plz solve both the questions!!
I'll mark the answer as brainliest!!
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1
they are in ap
So,
d=3p-7=7-(2p-1)
3p-7=7-2p+1
5p=15
p=3
the points are collinear
so,1/2[x1(y2-y1)+x2(y3-y1)+x3(y1-y2)]=0
2(-1+1)+p(3-1)+(-1)(1+1)=0
0+2p-2=0
2p=2
p=1
So,
d=3p-7=7-(2p-1)
3p-7=7-2p+1
5p=15
p=3
the points are collinear
so,1/2[x1(y2-y1)+x2(y3-y1)+x3(y1-y2)]=0
2(-1+1)+p(3-1)+(-1)(1+1)=0
0+2p-2=0
2p=2
p=1
danishk08:
pls mark it as brainliest and pls follow me for further help
hey friend, the 10th answer is wrong
Answered by
1
Hello friend, here are your answers.....
If helps then mark as brainliest.
If helps then mark as brainliest.
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a1 is 2p-1 tou wrote 2p+1. can u plz tell me the concept behind this
you are wrong
its because d= 7-(2p+1) = 7-2p-1
when we will open the bracket it will become negative
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