I'll do a)
you show me what you have done for the other two.
so you want the line from N to the midpoint of LP
midpoint of LP is Q( (1-3)/2, (-8+6)/2) = Q(-1,-1)
slope of NQ = (2-(-1))/(3-(-1) ) = 3/4
equation for NQ:
y-2 = (3/4)(x-3)
4y - 8 = 3x - 9
3x - 4y = 1
or
y = (3/4)x - 1/4
A triangle is formed from the points L(-3,6), N(3,2) and P(1,-8). Find the equation of the following lines:
a) the median from N
b) the right bisector of LP
c) the altitude from N
3 answers
(a) the midpoint M of LP is (-1,-1)
The slope of PM is 4/3
y-2 = 4/3 (x-3)
(b) the slope of the perpendicular bisector of LP (through M) is -3/4, so
y+1 = -3/4 (x+1)
(c) recall that the distance from (h,k) to the line Ax+By+C = 0 is
|Ah+Bk+C|/√(A^2+B^2)
So figure the equation of the line LP and plug it in.
The slope of PM is 4/3
y-2 = 4/3 (x-3)
(b) the slope of the perpendicular bisector of LP (through M) is -3/4, so
y+1 = -3/4 (x+1)
(c) recall that the distance from (h,k) to the line Ax+By+C = 0 is
|Ah+Bk+C|/√(A^2+B^2)
So figure the equation of the line LP and plug it in.
oops. go with Reiny on (a) and thus modify (b)