How do you write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (3,7) and perpendicular to 4x+5y=1?

1 Answer
Apr 21, 2018

Slope-intercept : y = 5/4 x + 13/4
Standard : -5x + 4y = 13

Explanation:

To write a linear equation in any form, you need at least two of the following three pieces of information: a point, a different point, and/or a slope. We are already given a single point, (3, 7).

Note that if we have a line y = mx + b in slope-intercept form, its slope is given by m. The line perpendicular to this line will have a slope of -1/m. So let's put our provided equation in slope-intercept form.

4x + 5y = 1
5y = 1 - 4x
y = 1/5 - 4/5 x = -4/5 x + 1/5

The slope of the given line is then -4/5. The slope of the line perpendicular to this must be 5/4. Now we have both a slope and a point.

To find our equation in slope-intercept form, we take the base form y = mx + b and plug in our slope of m = 5/4 and our point of (x, y) = (3, 7) to find b.

y = mx + b
y = 5/4 x + b
7 = 5/4 (3) + b
7 = 15/4 + b
7 - 15/4 = b
28/4 - 15/4 = b
13/4 = b.

Having found b, we now know our equation in slope-intercept form. That is, y = 5/4 x + 13/4.

The standard form of a linear equation is ax + by = c, where a, b, c are constants. It is easy to move from slope-intercept form into standard form.

y = 5/4 x + 13/4
4y = 5x + 13
-5x + 4y = 13

Thus, our answer in standard form is -5x + 4y = 13.