Determine the equation for each line.
A line parallel to the line 3x-2y-12=0 and has the same y intercept of the line of 4x-3y+8=0
To find the equation of a line parallel to the given line, we need to find the slope of the given line. The equation of a line can be written in slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept.
So, let's find the slope of the given line 3x - 2y - 12 = 0.
First, rearrange the equation to isolate y:
-2y = -3x + 12
y = (3/2)x - 6
The slope of this line is (3/2).
We want to find a line parallel to this line, so it will have the same slope.
Now, let's find the y-intercept of the line 4x - 3y + 8 = 0.
Rearranging the equation to isolate y:
-3y = -4x - 8
y = (4/3)x + (8/3)
The y-intercept of this line is (8/3).
The equation of the line parallel to the given line and with the same y-intercept is:
y = (3/2)x + (8/3)
b) A line perpendicular to the line y= 2x+3 and passing through (6,5).
To find the equation of a line perpendicular to the given line, we need to find the negative reciprocal of the slope of the given line. The equation of a line can be written in slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept.
The given line has a slope of 2. The negative reciprocal of 2 is -1/2.
We also know that the line passes through the point (6,5).
Using the point-slope form of a line, we can write the equation as:
y - y1 = m(x - x1)
where (x1, y1) is the given point.
Substituting the values into the equation:
y - 5 = (-1/2)(x - 6)
Expand and rearrange:
y - 5 = (-1/2)x + 3
y = (-1/2)x + 8
So, the equation of the line perpendicular to y = 2x + 3 and passing through (6,5) is y = (-1/2)x + 8.