In the xy-plane, A(1, 2), B(7, 4), and D(3, 8) are three vertices of rhombus ABCD. A line … — GRE General Test

In the $xy$ -plane, $A(1, 2)$ , $B(7, 4)$ , and $D(3, 8)$ are three vertices of rhombus $ABCD$ . A line $k$ passes through the origin $(0, 0)$ and bisects the area of rhombus $ABCD$ . What is the slope of line $k$ ?

$\frac{6}{5}$ Answer
B
$\frac{5}{6}$
C
$1$
D
$-1$
E
$\frac{3}{4}$

Answer

The slope of line

k

\frac{6}{5}

Any line that divides a parallelogram or rhombus into two equal areas must pass through its center of symmetry, which is the midpoint of its diagonals. The midpoint of diagonal

BD

with endpoints

(7, 4)

and

(3, 8)

\left(\frac{7+3}{2}, \frac{4+8}{2}\right) = (5, 6)

. Since line

k

passes through the origin

(0, 0)

and

(5, 6)

, its slope is

\frac{6 - 0}{5 - 0} = \frac{6}{5}

Step-by-Step Solution

Identify the key geometric property of area-bisecting lines for parallelograms and rhombuses.

Any line that bisects the area of a rhombus must pass through its center of symmetry (the intersection point of its diagonals).

A rhombus is centrally symmetric about the intersection point of its diagonals, so any line through this point divides the rhombus into two congruent regions.

Find the coordinates of the center of symmetry by calculating the midpoint of diagonal

BD

Midpoint

M = \left(\frac{7+3}{2}, \frac{4+8}{2}\right) = (5, 6)

Opposite vertices

B(7,4)

and

D(3,8)

define one of the diagonals of rhombus

ABCD

Calculate the slope of line

k

passing through the origin

(0, 0)

and center point

M(5, 6)

\text{Slope } m = \frac{6 - 0}{5 - 0} = \frac{6}{5}.

The slope formula between

(x_1, y_1)

and

(x_2, y_2)

\frac{y_2 - y_1}{x_2 - x_1}

Key Concept

Center of Symmetry and Area Bisectors of Quadrilaterals

Estimated Time:2m 0s

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