find the values of k for which points A(-6,10),B(-4,k) and C(3,-8) are collinear. also,find the ratio in which B divides AC and the length of AC.
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Answered by
21
A , B and C are colinear so area of triangle form by this points is zero,
now ,
area of triangle=1/2 {-6 (k+8)-4 (-8-10)+3 (10-k)}
0 =-6k-48+72+30-3k
0=-9k+54
k=6
now use section formula , for x axis
-4=(3m-6n)/(m+n)
-4 (m+n)=3m-6n
-4m-4n=3m-6n
-7m=-2n
m/n=2/7
hence ratio 2:7
length of AC=root {(-6-3)^2+(10+8)^2}
=root {81+324}=root (405)
now ,
area of triangle=1/2 {-6 (k+8)-4 (-8-10)+3 (10-k)}
0 =-6k-48+72+30-3k
0=-9k+54
k=6
now use section formula , for x axis
-4=(3m-6n)/(m+n)
-4 (m+n)=3m-6n
-4m-4n=3m-6n
-7m=-2n
m/n=2/7
hence ratio 2:7
length of AC=root {(-6-3)^2+(10+8)^2}
=root {81+324}=root (405)
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3
Answer:
given answer is absolutely right an error free
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