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!
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
1 answer