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True Value Rentals charges $20 to rent a moving
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Hello once again!
When you see a question like this, you need to find the equation of the straight line.
The formular used is y = mx + c
Where
m = slope
c = constant
First find the slope, since it's a straight line, any 2 coordinates can be used.
Now we need to substitude in the slope, and one of the coordinate you used to find the slope, to the formular to find the constant.
In this case i'm using the coordinate
(-2, 16)
y = mx + c
16 = -6(-2) + c
16 = 12 + c
c = 4
∴ The equation of the line is y = -6x + 4
The next step is to simply substitude in the x = 8 to the equation to find y.
y = -6(8) + 4
y = -48 + 4
y = -44
When you see a question like this, you need to find the equation of the straight line.
The formular used is y = mx + c
Where
m = slope
c = constant
First find the slope, since it's a straight line, any 2 coordinates can be used.
Now we need to substitude in the slope, and one of the coordinate you used to find the slope, to the formular to find the constant.
In this case i'm using the coordinate
(-2, 16)
y = mx + c
16 = -6(-2) + c
16 = 12 + c
c = 4
∴ The equation of the line is y = -6x + 4
The next step is to simply substitude in the x = 8 to the equation to find y.
y = -6(8) + 4
y = -48 + 4
y = -44
Since you're only asked the ordered pair of D'', it's much easier just to plot and reflect point D twice than to do that for all four points!
Remember that reflecting points is like putting a mirror at the line of reflection or flipping that point over at that line. The reflected point should be the same distance from the line of reflection as the original point.
1) Reflect D over the x-axis to get D'.
D is at (4,1). Draw a line that is perpendicular to the line of reflection and goes through D. D is as far from the line of reflection as D' should be on its other side (both are on that perpendicular line). Since D is 1 unit above the x-axis, that means D' is 1 unit below at (4, -1). See picture 1.
2) Reflect D' over <span>y=x+1 to get D''.
D' is at (4, -1). Draw </span>y=x+1 and the line perpendicular to it going through D''. D'' is the same distance from the line of reflection on the other side. See picture 2. D'' is at (-2, 5).
Answer: D'' is at (-2, 5)
Remember that reflecting points is like putting a mirror at the line of reflection or flipping that point over at that line. The reflected point should be the same distance from the line of reflection as the original point.
1) Reflect D over the x-axis to get D'.
D is at (4,1). Draw a line that is perpendicular to the line of reflection and goes through D. D is as far from the line of reflection as D' should be on its other side (both are on that perpendicular line). Since D is 1 unit above the x-axis, that means D' is 1 unit below at (4, -1). See picture 1.
2) Reflect D' over <span>y=x+1 to get D''.
D' is at (4, -1). Draw </span>y=x+1 and the line perpendicular to it going through D''. D'' is the same distance from the line of reflection on the other side. See picture 2. D'' is at (-2, 5).
Answer: D'' is at (-2, 5)