a) To find a line parallel to the line 3x - 2y - 12 = 0, we need to keep the same slope. The slope of the given line can be found by rearranging the equation in slope-intercept form (y = mx + b), where m is the slope:
3x - 2y - 12 = 0
-2y = -3x + 12
y = (3/2)x - 6
Since the slope of the given line is 3/2, the slope of the parallel line will also be 3/2. To find the equation of the parallel line, we can use the slope-intercept form and the given y-intercept of the line 4x - 3y + 8 = 0. Rearranging the equation in slope-intercept form:
4x - 3y + 8 = 0
-3y = -4x - 8
y = (4/3)x + 8/3
The y-intercept of this line is 8/3. Therefore, the equation of the line parallel to 3x - 2y - 12 = 0 and with the same y-intercept as 4x - 3y + 8 = 0 is:
y = (3/2)x + 8/3
b) To find a line perpendicular to the line y = 2x + 3, we need to find the negative reciprocal of the slope of the given line. The slope of the given line is 2, so the slope of the perpendicular line will be -1/2.
Using the point-slope form of a line (y - y₁ = m(x - x₁)), where (x₁, y₁) is a given point on the line, we can substitute the point (6, 5) and the slope -1/2 to find the equation of the perpendicular line:
y - 5 = (-1/2)(x - 6)
y - 5 = (-1/2)x + 3
y = (-1/2)x + 8
Therefore, the equation of the line perpendicular to y = 2x + 3 and passing through (6, 5) is:
y = (-1/2)x + 8
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
b) A line perpendicular to the line y= 2x+3 and passing through (6,5).
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