David W. answered • 03/02/16

Experienced Prof

The line: y-3 = 3(x+1) or y = 3x+4

You may (1) first find the equation of the perpendicular line that passes thru (5,-1), then convert it to Standard Form or (2) convert the equation to Standard Form, then find the equation of the line that passes thru (5,-1). See if you can follow both methods:

Important:

**If a line has slope m, then all perpendicular lines have a slope of (-1/m). That's called the negative reciprocal.**(1) find the equation in slope-intercept form, then convert to Standard Form

The line: y-3 = 3(x+1) or

**y = 3x+4** Perpendicular line: y = (-1/3)x + b [and we need point (5,-1) to find b

-1 = (-1/3)(5) + b

-1 = -5/3 + b

2/3 = b

The perpendicular line is: y = (-1/3)x + 2/3

The Standard Form of the equation of a line looks like: Ax+By=C

The perpendicular line is:

(1/3)x + y = 2/3

**x + 3y = 2**[by convention, coefficient of x is positive; x is first;

get rid of fractions]

(2) Convert to Standard Form, then find the specific equation

The line: y-3 = 3(x+1) or y = 3x+4

In Standard Form: 3x - y = 4 [Note: slope is (-A/B) = (-3)/(-1) = 3

Slope of all parallel lines: (-1/3) [that is, -A/B]

Equation of parallel line: x + 3y = D [and we need point (5,-1) to determine D]

5 + 3(-1) = D

2 = D

The parallel line, in Standard Form, is :

**x + 3y = 2**
Nicole V.

03/02/16