Question

This just isn’t making any sense to me please help before 11pm

8 AB has endpoints A(-5, 0) and B(4, 3). CD has endpoints C(-3, 9) and
D(1, -3). The equations of the lines containing AB and CD are x - 3y = -5
and 3x + y = 0, respectively.
a. How could you quickly check that these equations are correct?
b. Verify that the lines are perpendicular.
c. Find the point of intersection of AB and CD by solving the system
of equations.
d. Find the midpoints of AB and CD. Compare your results with Part c.
e. What kind of quadrilateral is ACBD? Explain your reasoning.

Answer 1

Part (a)

To verify a point is on the line, we plug the coordinates into that equation. If we get a true result at the end, then we've confirmed the point is on that line.

For point A(-5,0) we have x = -5 and y = 0 pair up together. Let's plug these coordinates into x - 3y = -5 to get the following steps shown below.

x - 3y = -5

-5 - 3(0) = -5

-5 - 0 = -5

-5 = -5

We get a true statement at the end, which confirms x-3y = -5 is true when x = -5 and y = 0. Therefore, point A is definitely on the line x - 3y = -5

Repeat similar steps for B(4,3) to show that this point is also on x - 3y = -5

x - 3y = -5

4 - 3(3) = -5

4 - 9 = -5

-5 = -5

This confirms point B is also on this line.

We have confirmed line AB is x-3y = -5

I'll let you check to see if C(-3,9) and D(1,-3) are on the line 3x+y = 0. The steps will follow the same outline as shown above.

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Part (b)

The standard form equation Ax+By = C has the slope m = -A/B, where B cannot be zero.

The equation x-3y = -5 has A = 1 and B = -3 to give us a slope of m = A/B = -1/(-3) = 1/3

Meanwhile the equation 3x+y = 0 gives the slope of A/B = -3/1 = -3

Notice that slopes 1/3 and -3 multiply to -1. This is one useful property of perpendicular lines. Their slopes always multiply to -1; this is assuming neither line is vertical, nor horizontal.

Put another way, the slopes are negative reciprocals of one another. We flip the fraction and flip the sign to go from 1/3 to -3, or vice versa.

I recommend using GeoGebra to plot out the points mentioned, and forming the lines AB and CD. This helps give a quick visual confirmation the lines are perpendicular.

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To summarize: We found the slopes of line AB and CD to be 1/3 and -3 respectively. The slopes multiply to -1 which is sufficient to conclude the lines are perpendicular.

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Part (c)

We have this system of equations

x - 3y = -5

3x + y = 0

To represent lines AB and CD respectively.

Let's triple everything in line CD to go from 3x+y = 0 to 9x+3y = 0

Therefore, an equivalent system is this

x - 3y = -5

9x + 3y = 0

The first equation hasn't changed. After this point, we can see the y terms add to 0 since -3y+3y = 0y = 0. Therefore, the y terms cancel which lets us solve for x.

10x = -5

x = -5/10

x = -0.5

Then we use this to find y

x-3y = -5

-0.5-3y = -5

-3y = -5+0.5

-3y = -4.5

y = -4.5/(-3)

y = 1.5

The point of intersection is (-0.5, 1.5)

Notes:

-0.5 = -1/2

1.5 = 3/2

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Part (d)

The x coordinates of points A and B are -5 and 4 in that order.

Add them up: -5+4 = -1

Divide the result in half: -1/2 = -0.5

This is the x coordinate of the midpoint of segment AB.

Repeat for the y coordinates

Add: 0+3 = 3

Divide in half: 3/2 = 1.5

The midpoint of segment AB is (-0.5, 1.5) which is exactly the result of part (c).

You should find that the midpoint of segment CD is also (-0.5, 1.5)

I'll let you do those steps.

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Part (e)

Quadrilateral ACBD is a kite because of the perpendicular diagonals. This was confirmed in part (b).

The sides aren't all the same length (which you can confirm with the distance formula), which means we don't have a rhombus.

Answer 2

a) See below.

b) See below.

`\textsf{c)} \quad \left(-(1)/(2),(3)/(2)\right)`

`\begin{aligned}\textsf{d)} \quad \textsf{Midpoint of $\overline{AB}$}&=\left(-(1)/(2),(3)/(2)\right)\\ \textsf{Midpoint of $\overline{CD}$}&=(-1,3)\end{aligned}`

e) Kite

Explanation:

Given endpoints:

• A = (-5, 0)
• B = (4, 3)
• C = (-3, 9)
• D = (1, -3)

Given equations of the lines:

• AB: x - 3y = -5
• CD: 3x + y = 0

Part a

To quickly check that the given equations are correct, input the x-value of a point on the line into the given equation of the line containing that point. If the y-value corresponds with the y-value of the point, the equation is correct.

Using point A to check equation AB:

`\begin{aligned}x=-5 \implies -5-3y&=-5\\-3y&=0\\y&=0\end{aligned}`

As point A is (-5, 0), the equation is correct.

Using point C to check equation CD:

`\begin{aligned}x=-3 \implies 3(-3)+y&=0\\-9+y&=0\\y&=9\end{aligned}`

As point C is (-3, 9), the equation is correct.

Part b

If two lines are perpendicular, their slopes are negative reciprocals.

The negative reciprocal of a number is -1 divided by the number.

Rearrange each equation to slope-intercept form, then compare slopes.

`\begin{aligned}\textsf{Line}\;AB: \quad x-3y&=-5\\-3y&=-x-5\\3y&=x+5\\y&=(1)/(3)x+(5)/(3)\end{aligned}`

Therefore, the slope of line AB is ¹/₃.

`\begin{aligned}\textsf{Line}\;CD: \quad 3x + y &= 0\\y&=-3x\end{aligned}`

Therefore, the slope of line CD is -3.

`\textsf{As} \;(-1)/(-3)=(1)/(3), \textsf{ the slopes are negative reciprocals}.`

Hence, the lines are perpendicular.

Part c

To find the point of intersection of AB and CD, solve the system of equations.

`\begin{cases}x - 3y = -5\\3x + y = 0\end{cases}`

Rearrange the second equation to isolate y:

`\implies y=-3x`

Substitute the found expression for y into the first equation and solve for x:

`\implies x-3(-3x)=-5`

`\implies x+9x=-5`

`\implies 10x=-5`

`\implies x=-(5)/(10)`

`\implies x=-(1)/(2)`

Substitute the found value of x into the expression for y and solve for y:

`\implies y=-3\left(-(1)/(2)\right)`

`\implies y=(3)/(2)`

Therefore, the solution to the system of equations is:

`\left(-(1)/(2),(3)/(2)\right)`

Part d

`\boxed{\begin{minipage}{7.4 cm}\underline{Midpoint between two points}\\\\Midpoint $=\left((x_2+x_1)/(2),(y_2+y_1)/(2)\right)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the endpoints.\\\end{minipage}}`

`\begin{aligned}\textsf{Midpoint of $\overline{AB}$}&=\left((x_B+x_A)/(2),(y_B+y_A)/(2)\right)\\&=\left((4+(-5))/(2),(3+0)/(2)\right)\\&=\left(-(1)/(2),(3)/(2)\right)\end{aligned}`

`\begin{aligned}\textsf{Midpoint of $\overline{CD}$}&=\left((x_D+x_C)/(2),(y_D+y_C)/(2)\right)\\&=\left((1+(-3))/(2),(-3+9)/(2)\right)\\&=\left(-1,3\right)\end{aligned}`

The midpoint of line segment AB is the solution to the system of equations.

Part e

Quadrilateral ABCD is a kite:

• The diagonals of a kite are perpendicular to each other, as proven in part b.
• The longer diagonal of a kite bisects the shorter diagonal. As the midpoint of line segment AB is the same as the solution to the system of equations, this proves that the longer diagonal (CD) bisects the shorter diagonal (AB).
• The two diagonals of a kite are not of the same length. As the longer diagonal bisects the shorter diagonal, and the midpoints of line segments AB and CD are not the same, this proves that the diagonals are not the same length.