Question

A quadrilateral has vertices located at
F
(
7
,
−
5
)
,

H
(
2
,
5
)
,

N
(
−
5
,
4
)
,
F(7,−5), H(2,5), N(−5,4), and
T
(
x
,
2
)
.
T(x,2). Determine the value of
x
x so that quadrilateral
N
H
T
F
NHTF will be a trapezoid and point
T
T is in the first quadrant.

Answer 1

To find the value of \( x \) for which the quadrilateral \( NHTF \) is a trapezoid with \( T(x, 2) \) in the first quadrant, we need to ensure that at least one pair of opposite sides is parallel.

The vertices are:
- \( F(7, -5) \)
- \( H(2, 5) \)
- \( N(-5, 4) \)
- \( T(x, 2) \)

First, we can find the slopes of the sides to determine when they are parallel.

1. **Finding the slope of side FH:**
\[
\text{slope of } FH = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-5)}{2 - 7} = \frac{10}{-5} = -2
\]

2. **Finding the slope of side NT:**
\[
\text{slope of } NT = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 4}{x - (-5)} = \frac{-2}{x + 5}
\]

For \( NHTF \) to be a trapezoid, we want \( FH \parallel NT \), which means their slopes must be equal:
\[
-2 = \frac{-2}{x + 5}
\]

Cross multiplying, we get:
\[
-2(x + 5) = -2
\]
Expanding:
\[
-2x - 10 = -2
\]
Adding 10 to both sides:
\[
-2x = 8
\]
Dividing by -2:
\[
x = -4
\]

Now, we verify whether \( T(-4, 2) \) is in the first quadrant. The point \( T(-4, 2) \) has a negative \( x \) coordinate, which means it is not in the first quadrant. We need to check if the other parallel sides can give us a \( T(x, 2) \) with a positive \( x \) value.

3. **Finding the slope of side NF:**
\[
\text{slope of } NF = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 4}{7 - (-5)} = \frac{-9}{12} = -\frac{3}{4}
\]

4. **Finding the slope of side HT:**
\[
\text{slope of } HT = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 5}{x - 2} = \frac{-3}{x - 2}
\]

Setting the slopes equal for \( NF \parallel HT \):
\[
-\frac{3}{4} = \frac{-3}{x - 2}
\]

Cross multiplying gives:
\[
-3(x - 2) = -12
\]

Solving this:
\[
3(x - 2) = 12
\]
Dividing by 3:
\[
x - 2 = 4
\]
Adding 2 to both sides:
\[
x = 6
\]

Now, \( T(6, 2) \) is indeed in the first quadrant, since both coordinates are valid for a point in that region.

Thus, the value of \( x \) such that quadrilateral \( NHTF \) is a trapezoid and point \( T \) is in the first quadrant is:
\[
\boxed{6}
\]