Question

A(7,-1), B(4,1), and C(-3,4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point p in AC; such that AP:CP =2:3

Answer 1

Given AP:CP=2:3
Coordinates of P are [{2*(-3)+3*7}/(2+3),{2*4+3*(-1)}/(2+3)]
=[(-6+21)/5,(8-3)/5]
=(15/5,5/5)
=(3,1)
Slope of BP=[(1-1)/(3-4)]
=0
Required equation of the line passing through points B and P is
y-y' = m(x-x')
y-1 = 0(x-3)
y-1=0
y=1

Answer 2

Explanation:

m1 =2, m2=3

AP:CP

2:3

m1x2+m2x1/m1+m2, m1y2+m2y1/m1+m2

2×(-3)+3×7/2+3,2×4+3×(-1)/2+3

-6+21/5, 8+(-3)/5

-6+21/5,8-3/5

15/5, 5/5

3,1

slope of BP= y2-y1/x2-x1

m = 1-1/4-3

m=0

equation of line= y-y1=m(x-x1)

y-1=0(x-3)

y-1=0-0

y-1=0

y=0+1

y=1 ...

y=1 ANSWER

HOPE THIS HELPS