=\sqrt{(y-8)^2+(0)^2}=y-8$

Distance between Q(-4,4) and O(0,y): $QO=\sqrt{(y-4)^2+(0-(-4))^2}=y-8$

$=\sqrt{(y-4)^2+16}$

As per given , O is equidistant from p(0,8) and Q(-4,4)..

⇒ PO=QO

$y-8=\sqrt{(y-4)^2+16}$

Squaring on both sides , we get

$(y-8)^2=(y-4)^2+16$

$y^2-16y+64=y^2-8y+16+16\ \ [\because\ (a-b)^2=a^2-2ab+b^2]$

$-16y+8y=32-64$

$-8y=-32\\\\\Rightarrow\ y=\dfrac{32}{8}=4$

Hence, the point on the y-axis is equidistant from P(0,8) and Q(-4,4) is (0,4).

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