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The Straight Lines — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsThe Straight Lines5 Marks
Question
Find the values of $\alpha$ so that the point $\text{p}(\alpha^2,\alpha)$ lies inside or on the triangle formed by the lines $x - 5y + 6 = 0, x - 3y + 2 = 0$ and $x - 2y - 3 = 0.$
✓

Answer

Let $\text{ABC}$ be the triangle of the equations whose sides $\text{AB, BC}$ and $\text{CA}$ are respectively $x - 5y + 6 = 0, x - 3y + 2 = 0$ and $x - 2y - 3 = 0$
The coordinates of the vertices are $A(9, 3), B(4, 2)$ and $C(13, 5).$
If the point $\text{p}(\alpha^2,\alpha)$ lies $n$ side the $\triangle\text{ABC},$ then
  1. $A$ and $P$ must be on the same side of $\text{BC.}$
  2. $B$ and $P$ must be on the same side of $\text{AC.}$
  3. $C$ and $P$ must be on the same side of $\text{AB.}$
Now,
$A$ and $P$ must be on the same side of $BC$ if,
$\big(9(1)+3(-3)+2\big)\Big(\alpha^2-3\alpha+2\Big)>0$
$(9-9+2)\big(\alpha^2-3\alpha+2\big)>0$
$\alpha^2-3\alpha+2>0$
$(\alpha-1)(\alpha-2)>0 $
$\alpha\in(-\infty,1)\cup(2,\infty) \ ...(\text{i})$
$B$ and $P$ must be on the same side of $AC$ if,
$\big(13(1)+5(-5)+6\big)\Big(\alpha^2-5\alpha+6\Big)>0$
$\Rightarrow(-6)\big(\alpha^2-5\alpha+6\big)>0$
$\Rightarrow\alpha^2-5\alpha+6<0$
$\Rightarrow(\alpha-2)(\alpha-3)<0 $
$\Rightarrow\alpha\in(2, 3) \ ...(\text{ii})$
$C$ and $P$ must be on the same side of $AB$ if,
$\big(4(1)+2(-2)-3\big)\Big(\alpha^2-2\alpha-3\Big)>0$
$(-3)\big(\alpha^2-2\alpha-3\big)>0$
$\alpha^2-2\alpha-3<0$
$(\alpha-3)(\alpha+1)<0 $
$\Rightarrow\alpha\in(-1, 3) \ ...(\text{iii})$
From $i, ii$, and $iii$
$\alpha\in[2, 3]$

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