1
The vertices of a triangle are A(0, 7)
and B(-5,-1) and c(-1, 2). Then the equation of
a bisector of angleABC is
Answers
Answer:If the bisector of angle B meets side AC at D then
AD/DC = BA/BC.
BA^2=(5--1)^2+(1--7)^2=36+64=100
BA=10
BC=(5-1)^2+(1-4)^2=16+9=25
BC=5
so
AD/DC=10/5=2.
The co-ordinates of D are
(2/3)C+(1/3)A=(2/3)(1,4)+(1/3)(-1,-7)
=(1/3)(1,1)
=(1/3,1/3).
The slope of DB is m=(1-1/3)/(5-1/3)=(2/3)/(14/3)=1/7.
The line BD passes through B so it has equation
y-1=(1/7)(x-5)
7y-7=x-5
7y=x+2.
ALTERNATIVELY:
The slope of AB is m1=(1--7)/(5--1)=8/6=4/3, the slope of BC is m2=(4-1)/(1-5)= -3/4. m1m2=-1 so AB is perpendicular to BC hence angle B is a right angle. If the slope of BC is m2=tanθ= -3/4 then the slope of the bisector of angle B is
tan(θ+45º)=(tanθ+tan45º)/(1-tanθtan45º)
= (-3/4+1)/(1-(-3/4)*1)
= (1/4)/(7/4)
= 1/7.
The bisector passes through B(5,1) so its equation is
y-1=(1/7)(x-5)
7y=x+2.
mark on brain
Answer:
y-1=(1/7)(x-5)
7y=x+2