Determine the slope and y intercept for d1: -2x+3y+1=0 d2: 2x-3y+1=0
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d1: -2x + 3y + 1 = 0
d2: 2x - 3y +1 = 0
For d1:
d1: -2x + 3y +1 = 0
==> y = (2/3)x + 1/2
Then the slope is m = 2/3
The y-intercept is where the function intersect with y-axis, then x should be 0:
==> y= (2/3)*0 + 1/3 = 1/3
Then y-intercept = 1/3
For d2:
d2= 2x+3y +1 = 0
==> y= (-2/3)x -1/3
Then the slope m = -2/3
The y-intercept = -1/3
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We can convert each equation to slope and intercept form:
y =mx+c , where m is slope and c is the y intercept of the line.
d1: -2x+3y+1 = 0 . add 2x-1 and rewrite as:
3y = 2x-1. Divide by 3.
y = (2/3) -1/3. So the slope of d1 is 2/3 and y intercept is -1/3.
d2:
2x-3y +1 = 0. Subtract 2x+1 .
-3y = -(2x+1) = -2x-1. Divide by -3.
y = (-2/3)x +(-1/3). So slope = -2/3 and y intercept -1/3.
We can express the equation ax+by+c=0 in the form y=(-a/b)x-c/b where -a/b is the slope and -c/b is the y intercept.
For the line d1 with the equation -2x+3y+1=0 the slope works out to 2/3 and the y-intercept is -1/3
For the line d2 with the equation 2x-3y+1=0 the slope works out to 2/3 and the y-intercept is 1/3
We'll put the equation of d1 in the standard form:
y=mx + n, where the coefficients represent:
- m - the slope
- n - y intercept
-2x+3y+1=0
We'll isolate 3y to the left side:
3y=2x-1
We'll divide by 3 both sides:
y=(2x/3) - 1/3
The standard form of the equation is:
y=(2x/3) - 1/3
The slope of the line d1 is m = 2/3 and the y intercept is n = -1/3.
We'll put the equation of d2 in the standard form:
y=mx + n, where the coefficients represent:
- m - the slope
- n - y intercept
2x-3y+1=0
We'll isolate -3y to the left side:
-3y=-2x-1
We'll divide by -3 both sides:
y=(2x/3) + 1/3
The standard form of the equation is:
y=(2x/3) + 1/3
The slope of the line d1 is m = 2/3 and the y intercept is n = 1/3.
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