Answer:
1. To find an equation of a line that passes through the point (3,17) and is parallel to -4y=8x+4, we can start by rearranging the given equation into slope-intercept form (y=mx+b), where m is the slope and b is the y-intercept. Dividing both sides of the equation by -4, we get y=-2x-1. Therefore, the slope of this line is -2.
Since the line we want is parallel to this line, it will have the same slope of -2. Using the point-slope form of the equation of a line, we can write:
y - 17 = -2(x - 3)
Simplifying and rearranging, we get the final equation of the line:
y = -2x + 23
2. To find an equation of a line that passes through the point (4,6) and is perpendicular to 3y=2x-9, we can start by rearranging the given equation into slope-intercept form (y=mx+b), where m is the slope and b is the y-intercept. Dividing both sides of the equation by 3, we get y=(2/3)x-3. Therefore, the slope of this line is 2/3.
Since the line we want is perpendicular to this line, its slope will be the negative reciprocal of 2/3, which is -3/2. Using the point-slope form of the equation of a line, we can write:
y - 6 = (-3/2)(x - 4)
Simplifying and rearranging, we get the final equation of the line:
y = -3/2x + 9