Question

6. What point on the y-axis is equidistant from P(0.8
and Q (-4, 4)?​

Answer 1

hope you got it the answer .........

Answer 2

Answer:

istance formula:

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

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

(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-8PO=

(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-8QO=

(y−4)

2

+(0−(−4))

2

=y−8

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

(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}y−8=

(y−4)

2

+16

Squaring on both sides , we get

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

2

−16y+64=y

2

−8y+16+16 [∵ (a−b)

2

=a

2

−2ab+b

2

]

-16y+8y=32-64−16y+8y=32−64

\begin{gathered}-8y=-32\\\\\Rightarrow\ y=\dfrac{32}{8}=4\end{gathered}

−8y=−32

⇒ y=

8

32

=4

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