Question

If the points A(x,y) ,B(3,6) and C(-3,4) are collinear,show that x-3y+15=0

Answer 1

As these points are collinear.
So, area of ABC=0
1/2(x(6-4)+3(4-y)-3(y-6))=0
2x+12-3y-3y+18=0
2x-6y+30=0
x-3y+15=0.
Hence proved

Answer 2

Answer:

As points are colinear ,

Area of ΔABC=0

Step-by-step explanation:

1/2{ x1(y2-y1) +x2(y3-y1) +x3(y1-y3) }  =0

=> 1/2 { x(6-4) +3(4-y) -3(y-6) }          =0

=> x(2) + 3(4-y) -3(y-6) =0 x 2/1

=> 2x+12-3y-3y+18 = 0

=> 2x-6y+30 = 0

=> 2[x-3y+15] =0

=> x-3y+15 = 0

Hence proved :)