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PS · Problem Solving

705-805

Functions

In the normal x-y coordinate plane there are 4 points A(-1,3), B(3,3), C(-1,7), and D(3,7). If a line passing through the origin bisects the area of the rectangle ABCD, what is the slope of the line?

Answer Choices

Correct answer marked below

A

5

B

6

7

Correct

D

8

E

9

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Official Explanation

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The rectangle ABCD has vertices A(-1,3), B(3,3), C(-1,7), D(3,7).

Width =

3 - (-1) = 4

Height =

7 - 3 = 4

So it is a square of side 4, area =

4 \times 4 = 16

.

We need a line through the origin that cuts the square into two equal areas of 8 each.

Let the line be

y = mx

.

The square lies in the region

-1 \le x \le 3

,

3 \le y \le 7

.

Since the line goes through the origin and the square is above and to the right of the origin, the line will enter the square at its bottom edge (y=3) and exit at its right edge (x=3) or top edge (y=7), depending on slope.

Find intersection with bottom edge y=3:

3 = m x \implies x = \frac{3}{m}

.

This point is on the bottom edge if

-1 \le \frac{3}{m} \le 3

.

Find intersection with right edge x=3:

y = m \cdot 3 = 3m

.

This point is on the right edge if

3 \le 3m \le 7

→

1 \le m \le \frac{7}{3}

.

For the line to bisect area, the region below the line inside the square must have area 8.

Consider two cases: exit through right edge or top edge.

Case 1: Exit through right edge (x=3). Then

1 \le m \le \frac{7}{3}

.

The line enters at

(\frac{3}{m}, 3)

and exits at

(3, 3m)

.

The region below the line inside the square is a trapezoid with vertices:

Bottom-left corner (-1,3), entry point

(\frac{3}{m}, 3)

, exit point (3,3m), and point on left edge? Actually, the left edge is x=-1, y from 3 to 7. The line may intersect left edge? Since origin is at (0,0) and slope positive, the line will not intersect left edge (x=-1) because at x=-1, y=-m, which is negative, so left edge is entirely above the line? Wait, check: at x=-1, y=-m (negative), so the line is below the entire left edge of the square. So the region below the line inside the square is bounded by: bottom edge from x=-1 to x=3/m, then the line segment from (3/m,3) to (3,3m), then right edge from (3,3m) down to (3,3)? No, right edge goes from (3,3) to (3,7). The portion from (3,3) to (3,3m) is part of the boundary.

Actually, easier: The area below the line inside the square = area under the line from x=3/m to x=3, minus the rectangle below y=3.

But careful: The square starts at y=3. So the region is a trapezoid with:

Left vertical side from y=3 to y on the line at x=3/m? Wait, at x=3/m, y=3 (entry). At x=-1, the line is at y=-m, which is below the square, so the region includes the entire bottom strip from x=-1 to x=3/m, y=3.

Thus the region consists of a rectangle from x=-1 to x=3/m, height 0? Actually, from y=3 to y=3, so zero area? That's just the bottom edge. The area comes from the part under the line from x=3/m to x=3, but above y=3.

So area = ∫ from x=3/m to 3 of (mx - 3) dx.

Compute: ∫ (mx - 3) dx =

\frac{m}{2}x^2 - 3x

from 3/m to 3.

At x=3:

\frac{m}{2}(9) - 9 = \frac{9m}{2} - 9

.

At x=3/m:

\frac{m}{2} \cdot \frac{9}{m^2} - 3 \cdot \frac{3}{m} = \frac{9}{2m} - \frac{9}{m} = \frac{9}{2m} - \frac{18}{2m} = -\frac{9}{2m}

.

Subtract:

(\frac{9m}{2} - 9) - (-\frac{9}{2m}) = \frac{9m}{2} - 9 + \frac{9}{2m}

.

Set equal to 8 (half area):

\frac{9m}{2} - 9 + \frac{9}{2m} = 8

Multiply by 2:

9m - 18 + \frac{9}{m} = 16

9m + \frac{9}{m} = 34

Multiply by m:

9m^2 + 9 = 34m

9m^2 - 34m + 9 = 0

Discriminant:

34^2 - 4\cdot9\cdot9 = 1156 - 324 = 832

, not a perfect square, so m not integer.

But options are integers 5,6,7,8,9. So this case doesn't yield integer slope.

Case 2: Exit through top edge (y=7). Then slope m > 7/3.

Intersection with top edge: y=7, so

7 = m x \implies x = \frac{7}{m}

.

This point is on top edge if

-1 \le \frac{7}{m} \le 3

. Since m>0,

\frac{7}{m} \le 3 \implies m \ge \frac{7}{3}

(already true). Also

\frac{7}{m} \ge -1

automatically for m>0.

The line enters at bottom edge at

(\frac{3}{m}, 3)

and exits at top edge at

(\frac{7}{m}, 7)

.

Now the region below the line inside the square consists of two parts:

Rectangle from x=-1 to x=3/m, height 4 (from y=3 to y=7)? Wait, not exactly. The region below the line includes all points in the square with y < mx.

Since the line goes from bottom to top, the region below it includes the entire left part of the square up to x=3/m? Actually, for x < 3/m, the line y=mx is above y=3? At x=3/m, y=3. For x < 3/m, since m>0, y=mx < 3. So below the bottom edge of the square. So for x < 3/m, the square's bottom is at y=3, which is above the line. So the region below the line inside the square for x < 3/m is empty because the square starts at y=3, and the line is below that.…

In the normal x-y coordinate plane there are 4 points A(-1,3), B(3,3), C(-1,7), and D(3,7). If a line passing through the origin bisects the area of the rectangle ABCD, what is the slope of the line?

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