find the equation of a line passing through the point (2,1) and which is perpendicular to the line

3x+2y-5=0

find the equation of a line passing through the point (2,1) and which is perpendicular to the line

3x+2y-5=0

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Saugus, MA

3x+2y-5=0

2y=-3x+5

y=(-3/2)x+(5/2)

slope of this line is -3/2

a line perpendicular to this line will have a slope that is the negative reciprocal of (-3/2)

the slope of the perpendicular line is 2/3

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

y-1=(2/3)x-(4/3)

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

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

or you can use y=mx+b

y=(2/3)x+b

1=(2/3)(2)+b

1=(4/3)+b

1-(4/3)=b

-1/3=b

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

Middletown, CT

Hi Dalia;

3x+2y-5=0

We need the equation in the format of y=mx+b for which m=slope, and b is the y-intercept, the value of y when x=0.

3x+2y-5=0

Let's add 5 to both sides...

5+3x+2y-5=0+5

3x+2y=5

Let's subtract 3x from both sides...

-3x+3x+2y=-3x+5

2y=-3x+5

Let's divide both sides by 2...

(2y)/2=[(-3x)/2]+(5/2)

y=(-3/2)x+(5/2)

The slope of this line is -3/2.

The slope of the line perpendicular to such is positive 2/3.

y=2/3x+b

The y-intercept, b, is unknown at this time.

Let's plug-in the one coordinate provided to establish such; (2,1).

1=[(2/3)(2)]+b

1=(4/3)+b

Let's subtract 4/3 from both sides...

-4/3+1=(4/3)+b-(4/3)

-1/3=b

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

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