InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
In each of the following, check how the points A,B and C are situated. (i) `A(-2,2),B(8,-2),C(-4,-3)` (ii) `A(-a,-b),B(a,b),C(a^2,ab),agt1`(iii) `A(4,0),B(-1,-1),C(3,5)` |
|
Answer» Correct Answer - (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. |
|