Question

On joining the points $(1,1)(5,-5)$ and $(-1,4)$ we get a/an

• A. Isosceles triangle
• B. equilateral triangle
• C. Scalene triangle
• D. Straight line

Answer 1

The correct option is D Straight line

Given $A(1,1)B(5,-5)C(-1,4)$

$\left|AB\right|=\sqrt{{(5-1)}^2+{(-5-1)}^2}=\sqrt{4^2+6^2}=\sqrt{52}\left|BC\right|=\sqrt{{(5+1)}^2+{(-5-4)}^2}=\sqrt{36+81}=\sqrt{117}\left|AC\right|=\sqrt{{(1+1)}^2+{(1-4)}^2}=\sqrt{2^2+3^2}=\sqrt{13}\left|AB\right|\ne \left|BC\right|\ne \left|AC\right|$

Now lets see if these points are collinear .

If we have three points $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ then they are collinear if

$\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=0$

Using this

$\left|\begin{array}{ccc}1& 1& 1\\ 5& -5& 1\\ -1& 4& 1\end{array}\right|=1\left|\begin{array}{cc}-5& 1\\ 4& 1\end{array}\right|-1\left|\begin{array}{cc}5& 1\\ -1& 1\end{array}\right|+1\left|\begin{array}{cc}5& -5\\ -1& 4\end{array}\right|=-9-6+(20-5)=-15+15=0$

$\therefore A,B,C$ lie on a straight line.