Question

Prove that point $A\equiv ({\csc }^2\theta ,0),B\equiv (0,{\sec }^2\theta )$ and $C\equiv (1,1)$ are collinear by distance formula.

Answer 1

Distance formula is given by,

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$A\equiv ({\csc }^2\theta ,0),B\equiv (0,{\sec }^2\theta ),C\equiv (1,1)$

$(AB{)}^2=({\csc }^2\theta -0{)}^2+(0-{\sec }^2\theta {)}^2$

$=(1+{\cot }^2\theta {)}^2+(1+{\tan }^2\theta {)}^2$

upon simplification, we get

$AB=\frac{{\sec }^2\theta }{{\tan }^2\theta }\sqrt{1+{\tan }^4\theta }$...(1)

$(AC{)}^2=({\csc }^2\theta -1{)}^2+(0-1{)}^2$

$(AC{)}^2=({\cot }^2\theta {)}^2+1$

upon simplification, we get,

$AC=\frac{\sqrt{1+{\tan }^4\theta }}{{\tan }^2\theta }$...(2)

$(CB{)}^2=(0-1{)}^2+({\sec }^2\theta -1{)}^2$

$(CB{)}^2=1+{\tan }^4\theta$

$CB=\sqrt{1+{\tan }^4\theta }$...(3)

Adding (2) and (3), we get,

$AC+CB=\frac{\sqrt{1+{\tan }^4\theta }}{{\tan }^2\theta }+\sqrt{1+{\tan }^4\theta }$

$=\frac{\sqrt{1+{\tan }^4\theta }+{\tan }^2\theta \sqrt{1+{\tan }^4\theta }}{{\tan }^2\theta }$

$=\frac{\sqrt{1+{\tan }^4\theta }}{{\tan }^2\theta }(1+{\tan }^2\theta )$

$=\frac{{\sec }^2\theta }{{\tan }^2\theta }\sqrt{1+{\tan }^4\theta }$

$=AB$

Therefore A,B,C are collinear and hence proved.