How would you solve...
In the xy-plane, the graph of which of the following linear equations is perpendicular to the graph of the linear equation 3x-4y=0?
How would you solve...
In the xy-plane, the graph of which of the following linear equations is perpendicular to the graph of the linear equation 3x-4y=0?
What lines?
Any line perpendicular to the given line looks like:
$\displaystyle 4x+3y=K$.
A) y= -4/3x + 12
B) y= -3/4x + 12
C) y= 3/4x + 12
D) y= 4/3x + 12
I don't know how to find a perpendicular line and when I tried it on a graphing calculator I got it wrong. They say the answer is A but I didn't know how they got that.
A) y= -4/3x + 12
B) y= -3/4x + 12
C) y= 3/4x + 12
D) y= 4/3x + 12
I don't know how to find a perpendicular line and when I tried it on a graphing calculator I got it wrong. They say the answer is A but I didn't know how they got that.
Put the answer I gave into the slope-intercept form. You will see it.