perpendicular lines

wrestling125

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Mar 20, 2005
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i am having a hard time figuring out this question

given two lines whose equations are 3x+y-8=0 and -2xby+9=0 determine the value of b such that the two lines will be perpendicular.
 
Hello, wrestling125!

You're expect to be familiar with the slope-intercept form: . y = mx + b
. . . where <u>m</u> (the coefficient of x) is the slope, <u>b</u> is the y-interecept.

You should also know when two slopes are perpendicular:
. . . they are <u>negative</u> <u>reciprocals</u> of each other.
That is, they have opposite signs and one is the "flip over" of the other.

Given two lines whose equations are 3x + y - 8 = 0 and -2x + by + 9 = 0,
determine the value of b such that the two lines will be perpendicular.
Solve the equations for y: . 3x + y - 8 = 0 . ---> , y - -3x + 8
. . . . . . . . . . . . . . . . . . . . -2x + by + 9 = 0 . ---> . y = (2/b)x - (9/b)

The slope of the first line is: . m<sub>1</sub> .= .-3
. . . The slope perpendicular to it would be: . m .= .+(1/3)

The slope of the second line is: . m<sub>2</sub> .= .2/b
. . . and we want this to equal +1/3

So we have: . 2/b .= .1/3 . ---> . b = 6
 
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