Question

please Help me by answering this question.

Answer 1

Answer:

are of triangle =1/2 x1(y2-y3)+x2(y3-y1)+x3(y1-y2)=0 then the points are coliner

1/2*a(1-5)+2(5-a)+5(a-1)=0

then a=5

Answer 2

if the points are collinear then the area a of triangle formed by them must be equal to zero.

area of triangle ABC= 0

1/2 [x1(y2-y3)+x2 (y3-y1)+x3 (y1-y2)]=0

1/2 [a (1-a)+2 (a-3)+5 (3-1)]=0

1/2 [a-a^2+2a-6+10]=0

1/2 [-a^2+3a+4]=0

-a^2+3a+4=0/2

-a^2+3a+4=0

a^2-3a-4=0

on comparing with ax^2+bx+c

we get a=1,b=-3a,c=-4

now D=b^2-4ac

D=(-3)^2-4×1×(-4)

D=9+16

D=25

now using quadratic formula

$x = \frac{ - b + \sqrt{d} }{2a} and \: x = \frac{ - b - \sqrt{d} }{2a} \\ = > a= \frac{ - ( - 3) + \sqrt{25} }{2 \times 1} and \: a = \frac{ - ( - 3) - \sqrt{25} }{2 \times 1} \\ = > a = \frac{3 + 5}{2} and \:a= \frac{3 - 5}{2} \\ = > a = \frac{8}{2} and \: a= \frac{ - 2}{2} \\ = > a = 4 \: and \: a = - 1(which \: is \: neglected)$

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