Find the equation of the line that passes through (1,3) and is perpendicular to the line 2x+3y+5=0
A. 3x-2y+3=0
B. 2x+3y-11=0
C. 2x+3y-9=0
D. 3x-2y-7=0
E. None of these
I got E?
the slope is 3/2, so
y-3 = 3/2 (x-1)
2y-6 = 3x-3
3x-2y+3 = 0
Looks like (A) to me
To find the equation of a line that is perpendicular to the given line, we need to determine the slope of the given line and then use its negative reciprocal as the slope of the perpendicular line.
The given line has the equation 2x + 3y + 5 = 0. To find its slope, we rearrange the equation in the slope-intercept form y = mx + b, where m is the slope:
2x + 3y + 5 = 0
3y = -2x - 5
y = -2/3x - 5/3
The slope of the given line is -2/3.
The negative reciprocal of -2/3 is 3/2, which will be the slope of the perpendicular line.
Now, we can use the point-slope form of a line to find the equation of the line passing through the point (1,3) with slope 3/2:
y - y1 = m(x - x1)
y - 3 = 3/2(x - 1)
y - 3 = 3/2x - 3/2
y = 3/2x - 3/2 + 6/2
y = 3/2x + 3/2
This equation can be rearranged to have integer coefficients:
2y = 3x + 3
3x - 2y + 3 = 0
So, the equation of the line that passes through (1,3) and is perpendicular to the line 2x + 3y + 5 = 0 is 3x - 2y + 3 = 0.
Therefore, the correct answer is option A.