a.) x - 2y - 15 = 0
b.) 2x - y - 15 = 0
c.) -x + 2y + 15 = 0
d.) - 2x + y - 15 = 0
b.) 2x - y - 15 = 0
c.) -x + 2y + 15 = 0
d.) - 2x + y - 15 = 0
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By point-slope form, we see that the equation of a line with a slope of 2 and passes through (5, -5) is:
y - (-5) = 2(x - 5) ==> y = 2x - 15.
Putting this in standard form by bringing the y over to the other side gives:
2x - y - 15 = 0.
Therefore, the answer is B).
I hope this helps!
y - (-5) = 2(x - 5) ==> y = 2x - 15.
Putting this in standard form by bringing the y over to the other side gives:
2x - y - 15 = 0.
Therefore, the answer is B).
I hope this helps!
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The "slope-intercept" format would look like this:
y = mx + c
where m is the slope and c is the "intercept" (the value of y, when x=0)
In your problem, you are given this information:
The slope is 2 (m = 2)
The point (x, y) = (5, -5) is on the line.
This means that the equation will look like this:
y = 2x + c
-2x + y - c = 0
AND, if you use the values x=5 and y = -5, the equation should be valid.
For example, D looks promising, so let's try it:
-2x + y - 15 = 0
Does it have the proper slope (the coefficient of x is -2 when the coefficient of y is +1, so yes, it has the proper slope).
Does it balance when we use x = 5 and y = -5 ? (this is the same as asking: is the point (5, -5) on the line?)
-2(5) + (-5) - 15 = -10 - 5 - 15 = -30
is NOT equal to zero, therefore it does not balance (meaning: the point is not on that line).
y = mx + c
where m is the slope and c is the "intercept" (the value of y, when x=0)
In your problem, you are given this information:
The slope is 2 (m = 2)
The point (x, y) = (5, -5) is on the line.
This means that the equation will look like this:
y = 2x + c
-2x + y - c = 0
AND, if you use the values x=5 and y = -5, the equation should be valid.
For example, D looks promising, so let's try it:
-2x + y - 15 = 0
Does it have the proper slope (the coefficient of x is -2 when the coefficient of y is +1, so yes, it has the proper slope).
Does it balance when we use x = 5 and y = -5 ? (this is the same as asking: is the point (5, -5) on the line?)
-2(5) + (-5) - 15 = -10 - 5 - 15 = -30
is NOT equal to zero, therefore it does not balance (meaning: the point is not on that line).
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The equation of a line is: y=mx + b. b = the place where the line intersects with the y-axis. m = the slope of the line. So you already know what m is. m=2.
You also have a point on the line. so you know what x and y are. So let's plug all this in an solve for b.
(5, -5) means x = 5, y = -5
-5 = 2(5) + b. So: -5 = 10 + b. This becomes: b = -5 - 10. That means b=-15
So the equation is; y = 2x + (-15) or y = 2x -15. Let's move the terms around: -2x + y + 15 = 0. But for standard form, you want to get rid of that negative sign in front of the x term, so multiply through by negative -1. This gives 2x - y - 15 = 0.
The answer is B.
You also have a point on the line. so you know what x and y are. So let's plug all this in an solve for b.
(5, -5) means x = 5, y = -5
-5 = 2(5) + b. So: -5 = 10 + b. This becomes: b = -5 - 10. That means b=-15
So the equation is; y = 2x + (-15) or y = 2x -15. Let's move the terms around: -2x + y + 15 = 0. But for standard form, you want to get rid of that negative sign in front of the x term, so multiply through by negative -1. This gives 2x - y - 15 = 0.
The answer is B.
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b