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## find the equation of a line

find the equation of a line passing through the point (2,1) and which is perpendicular to the line
3x+2y-5=0

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)
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)