Line AB passes through points A(–6, 6) and B(12, 3). If the equation of the line is written in slope-intercept form, y = mx + b, then m = –1/6 and b =
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LET P DIVIDES AB IN K:1
X=12K-6/K+1
and
Y=3K+6/K+1
PUT VALUE OF X AND Y IN Y=MX+B
3K+6/K+1 =-1(12K-6)/6(K+1)+ B
3K+6/K+1=-1×6(2K-1)/6(K+1)+B
3K+6/K+1 +2K-1/K+1= B
5K+5/K+1=B
5(K+1)/K+1=B
B= 5
X=12K-6/K+1
and
Y=3K+6/K+1
PUT VALUE OF X AND Y IN Y=MX+B
3K+6/K+1 =-1(12K-6)/6(K+1)+ B
3K+6/K+1=-1×6(2K-1)/6(K+1)+B
3K+6/K+1 +2K-1/K+1= B
5K+5/K+1=B
5(K+1)/K+1=B
B= 5
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