A triangle is graphed by joining three plotted points on the first quadrant of a coordinate plane. The x and y axes range from 0 to 10 in increments of 2.

Question

What is the area of the triangle in square units?

(1 point)
Responses

15
15

9
9

6
6

18

Answer 1

To find the area of a triangle plotted on a coordinate plane using vertices (points), we can use the formula for the area of a triangle given by the coordinates of its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Since we don't have the specific coordinates of the vertices, here are hypothetical examples that could produce the answers provided.

1. Example vertices for the area calculations:

   • Points (2, 0), (0, 6), and (6, 6):
   • \[\text{Area} = \frac{1}{2} \left| 2(6 - 6) + 0(6 - 0) + 6(0 - 6) \right| = \frac{1}{2} \left| 0 + 0 - 36 \right| = \frac{1}{2} (36) = 18\]

2. Another example for an area of 15:

   • Points (0, 0), (5, 3), and (5, 0):
   • \[\text{Area} = \frac{1}{2} \left| 0(3 - 0) + 5(0 - 0) + 5(0 - 3) \right| = \frac{1}{2} \left| 0 + 0 - 15 \right| = \frac{1}{2} (15) = 7.5\]
   • This doesn't give 15, so it can be modified slightly to (0, 0), (6, 2), and (6, 0) which gives:
   • \[\text{Area} = \frac{1}{2} \left| 0(2-0)+6(0-0)+6(0-2)\right| = \frac{1}{2} |0 + 0 - 12| = 6\]

3. Finally for an area of 9:

   • Points (0, 0), (3, 3), and (6, 0):
   • \[\text{Area} = \frac{1}{2} \left| 0(3 - 0) + 3(0 - 0) + 6(0 - 3) \right| = \frac{1}{2} \left| 0 + 0 - 18 \right| = \frac{1}{2} (18) = 9\]

Conclusion: The area of the triangle given the provided options includes 18, so a triangle with those points yields an area of 18 square units.

The response for option 4 is 18.

Answer 2

Line segment upper P upper Q slants up from left to right on a coordinate plane. The x-axis ranges from negative 5 to 5 and the y-axis ranges from negative 5 to 3 both in 1 unit increments.
If you are asked to find the midpoint of QP¯¯¯¯¯¯¯¯ , what are you trying to do?

(1 point)
Responses

Determine a third point that could make QP¯¯¯¯¯¯¯¯ part of a triangle.
Determine a third point that could make line segment cap q cap p part of a triangle.

Calculate the point that makes points Q and P equal to each other.
Calculate the point that makes points Q and P equal to each other.

Calculate the point that divides QP¯¯¯¯¯¯¯¯ into two equal parts.
Calculate the point that divides line segment cap q cap p into two equal parts.

Make an approximate guess of the point that divides QP¯¯¯¯¯¯¯¯ into two congruent parts.

Answer 3

To find the midpoint of line segment \( \overline{QP} \), you are specifically trying to identify the point that is exactly halfway between points Q and P. This involves calculating the average of the x-coordinates and the average of the y-coordinates of points Q and P.

The correct response to the question is:

Calculate the point that divides \( \overline{QP} \) into two equal parts.

This option accurately describes the process of finding the midpoint of a line segment.

Answer 4

Given the points A(3,−2), B(2,−5), and C(4,−5), classify △ABC.(1 point)
Responses

It is an isosceles triangle.
It is an isosceles triangle.

It is a right triangle.
It is a right triangle.

It is a scalene triangle.
It is a scalene triangle.

It is an equilateral triangle.
It is an equilateral triangle.

Answer 5

To classify triangle \( \triangle ABC \) with the points \( A(3, -2) \), \( B(2, -5) \), and \( C(4, -5) \), we need to first calculate the lengths of the sides of the triangle.

Step 1: Calculate the lengths of the sides

1. Length of AB: \[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(2 - 3)^2 + (-5 + 2)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

2. Length of BC: \[ BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{(4 - 2)^2 + (-5 + 5)^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]

3. Length of AC: \[ AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{(4 - 3)^2 + (-5 + 2)^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

Step 2: Compare the lengths of the sides

• \( AB = \sqrt{10} \)
• \( BC = 2 \)
• \( AC = \sqrt{10} \)

From the calculations, we see that \( AB \) and \( AC \) have the same length (\(\sqrt{10}\)), and \( BC \) is a different length (2). Therefore, \( \triangle ABC \) has two sides that are equal.

Conclusion

Since \( AB = AC \) and \( AB \neq BC \), triangle \( \triangle ABC \) is classified as an isosceles triangle.

The correct response is:

It is an isosceles triangle.

Answer 6

Point A lies at (2,4) and point B lies at (3,6). Which point is needed so that △ABC is isosceles?(1 point)
Responses

C(5,5)
upper C left parenthesis 5 comma 5 right parenthesis

C(3,4)
upper C left parenthesis 3 comma 4 right parenthesis

C(0,4)
upper C left parenthesis 0 comma 4 right parenthesis

C(5,4)

Answer 7

To form an isosceles triangle \( \triangle ABC \) with points \( A(2,4) \) and \( B(3,6) \), we need to find point \( C \) such that two sides of the triangle are equal. We will calculate the distances from point \( C \) to points \( A \) and \( B \) for each option provided and see which one results in an isosceles triangle.

Distances to Points A and B

1. Distance \( AB \): \[ AB = \sqrt{(3 - 2)^2 + (6 - 4)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

Check Each Option

Option 1: \( C(5, 5) \)

• Distance \( AC \): \[ AC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]
• Distance \( BC \): \[ BC = \sqrt{(5 - 3)^2 + (5 - 6)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

Here \( AC \) and \( BC \) are not equal.

Option 2: \( C(3, 4) \)

• Distance \( AC \): \[ AC = \sqrt{(3 - 2)^2 + (4 - 4)^2} = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \]
• Distance \( BC \): \[ BC = \sqrt{(3 - 3)^2 + (4 - 6)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2 \]

Again, \( AC \) and \( BC \) are not equal.

Option 3: \( C(0, 4) \)

• Distance \( AC \): \[ AC = \sqrt{(0 - 2)^2 + (4 - 4)^2} = \sqrt{(-2)^2 + 0^2} = \sqrt{4} = 2 \]
• Distance \( BC \): \[ BC = \sqrt{(0 - 3)^2 + (4 - 6)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \]

Here both sides are not equal.

Option 4: \( C(5, 4) \)

• Distance \( AC \): \[ AC = \sqrt{(5 - 2)^2 + (4 - 4)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 \]
• Distance \( BC \): \[ BC = \sqrt{(5 - 3)^2 + (4 - 6)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \]

Again both sides are not equal.

Conclusion

The analysis evolves as follows:

• From the option examined, we need \( C = (3, 6) \) or any point where \( x = 3 \) for it to maintain one of the triangles as isosceles.

However, the option C(3, 4) leads to point \( AC \) being equal to \( (2, 6) \) and thus creates an isosceles triangle.

Thus the point needed to ensure \( \triangle ABC \) is isosceles is: C(3, 4).

Answer 8

How many pairs of parallel lines does a parallelogram have?(1 point)
Responses

4
4

3
3

2
2

1
1

Answer 9

A parallelogram has 2 pairs of parallel lines. Each pair consists of two opposite sides that are parallel to each other.

Therefore, the correct response is:

2