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Question

The coordinates of points A,B,C and P are (2,3),(1,2),(4,3) and (t,t2) respectively, where t>0. If the area of ABC is twice the area of PAB, then the number of value(s) of t is

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Solution

Given coordinates are A=(2,3),B=(1,2),C=(4,3) and P=(t,t2)

Area of ABC
=1221423233=12|15+6|=1

Area of PAB
=12t21tt232t2=12|3t2t2+1+t22t|=12|t2+t+1|

Given, the area of ABC is twice that of PAB, so
1=2×12|t2+t+1|t2+t+1=±1t2t1=±1t2t=0,2

Now, t2t=0
t=0,1

Also, t2t2=0
(t2)(t+1)=0t=1,2

Hence, t=1,2 as t>0

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