Expert: Josh - 11/27/2007

Question
I have read a number of ways to try and work out this question but i just cant do it! Need help quick. Thank you!


Find the equation of the line that passes through the point (1,2) and is perpendicular to the line 3x + 2y =5.

Answer
Katie,

First thing is to work out the gradient (slope) of the line from the equation 3x+2y=5. We need this information in order to work out the slope of the perpendicular line.

We rearrange the equation into the form y=mx+b, where the symbol "m" carries the meaning of the slope of a straight line; "b" represents the point where the line cuts the y-axis.

3x+2y=5 ...subtract 3x from both sides
2y=-3x+5 ...then divide throughout by 2
y=(-3/2)x+5/2

Comparing this with y=mx+b, the slope "m"=-3/2 and b=5/2.

Remember this result:
The slope of the given line "m" multiplied by the slope of the perpendicular line "n" equals -1.

Applying this result:
m*n=-1
-(3/2)*n=-1
n=2/3
This is the slope of the perpendicular line.

So, the perpendicular line must be y=nx+c, where c represents the y-intercept (as before) and its value is to be worked out. We already know n=2/3.

We are looking at y=(2/3)x+c. If the point (1,2) lies on the perpendicular, x=1, y=2 must satisfy the equation. Putting these values into the equation,

y=(2/3)x+c
2=(2/3)*1+c
c=2-2/3
c=4/3.

Final equation for perpendicular is y=(2/3)x+(4/3). If you prefer, you can multiply both sides by 3 to get 3y=2x+4.