If you
can find the midpoint of a line segment
know that the slope of the perpendicular is -1/(slope of line segment)
remember the point-slope form of a line
then all of these problems are straighforward. Review those topics, give it a try, show where you get stuck, if you do.
1. Determine the equation of the right bisector of the line segment E(2, 6) and F(4, -2). Draw the diagram.
2. Given three points D(2, 5) E(-2, -3) and F(4, -6) determine whether or not the line through D and E is perpendicular to the line through E and F.
3. A triangle is formed from the points L(-3, 6), N(3, 2) and P(1, -8). Find the equation of the following lines:
the median from N
the right bisector of LP
the altitude from N
2 answers
1. E(2,6), M(x,y), F(4.-2).
m1 = (-2-6)/(4-2) = -8/2 = -4
m2 is the negative reciprocal of m1:
m2 = 1/4
x = (2+4)/2 = 3
y = (6+(-2))/2 = 2
Y = mx + b = 2
(1/4)*x + b = 2
3/4 + b = 2
b = 2-3/4 = 5/4
Eq: Y = x/4 + 5/4
2. D(2,6), E(-2,-3), F(4,-6).
m1 = (-3-6)/(-2-2) = -9/-4 = 9/4.
m2 = (-6-(-3))/(4-(-2)) = -3/6 = -1/2.
m2 is NOT the negative reciprocal of m1,
therefore, the lines are NOT perpendicular.
m1 = (-2-6)/(4-2) = -8/2 = -4
m2 is the negative reciprocal of m1:
m2 = 1/4
x = (2+4)/2 = 3
y = (6+(-2))/2 = 2
Y = mx + b = 2
(1/4)*x + b = 2
3/4 + b = 2
b = 2-3/4 = 5/4
Eq: Y = x/4 + 5/4
2. D(2,6), E(-2,-3), F(4,-6).
m1 = (-3-6)/(-2-2) = -9/-4 = 9/4.
m2 = (-6-(-3))/(4-(-2)) = -3/6 = -1/2.
m2 is NOT the negative reciprocal of m1,
therefore, the lines are NOT perpendicular.