|
Question 55458: Find the equation of the line that is parallel to 2x-y=4 and contains the point (2,-3).
-y = -2x + 4
-y = 1/2(x + 4)
-y = 1 + 2
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Since you don't specify in which form you want your equation, let's use the slope-intercept form. y = mx+b
First, remember that parallel lines have the same slope, so let's find the slope, m, of the given line by putting it into the slope-intercept form:
2x-y = 4 Add y to both sides of the equation.
2x = y+4 Now subtrct 4 from both sides.
2x-4 = y or y = 2x-4 Comparing this with the general form of the slope-intercept equation: y = mx+b, you can see that the slope, m, = 2. So now for your new equation, you can write:
y = 2x+b But you still need to find the value of b, the y-intercept of the new equation. You can use the coordinates of the given point (2, -3) to do this. Substitute the x- and y-coordinates of the given point into the equation: y = 2x+b, then solve for b.
-3 = 2(2)+b
-3 = 4+b Subtract 4 from both sides.
-7 = b or b = -7 So now you can write the equation of the new line.
y = 2x-7 This is the equation of a line that is parallel to the line 2x-y = 4 and which passes through the point (2, -3).
|
|
|
| |
