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Distance Formula

Prove that th...

Question

Prove that the points $(a t_{1}^{2}, 2 a t_{1})$ , $(a t_{2}^{2}, 2 a t_{2})$ and $(0, a)$ will be collinear, if $t_{1} t_{2} = - 1$ .

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Solution

$A (a t_{1}^{2}, {2 a t}_{1})$ and $B (a t_{2}^{2}, {2 a t}_{2})$ and $C (0, a)$
=>Area of $△ A B C = 0$ for collinearity
=>Area of $△ A B C = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a t_{1}^{2} & {2 a t}_{1} a t_{2}^{2} & {2 a t}_{2} 0 & a a t_{1}^{2} & {2 a t}_{1} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ 0 = \frac{1}{2} ∣ ∣ 2 a^{2} t_{1}^{2} t_{2} - 2 a^{2} t_{1} t_{2}^{2} + a^{2} t_{2}^{2} - a^{2} t_{1}^{2} ∣ ∣ 0 = \frac{1}{2} ∣ ∣ a^{2} (t_{1} t_{2} ({2 t}_{1} - {2 t}_{2})) + a^{2} (t_{2} - t_{1}) (t_{2} + t_{1}) ∣ ∣ 0 = \frac{1}{2} ∣ ∣ a^{2} (t_{1} t_{2} + 1) (t_{1} + t_{2}) ∣ ∣ => (t_{1} t_{2} + 1) = 0 o r t_{1} - t_{2} = 0$
As $t_{1} - t_{2}$ cannot be zero,if so they will coincide.
So $t_{1} t_{2} = - 1$

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