Class 11: The Straight Line

$\displaystyle \textbf{Question 1: }\text{Find the values of }\alpha\text{ so that the point }P(\alpha^2,\alpha)\text{ lies inside or on the}$
$\displaystyle \text{triangle formed by the lines }x-5y+6=0,\ x-3y+2=0\text{ and }x-2y-3=0.$
$\displaystyle \text{Answer:}$ 2021-01-30_19-53-05
$\displaystyle \text{Let the triangle be }\triangle ABC,\text{ where}$
$\displaystyle AB:x-5y+6=0\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle BC:x-3y+2=0\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle CA:x-2y-3=0\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle \text{Solving the equations in pairs, we obtain }A(9,3),\ B(4,2)\text{ and }C(13,5).$
$\displaystyle \text{Given point }P(\alpha^2,\alpha).$
$\displaystyle \text{For }P\text{ to lie inside or on }\triangle ABC,\text{ the following conditions must hold simultaneously:}$
$\displaystyle \text{(i) }A\text{ and }P\text{ must lie on the same side of }BC.$
$\displaystyle [9-3(3)+2](\alpha^2-3\alpha+2)\geq0$
$\displaystyle 2(\alpha^2-3\alpha+2)\geq0$
$\displaystyle (\alpha-1)(\alpha-2)\geq0$
$\displaystyle \therefore \alpha\in(-\infty,1]\cup[2,\infty)\qquad\ldots\ldots\ldots\text{(iv)}$
$\displaystyle \text{(ii) }B\text{ and }P\text{ must lie on the same side of }CA.$
$\displaystyle [4-2(2)-3](\alpha^2-2\alpha-3)\geq0$
$\displaystyle -3(\alpha^2-2\alpha-3)\geq0$
$\displaystyle (\alpha-3)(\alpha+1)\leq0$
$\displaystyle \therefore \alpha\in[-1,3]\qquad\ldots\ldots\ldots\text{(v)}$
$\displaystyle \text{(iii) }C\text{ and }P\text{ must lie on the same side of }AB.$
$\displaystyle [13-5(5)+6](\alpha^2-5\alpha+6)\geq0$
$\displaystyle -6(\alpha^2-5\alpha+6)\geq0$
$\displaystyle (\alpha-2)(\alpha-3)\leq0$
$\displaystyle \therefore \alpha\in[2,3]\qquad\ldots\ldots\ldots\text{(vi)}$
$\displaystyle \text{Taking the intersection of (iv), (v) and (vi), we get}$
$\displaystyle \therefore \alpha\in[2,3].$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the values of the parameter }a\text{ so that the point }(a,2)\text{ is an interior}$
$\displaystyle \text{point of the triangle formed by the lines }x+y-4=0,\ 3x-7y-8=0\text{ and }4x-y-31=0.$
$\displaystyle \text{Answer:}$ 2021-01-30_19-53-20
$\displaystyle \text{Let the triangle be }\triangle ABC,\text{ where}$
$\displaystyle AB:x+y-4=0\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle BC:3x-7y-8=0\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle CA:4x-y-31=0\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle \text{Solving the equations in pairs, we obtain}$
$\displaystyle A(7,-3),\qquad B\left(\frac{18}{5},\frac{2}{5}\right),\qquad C\left(\frac{209}{25},\frac{61}{25}\right).$
$\displaystyle \text{Given point }P(a,2).$
$\displaystyle \text{For }P\text{ to lie inside }\triangle ABC,\text{ the following conditions must hold simultaneously:}$
$\displaystyle \text{(i) }A\text{ and }P\text{ must lie on the same side of }BC.$
$\displaystyle [3(7)-7(-3)-8](3a-7(2)-8)>0$
$\displaystyle 34(3a-22)>0$
$\displaystyle 3a-22>0$
$\displaystyle \therefore a>\frac{22}{3}\qquad\ldots\ldots\ldots\text{(iv)}$
$\displaystyle \text{(ii) }B\text{ and }P\text{ must lie on the same side of }CA.$
$\displaystyle \left[4\left(\frac{18}{5}\right)-\frac{2}{5}-31\right](4a-2-31)>0$
$\displaystyle (-17)(4a-33)>0$
$\displaystyle 4a-33<0$
$\displaystyle \therefore a<\frac{33}{4}\qquad\ldots\ldots\ldots\text{(v)}$
$\displaystyle \text{(iii) }C\text{ and }P\text{ must lie on the same side of }AB.$
$\displaystyle \left[\frac{209}{25}+\frac{61}{25}-4\right](a+2-4)>0$
$\displaystyle \frac{34}{5}(a-2)>0$
$\displaystyle a-2>0$
$\displaystyle \therefore a>2\qquad\ldots\ldots\ldots\text{(vi)}$
$\displaystyle \text{Taking the intersection of (iv), (v) and (vi), we get}$
$\displaystyle \therefore a\in\left(\frac{22}{3},\frac{33}{4}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Determine whether the point }(-3,2)\text{ lies inside or outside the triangle}$
$\displaystyle \text{whose sides are }x+y-4=0,\ 3x-7y+8=0\text{ and }4x-y-31=0.$
$\displaystyle \text{Answer:}$ 2021-01-30_19-53-34
$\displaystyle \text{Let the triangle be }\triangle ABC,\text{ where}$
$\displaystyle AB:x+y-4=0\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle BC:3x-7y+8=0\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle CA:4x-y-31=0\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle \text{Solving the equations in pairs, we obtain}$
$\displaystyle A\left(\frac{209}{25},\frac{61}{25}\right),\qquad B\left(\frac{18}{5},\frac{2}{5}\right),\qquad C(7,-3).$
$\displaystyle \text{Given point }P(-3,2).$
$\displaystyle \text{The point }P\text{ lies inside or on }\triangle ABC\text{ only if the following three conditions hold simultaneously:}$
$\displaystyle \text{(i) }A\text{ and }P\text{ lie on the same side of }BC.$
$\displaystyle \left[3\left(\frac{209}{25}\right)-7\left(\frac{61}{25}\right)+8\right]\left[3(-3)-7(2)+8\right]>0$
$\displaystyle \left(\frac{34}{5}\right)(-15)=-102<0.$
$\displaystyle \text{Hence, this condition is FALSE.}$
$\displaystyle \text{(ii) }B\text{ and }P\text{ lie on the same side of }CA.$
$\displaystyle \left[4\left(\frac{18}{5}\right)-\frac{2}{5}-31\right]\left[4(-3)-2-31\right]>0$
$\displaystyle (-17)(-45)=765>0.$
$\displaystyle \text{Hence, this condition is TRUE.}$
$\displaystyle \text{(iii) }C\text{ and }P\text{ lie on the same side of }AB.$
$\displaystyle [7+(-3)-4][(-3)+2-4]>0$
$\displaystyle (0)(-5)=0.$
$\displaystyle \text{Since vertex }C\text{ lies on }AB,\text{ this condition is satisfied.}$
$\displaystyle \text{As all three conditions are not satisfied simultaneously,}$
$\displaystyle \therefore \text{the point }(-3,2)\text{ lies outside }\triangle ABC.$
$\displaystyle \\$