BRAINLIEST Find an equation of the line that is perpendicular to 2x-3y=3 that passes through the point (1,2)
Mathematics · College · Thu Feb 04 2021
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We must must transform the standard form equation 2x-3y=3 into a slope-intercept form that is y=mx+b to find its slope.
2x-3y=3
2x-3=3y
3y=2x-3
y=⅔x-3/3
y=⅔x-1
by comparing y=⅔x-1 with slope intercept form, the slope of our first line = ⅔
Perpendicular lines slopes product is -1that means m1*m2=-1
here m1 is slope of first line that is ⅔ and m2 is slope of another line that is :-
m1*m2=-1
⅔*m2=-1
m2=-1*3/2
m2=-3/2
Since the equation of line passing through the point (1,2), therefore the equation of second line is y-y1=m(x-x1)
(x1,y1)=(1,2) put in this equation:-
y-2=-3/2(x-1)
y-2=-3x/2+3/2
y+3x/2=3/2+2
y+3x/2=7/2
multiple this equation by 2 to remove denominator
2y+3x=7( equation of the line that is perpendicular to 2x-3y=3 that passes through the point (1,2) )