In the standard (x, y) coordinate plane, a line segment has endpoints A(-3, k) and B(5, 3)… — ACT

In the standard $(x, y)$ coordinate plane, a line segment has endpoints $A(-3, k)$ and $B(5, 3)$ . If the midpoint of segment $AB$ lies on the $x$ -axis, what is the length of segment $AB$ ?

Cevap: 10

Cevap

The length of segment

AB

is 10.

The midpoint of segment

AB

with endpoints

A(-3, k)

and

B(5, 3)

\left(1, \frac{k + 3}{2}\right)

. Since the midpoint lies on the

x

-axis, its

y

-coordinate must be

0

. Solving

\frac{k + 3}{2} = 0

gives

k = -3

. This means the endpoints are

A(-3, -3)

and

B(5, 3)

. The distance between these two points is

\sqrt{(5 - (-3))^2 + (3 - (-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10

Adım Adım Çözüm

Set up the equation for the

y

-coordinate of the midpoint.

\frac{k + 3}{2} = 0

The midpoint of a segment with endpoints

(x_1, y_1)

and

(x_2, y_2)

has a

y

-coordinate of

\frac{y_1 + y_2}{2}

. Since the midpoint lies on the

x

-axis, its

y

-coordinate must be

0

Solve the equation for

k

k = -3

Multiply both sides of the equation by

2

to get

k + 3 = 0

, then subtract

3

from both sides.

Substitute

k = -3

to find the coordinates of point

A

A(-3, -3)

This provides both complete endpoints,

A(-3, -3)

and

B(5, 3)

, which are needed to find the distance.

Apply the distance formula to find the length of segment

AB

AB = 10

The distance formula is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

. Substituting the coordinates gives

AB = \sqrt{(5 - (-3))^2 + (3 - (-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10

Anahtar Kavram

Distance and Midpoint Formulas

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