In the standard (x, y) coordinate plane, quadrilateral ABCD is an isosceles trapezoid wher… — ACT

In the standard $(x, y)$ coordinate plane, quadrilateral $ABCD$ is an isosceles trapezoid where $AB$ is parallel to $CD$ and the length of $AD$ equals the length of $BC$ . The coordinates of three of the vertices are $A(0, 0)$ , $B(16, 12)$ , and $C(9, 13)$ . If $ABCD$ is not a parallelogram, what is the $y$ -coordinate of vertex $D$ ?

A
$-1$
B
$1$
C
$5$
$7$ Answer
E
$13$

Answer

The correct

y

-coordinate of vertex

D

7

The correct

y

-coordinate is

7

. The slope of base

AB

\frac{12-0}{16-0} = \frac{3}{4}

, meaning the parallel line containing base

CD

y = \frac{3}{4}x + \frac{25}{4}

. The length of the leg

BC

\sqrt{(9-16)^2 + (13-12)^2} = \sqrt{50}

. Setting the distance of leg

AD

equal to

\sqrt{50}

gives the equation

x^2 + y^2 = 50

. Substituting the line equation yields

x^2 + (\frac{3x+25}{4})^2 = 50

, which simplifies to the quadratic

x^2 + 6x - 7 = 0

. Solving for

x

gives

x = 1

x = -7

. If

x = -7

, then

y = 1

, which makes

ABCD

a parallelogram. If

x = 1

, then

y = 7

, which successfully forms a non-parallelogram isosceles trapezoid.

Step-by-Step Solution

Calculate the slope of the base

AB

Slope

m = \frac{12 - 0}{16 - 0} = \frac{3}{4}

Since the trapezoid has bases

AB \parallel CD

, the line containing segment

CD

must also have a slope of

\frac{3}{4}

Determine the equation of the line containing segment

CD

passing through

C(9, 13)

y - 13 = \frac{3}{4}(x - 9) \implies y = \frac{3}{4}x + \frac{25}{4}

Using the point-slope form with vertex

C

and the parallel slope allows us to express the coordinates of

D

(x, \frac{3}{4}x + \frac{25}{4})

Calculate the length of the leg

BC

using the distance formula.

BC = \sqrt{(9 - 16)^2 + (13 - 12)^2} = \sqrt{(-7)^2 + 1^2} = \sqrt{50}

Since the trapezoid is isosceles with

AD = BC

, the distance from the origin

A(0,0)

to vertex

D(x, y)

must also satisfy

AD = \sqrt{50}

Set up the distance equation for

AD

and substitute the line equation for

y

x^2 + y^2 = 50 \implies x^2 + \left(\frac{3}{4}x + \frac{25}{4}\right)^2 = 50

Substituting the relation for

y

in terms of

x

allows us to solve for the

x

-coordinate of vertex

D

Solve the quadratic equation for

x

16x^2 + (9x^2 + 150x + 625) = 800 \implies 25x^2 + 150x - 175 = 0 \implies x^2 + 6x - 7 = 0 \implies (x + 7)(x - 1) = 0 \implies x = 1 \text{ or } x = -7

The solutions to this quadratic equation yield two potential coordinates for vertex

D

Find the corresponding

y

-coordinates and verify the non-parallelogram condition.

x = -7

, then

y = 1

, which makes

ABCD

a parallelogram. If

x = 1

, then

y = 7

, which makes

ABCD

a non-parallelogram isosceles trapezoid.

The question specifies that

ABCD

is not a parallelogram, so we choose the solution

D(1, 7)

, giving a

y

-coordinate of

7

Key Concept

Using coordinate geometry (slopes and distances) to determine the properties and vertices of a quadrilateral.

Alternative Method

In an isosceles trapezoid, the perpendicular bisector of the base is the axis of symmetry. The midpoint of base

AB

M(8, 6)

. Since the slope of

AB

\frac{3}{4}

, the slope of the perpendicular bisector is the negative reciprocal,

-\frac{4}{3}

. The equation of this perpendicular bisector is

y - 6 = -\frac{4}{3}(x - 8) \implies 4x + 3y - 50 = 0

. Reflecting vertex

C(9, 13)

across this line yields vertex

D

. The projection of

C

onto the bisector is found by intersecting it with the parallel base line

3x - 4y + 25 = 0

, giving the intersection point

P(5, 10)

. Reflecting

C

across

P

gives

D = 2P - C = (2(5) - 9, 2(10) - 13) = (1, 7)

, which confirms the

y

-coordinate is

7

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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