How do you write an equation of a line Perpendicular to $3x+4y=12$ , through (7, 1)?

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Answer 1 · 2016-11-14T10:45:29

$4x-3y-25=0$

Product of the slopes of two lines perpendicular to each other is $-1$ . Let the slope of desired line be $m$ .

Writing the equation of given line $3x+4y=12$ in slope intercept form, we get

$4y=-3x+12$ or $y=-3/4x+3$

Hence slope of this line is $-3/4$ .

As such we have $(-3)/4 xxm=-1$

or $m=-1xx4/(-3)=4/3$

Hence, slope of desired line is $4/3$ and as it passes through $(7,1)$

we can have equation of desired line using point slope form of equation i.e.

$(y-1)=4/3(x-7)$

or $3((y-1)=4(x-7)$

or $3y-3=4x-28$

or $4x-3y-25=0$
graph{(3x+4y-12)(4x-3y-25)=0 [-6.21, 13.79, -3.32, 6.68]}

How do you write an equation of a line Perpendicular to 3x+4y=123x+4y=12, through (7, 1)?