Find the points (3,-8);(4,-11);(5,-k) are collinear then find k
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Sol.A(3,−8)B(4,−11)C(5,−k)
If this points are collinear
then they lie on same line
so they will form a triangle
whose area is O Ar.△=0
⇒
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]=0
x
1
=3 y
1
=−8
x
2
=4 y
2
=−11
x
3
=5 y
3
=−K
⇒
2
1
[3(−11+k)+4(−k−(−8))+5(−8−(−11))]=0
⇒
2
1
[3(+k−11)+4(−k+8)+5(−8+11)]=0
⇒
2
1
[3k−33+32−4k+5(3)]=0
[−33+32+15−4k+3k]=0
[14−k]=0
k=14
k=14
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