In the standard (x, y) coordinate plane, the midpoint of the line segment with endpoints A… — ACT

In the standard $(x, y)$ coordinate plane, the midpoint of the line segment with endpoints $A(1, 3)$ and $B\left(7, \frac{x}{2}\right)$ is $M$ . If the distance from the origin $(0, 0)$ to $M$ is $\sqrt{\frac{89}{4}}$ units, and $x$ is a positive number, what is the value of $x$ ?

A
2
4Answer
C
8
D
14
E
16

Answer

The correct value of

x

4

The midpoint

M

A(1, 3)

and

B\left(7, \frac{x}{2}\right)

has coordinates

\left(\frac{1+7}{2}, \frac{3 + \frac{x}{2}}{2}\right) = \left(4, \frac{x+6}{4}\right)

. The distance from

(0, 0)

M

\sqrt{4^2 + \left(\frac{x+6}{4}\right)^2} = \sqrt{\frac{89}{4}}

. Squaring both sides yields

16 + \frac{(x+6)^2}{16} = \frac{89}{4}

. Multiplying through by 16 gives

256 + (x+6)^2 = 356

, which simplifies to

(x+6)^2 = 100

. Taking the square root of both sides gives

x+6 = 10

(since

x

is positive), which simplifies to

x = 4

Step-by-Step Solution

Find the coordinates of the midpoint

M(x_M, y_M)

of the segment

AB

M = \left(4, \frac{x+6}{4}\right)

By the midpoint formula,

x_M = \frac{1 + 7}{2} = 4

and

y_M = \frac{3 + \frac{x}{2}}{2} = \frac{\frac{6+x}{2}}{2} = \frac{x+6}{4}

Set up the distance equation from the origin

(0, 0)

M\left(4, \frac{x+6}{4}\right)

using the distance formula.

\sqrt{4^2 + \left(\frac{x+6}{4}\right)^2} = \sqrt{\frac{89}{4}}

The distance between the origin

(0, 0)

and any point

(x, y)

is given by

d = \sqrt{x^2 + y^2}

Solve the distance equation for the positive variable

x

x = 4

Squaring both sides gives

16 + \frac{(x+6)^2}{16} = \frac{89}{4}

. Multiplying by 16 yields

256 + (x+6)^2 = 356

, which simplifies to

(x+6)^2 = 100

. Since

x

is positive,

x+6 = 10 \implies x = 4

Key Concept

Solving coordinate geometry problems by combining the midpoint formula and the distance formula.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic