Question

If the distance between the points (7,4 )and ( 3 ,y) is 4 units find the value of y

Answer 1

Answer:

see this pic of the sum and under stand

Answer 2

The value of "y" is "4". Further explanation is given below.

Step-by-step explanation:

Given :

(x1,y1) = (7,4)

(x1,y2) = (3,y)

Distance = 4 units

As we know,

Distance = $\sqrt{(x2-x1)^2+(y2-y1)^2}$

On putting the values, we get

⇒  4 = $\sqrt{(3-7)^2+(y-4)^2}$

⇒  4 = $\sqrt{-4^2+(y-4)^2}$

⇒  4 = $\sqrt{16+(y-4)^2}$

On squaring both sides, we get

⇒  16 = 16+(y-4)²

On subtracting "16" from both sides, we get

⇒  16-16 = 16+(y-4)²-16

⇒  0 = (y-4)²

⇒  0 = y²+16-8y

⇒  y²-8y+16 = 0

⇒  y²-4y-4y+16 = 0

⇒  y(y-4)-4(y-4) = 0

⇒  (y-4) (y-4) = 0

⇒  y-4 = 0

⇒  y = 4

Or,

⇒  y-4 = 0

⇒  y = 4

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