Question

The possible values of a for which the point $(a,a^2)$ lies inside the triangle formed by the straight lines 2x + 3y – 1 = 0, x + 2y = 3 and 5x – 6y – 1 = 0 is

• A. $(-3,2)\cup (3,6)$
• B. $(\frac{-3}{2},-1)\cup (\frac{1}{2},1)$
• C. $(-3,1)\cup (1,2)$
• D. $(-1,1)\cup (1,3)$

Answer 1

The correct option is B

$(\frac{-3}{2},-1)\cup (\frac{1}{2},1)$

coordinate of triangle will be $A(\frac{5}{4},\frac{7}{8}),B(-7,5)$ and $C(\frac{1}{3},\frac{1}{9}).P(a,a^2)$ lies inside the triangle. So, P and A must lie on same side of BC. P and B must be on same side of CA and C and P must lies on the same side of AB.

Hence $(\frac{5}{2}+\frac{21}{8}-1)(2a+3a^2-1)>0$

$aϵ(-\infty ,-1)\cup (\frac{1}{3},\infty )$

and $(-35-30-1)(5a-6a^2-1)>0$

$aϵ(-\infty ,\frac{1}{3})\cup (\frac{1}{2},\infty )$

Also, $(\frac{1}{3}+\frac{2}{9}-3)(a+2a^2-3)>0$

$aϵ(-\frac{3}{2},1)$

So, $aϵ(-\frac{3}{2},-1)\cup (\frac{1}{2},1)$