How do you write an equation of a line passing through (3, -5), perpendicular to # 2x + 3y = -5#?

Question

1358 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-07T09:01:23

#=> y=3/2x-19/2#

Let the given point be #P_2->(x_2,y_2)=(3,-5)#

Given equation;#" "2x+3y=-5#

Write this in standard form of: #y=mx+c#
where m is the gradient (slope).

#color(brown)(2x+3y=-5)color(blue)(" "->" "y=-2/3x-5/3)#

So for this line the gradient (m) is #-2/3#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The gradient of the line perpendicular to this is #-1/m#

So #-1/m = +3/2#

Thus the perpendicular line equation is:#" "y=3/2x+c#

We know that this line passes through the point #P_2#

So #P_2->y_2=3/2x_2+c#

#=>-5=3/2(3)+c#

#=> c=-19/2#

#=> y=3/2x-19/2 larr" perpendicular line"#