Find the value of k, if the points A (2, 3), B (4, k) and C (6,-
3) are collinea
Answers
Answered by
2
k =0 …...…................
2,4,6,are in,AP
3,k,-3 are in AP
k=0
mkrishnan:
wow
super
i just know Tamil and English
Malayalam
are u in kerala
no i am in Kolkata
which standard do u study
I am in class 11th with pcb and math in extra subject
but i am higher secondary maths teacher
wow
Answered by
4
Answer:
Let
X1 =2
X2 = 4
X3= 6
Y1= 3
Y2=k
Y3=-3
Step-by-step explanation:
Hence it collinea therefore area of triangles made by these points will be zero
So
X1(y2-y3)+x2(y3-y1)+x3(y1-y2)=0
Substituting the values we obtain
1/2[2(k+3)+ 4(-3-3)+6(3-k)]=0
1/2[2k+6+4(-6)+18-6k]=0
1/2(2k-6k+6+18-24)=0
1/2(-4k+24-24)=0
(1/2)(-4k+0)=0
-4k=0×2=0
K=0
Thx
Let
X1 =2
X2 = 4
X3= 6
Y1= 3
Y2=k
Y3=-3
Step-by-step explanation:
Hence it collinea therefore area of triangles made by these points will be zero
So
X1(y2-y3)+x2(y3-y1)+x3(y1-y2)=0
Substituting the values we obtain
1/2[2(k+3)+ 4(-3-3)+6(3-k)]=0
1/2[2k+6+4(-6)+18-6k]=0
1/2(2k-6k+6+18-24)=0
1/2(-4k+24-24)=0
(1/2)(-4k+0)=0
-4k=0×2=0
K=0
Thx
Is my answer right
no k=0
you take x3=5 but x3=6
nice
Formula is 1/2 missed
when equal to zero. 1/2 need not
why didn't give atleast thank for me
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