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Check how the points A,B and C are situa...

Check how the points A,B and C are situated where `A(4,0),B(-1,-1),C(3,5)` .

Text Solution

Verified by Experts

The correct Answer is:
(i) ABC is triangle right angled at A
(ii) collinear points
(iii) ABC is right angled isoceles triangle
(i) `A(-2,2), B((8,-2),C(-4,-3)`
`AB=sqrt((8-(-2))^2+(-2-2)^(2))`
`=sqrt(100+16)=2sqrt(29)`
`BC=sqrt((8-(-4))^2+(-2-(-3))^(2))`
`=sqrt(144+1)=sqrt(5)=sqrt(29)`
`CA=sqrt((-2-(-4))^2+(2-(-3))^(2))`
`=sqrt(4+25)=sqrt(29)`
Thus, `AB^2+CA^2=BC^2`
So, triangle is right angled at A.
(ii) `A(-a,-b),B(a,b),C(a^2,ab)`
`AB=sqrt((2a)^2+(2b)^(2))=2sqrt(a^2+b^(2))`
`BC=sqrt((a^2-a)^2+b^2(a-1)^(2))=(a-1)sqrt(a^2+b^(2)`
`AC=sqrt((a^2+a)^2+b^2(a+1)^(2))=(a+1)sqrt(a^2+b^2)`
Thus, `AB+BC=AC`.
So, points are collinear.
(iii) `A(4,0),B(-1,-1),C(3,5)`.
`AB=sqrt((-1-4)^2+b^2(-1-0)^(2))=sqrt(25+1)=sqrt(26)`
`BC=sqrt((3+1)^2+b^2(5+1)^(2))=sqrt(16+36)=sqrt(52)`
`CA=sqrt((4-3)^2+b^2(0-5)^(2))=sqrt(1+25)=sqrt(26)`
Thus, `AB=CA and BC62=AB^2+CA62`
So, triangle ABC is right angled isoceles.
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