Find the equation of the line which passes through the point (6, -4) and is parallel to to the line 2x-3y+3=0.
Please show working out as well, thanks.
3y = 2 x + 3
is
y = (2/3)x + 1
slope = 2/3
so
y = (2/3)x + b if parallel
-4 = (2/3)6 + b
b = -8
so
y = (2/3)x-8
3 y = 2 x - 24
The slope of 2x-3y+3=0 is 2/3.
The parallel line you want has a slope of 2/3 and passes through (6,-4)
So, using the point-slope form of the line,
y+4 = 2/3 (x-6)
To find the equation of a line parallel to another line, we need to determine the slope of the given line. Once we have the slope, we can use the point-slope form of a line to find the equation of the parallel line.
The given line has the equation 2x - 3y + 3 = 0. To put it in slope-intercept form (y = mx + b), we need to isolate y. Subtracting 2x from both sides, we get -3y = -2x - 3. Dividing by -3, we have y = (2/3)x + 1.
The slope of the given line is the coefficient of x in this equation, which is 2/3. Since the parallel line has the same slope, the equation of the parallel line will have the form y = (2/3)x + b, where b is the y-intercept.
To determine the value of b, we can use the given point (6, -4) that the parallel line passes through. Plugging in these values into the equation, we get -4 = (2/3)(6) + b. Simplifying, -4 = 4 + b. Subtracting 4 from both sides, we find that b = -8.
Now we can write the equation of the parallel line using the slope-intercept form: y = (2/3)x - 8.