Find the equation, in standard form, of the line perpendicular to 2x+3y=−5 and passing through (3, -5)

honigtropfenvi 2022-09-12 Answered
Find the equation, in standard form, of the line perpendicular to 2x+3y=−5 and passing through (3, -5)
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Marie Horn
Answered 2022-09-13 Author has 12 answers
The first information that we have is that we're looking for a line passing through (3,−5). The family of lines passing through a given point ( x 0 , y 0 ) is y - y 0 = m ( x - x 0 ) , where m is the slope of the line.
So, the lines passing through (3,−5) are of the form
y + 5 = m p ( x - 3 ) , where i called the slope m p for perpendicular.
To find m p , let's find the slope m of the original line before: bringing into standard form the equation 2x+3y=−5, we get y = - 2 x - 5 3 = - 2 3 x - 5 3 . So, the original line has slope - 2 3 .
Now, two lines are perpendicular if their slopes m and m' are in the relation m = - 1 m .
From this relation, we have m p = 3 2 , and the solution is thus
y + 5 = 3 2 ( x - 3 )
Here's a link where you can check that the two lines are perpendicular, and that the second one passes through (3,5).

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