Find the equation of a line which is

perpendicular to 2x + 3y = 4 passing through ( -5, 1)

Can you show mw how to work this out step by step please?

1 answer

As an example:
3x + 2y = 5 passing through ( -3, 2)

The coordinates of the point is x1=-3, and y1=2.
The required equation is
2(x-x1)-3(y-y1)=0, i.e. ignore the constant term.
2(x-(-3))-3(y-2)=0
Notice how the coefficients of x and y have been switched around and the sign of one of the coefficients changed. This is done to get a line perpendicular to the original one.
Now simplify:
2(x-(-3))-3(y-2)=0
2x+6-3y+6=0
2x-3y=-12
Check by substituting x1 for x, and y1 for y:
2(-3)-3(2)=-12 OK

When the equations are in the normalized form : Ax+By+C=0
check for perpendicularity between 2x+3y=5 and 3x-2y=-12
the sum of the coefficients of x and the coefficients of y should be zero:
2*3+3*(-2)=0 the lines are perpendicular.

When the equations are of the form
y=mx+b and y=m1x+b1
then m*m1 should equal -1.
The problem with this form is that it does not work with vertical lines.

There should be enough information for you to work out your homework problem!