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 $(x_3,y_3)$:

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

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):
   • $$\text{Area}=\frac{1}{2}|2(6-6)+0(6-0)+6(0-6)|=\frac{1}{2}|0+0-36|=\frac{1}{2}(36)=18$$

2. Another example for an area of 15:

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

3. Finally for an area of 9:

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

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 $\mathrm{△}ABC$ 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, $\mathrm{△}ABC$ has two sides that are equal.

Conclusion

Since $AB=AC$ and $AB\ne BC$, triangle $\mathrm{△}ABC$ 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 $\mathrm{△}ABC$ 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 $\mathrm{△}ABC$ 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