Expert: Josh - 10/19/2009

Question
Write the slope-intercept form for an equation of the line perpendicular to the graph of the given equation and passing through the given point   8x-3y=7, (4,5)

Answer
Hi Evan,

The slope-intercept form for a straight line equation is y=s*x+b.

Parameters:
s represents the slope of the line (positive if it rise from left to right like /, negative if it falls from left to right like \)
b is called the y-intercept because this is the point where the line crosses the y-axis (the value of y when x=0).

Remember this: If "m" is the slope (aka gradient) of a straight line, then any line perpendicular to it must have a slope "n", such that m*n = -1. In other words, the perpendicular line has a slope of n=-1/m.

Step 1: We need to know the gradient of the given line 8x-3y=7.
Rearranging the equation as 3y=8x-7, then dividing both 3 on both sides, we obtain an equivalent equation in the slope-intercept form y=(8/3)x-(7/3). The interpretation is now quite simple, comparing with y=m*x+b, m=(8/3). This gives the slope.

Step 2: Using the fact that n=-1/m, n=-3/8. Thus, the perpendicular line is an equation of the form y=n*x+b.

Step 3: If (x=4,y=5) is a point on the perpendicular line, it must satisfy the equation y=n*x+b, where n=-3/8. So, we work out the value of "b". b = y-n*x = 5-(-3/8)*4 = 6.5

Final answer: y=-(3/8)x+6.5 is the equation of the perpendicular line in slope-intercept form.