Question

for what value of a , the points (a,2-2a), (-a+1,2a) and (-4-a, 6-2a) are collinear

Answer 1

given points are collinear

then

X1 (Y2-Y3)+X2 (Y3-Y1)+X3 (Y1-Y2)=0

X1=a , X2=-a+1 , X3=-4-a

Y1=2-2a ,Y2=2a ,Y3=6-2a

now put the values in formula

a (2a-(6-2a))+(-a+1)(6-2a-(2-2a ))+(-4-a )(2-2a-(2a))=0

a (2a-6+2a)-a+1(6-2a-2+2a)-4-a (2-2a-2a)=0

a (4a-6)-a+1 (4)-4-a (2-4a)=0

4a^2-6a-4a+4-8+16a-2a+4a^2=0

8a^2+4a-4=0

8a^2+8a-4a-4=0

8a (a+1)-4 (a+1)=0

(8a-4)(a+1)=0

a=4/8 ,-1

a=1/2 ,-1

Answer 2

Answer:

a =1\2,-1 is the answer

HOPE IT helps you