# What is the abscissa of endpoint N(x,8) if the length of MN=5 and M(3,5)?

neela | Student

Given MN = 5,  the coordinates of M is (3,5) and the coordinates of N is (x,8).

So MN = sqrt {(xN-xM)^2+(yN-yM)^2}

Therefore MN = sqrt{(x-3)^2+(8-5)^2}.

MN = 5 = sqrt{(x-3)^2+3^3}.

5^2 = (x-3)^2+9.

25 - 9 = (x-3)^2.

16 = (x-3)^2.

x-3 = sqrt16 , or x- 3 = -sqrt16.

So x= 3+4, Or x= 3-4.

So x= 7, or x= 3-4 = -1.

Therefore the abscissa or the x coordinate of N is  7 or -1.

giorgiana1976 | Student

We'll write the formula of the length of the segment MN:

d = sqrt[(xN - xM)^2 + (yN - yM)^2]

We'll substitute the coordinates for M and N in the formula:

M(3,5) and N(x,8)

d = sqrt[(x - 3)^2 + (8 - 5)^2]

We'll substitute the length of the segment MN by 5:

5 =  sqrt[(x - 3)^2 + (3)^2]

We'll raise to square both sides:

25 = (x-3)^2 + 9

We'll subtract 9 both sides:

25 - 9 = (x-3)^2

16 = (x-3)^2

We'll extract the square root:

x-3 = +4

x = 7

or

x - 3 = -4

x = -1

The abscissa of the endpoint N is x = -1 or x = 7.