SATHEE: Straight Line and Pair of Straight Lines 1 Question 56

Straight Line and Pair of Straight Lines 1 Question 56

56. Determine all values of $α$ for which the point $(α, α^{2})$ lies inside the triangles formed by the lines $2 x + 3 y - 1 = 0$ ,

$x + 2 y - 3 = 0, 5 x - 6 y - 1 = 0$

(1992, 6M)

Show Answer

Solution:

Given lines are $2 x + 3 y - 1 = 0$

On solving Eqs. (i), (ii) and (iii), we get the vertices of a triangle are $A (- 7, 5), B \frac{1}{3}, \frac{1}{9}$ and $C \frac{5}{4}, \frac{7}{8}$ .

Let $P (α, α^{2})$ be a point inside the $△ A B C$ . Since, $A$ and $P$ are on the same side of $5 x - 6 y - 1 = 0$ , both $5 (- 7) - 6 (5) - 1$ and $5 α - 6 α^{2} - 1$ must have the same sign, therefore

$\begin{array}{cc} 5 α - 6 α^{2} - 1 < 0 \\ \Rightarrow & 6 α^{2} - 5 α + 1 > 0 \\ \Rightarrow & (3 α - 1) (2 α - 1) > 0 \\ α < \frac{1}{3} or α > \frac{1}{2} \end{array}$

Also, since $P (α, α^{2})$ and $C \frac{5}{4}, \frac{7}{8}$ lie on the same side of $2 x + 3 y - 1 = 0$ , therefore both $2 \frac{5}{4} + 3 \frac{7}{8} - 1$ and $2 α + 3 α^{2} - 1$ must have the same sign.

Therefore,

$2 α + 3 α^{2} - 1 > 0$

$\begin{aligned} \Rightarrow & (α + 1) & α - \frac{1}{3} > 0 \\ \Rightarrow & α < - 1 \cup α > 1 / 3 \end{aligned}$

and lastly $\frac{1}{3}, \frac{1}{9}$ and $P (α, α^{2})$ lie on the same side of the line therefore, $\frac{1}{3} + 2 \frac{1}{9} - 3$ and $α + 2 α^{2} - 3$ must have the same sign. Therefore, $2 α^{2} + α - 3 < 0$