Slope of original line: 2/5
Slope of perpendicular line: -5/2 [negative reciprocal]
Slope of new line should be: y=-5/2x + 5/2
Find the equation of the line perpendicular to line 2x−5y−10=0 and containing point (−1, 5).
6 answers
Re-arrange the equation in y=mx + b
then look at the slope (m)
then use the perpendicular slope (-5/2), and the point (-1, 5) and solve for b : )
then look at the slope (m)
then use the perpendicular slope (-5/2), and the point (-1, 5) and solve for b : )
Lyla is incorrect... she did not sub in her point and solve for b : (
BUT here perpendicular slope is correct : ) a good starting point : )
BUT here perpendicular slope is correct : ) a good starting point : )
Easy way to do these:
Since the lines are perpendicular, their slopes must be negative reciprocals of each other.
So just interchange the x and y coefficients, and change the sign on one of them,
then sub in the given point to find the constant.
2x−5y−10=0 ---> 5x + 2y = C
using (−1, 5) ---> 5(-1) + 2(5) = C
C = 5
new equation: 5x + 2y = 5
Since the lines are perpendicular, their slopes must be negative reciprocals of each other.
So just interchange the x and y coefficients, and change the sign on one of them,
then sub in the given point to find the constant.
2x−5y−10=0 ---> 5x + 2y = C
using (−1, 5) ---> 5(-1) + 2(5) = C
C = 5
new equation: 5x + 2y = 5
Oh my! Which gives the exact same answer Layla replied with!
I had a negative sign incorrect!
Great job team!
I had a negative sign incorrect!
Great job team!
I'm sorry Ms. Pi, but you didn't do anything. You aren't part of the team.
1 answer