Class 12: Determinants – Exercise 6.3

$\displaystyle \textbf{Question 1: }\text{Find the area of the triangle with vertices at the points:}$
$\displaystyle \text{(i) }(3,8),\;(-4,2),\;(5,-1)$
$\displaystyle \text{(ii) }(2,7),\;(1,1)\text{ and }(10,8)$
$\displaystyle \text{(iii) }(-1,-8),\;(-2,-3)\text{ and }(3,2)$
$\displaystyle \text{(iv) }(0,0),\;(6,0)\text{ and }(4,3)$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}3&8&1\\-4&2&1\\5&-1&1\end{vmatrix}\right|$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}3&8&1\\-7&-6&0\\2&-9&0\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|\begin{vmatrix}-7&-6\\2&-9\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|63+12\right|$
$\displaystyle =\frac{75}{2}\text{ square units}$

$\displaystyle \text{(ii)}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}2&7&1\\1&1&1\\10&8&1\end{vmatrix}\right|$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}2&7&1\\-1&-6&0\\8&1&0\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|\begin{vmatrix}-1&-6\\8&1\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|-1+48\right|$
$\displaystyle =\frac{47}{2}\text{ square units}$

$\displaystyle \text{(iii)}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}-1&-8&1\\-2&-3&1\\3&2&1\end{vmatrix}\right|$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}-1&-8&1\\-1&5&0\\4&10&0\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|\begin{vmatrix}-1&5\\4&10\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|-10-20\right|$
$\displaystyle =15\text{ square units}$

$\displaystyle \text{(iv)}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}0&0&1\\6&0&1\\4&3&1\end{vmatrix}\right|$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}0&0&1\\6&0&0\\4&3&0\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|\begin{vmatrix}6&0\\4&3\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|18-0\right|$
$\displaystyle =9\text{ square units}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Using determinants, show that the following points are collinear:}$
$\displaystyle \text{(i) }(5,5),\;(-5,1)\text{ and }(10,7)$
$\displaystyle \text{(ii) }(1,-1),\;(2,1)\text{ and }(4,5)$
$\displaystyle \text{(iii) }(3,-2),\;(8,8)\text{ and }(5,2)$
$\displaystyle \text{(iv) }(2,3),\;(-1,-2)\text{ and }(5,8)$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$
$\displaystyle \Delta=\begin{vmatrix}5&5&1\\-5&1&1\\10&7&1\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}5&5&1\\-10&-4&0\\5&2&0\end{vmatrix}$
$\displaystyle =\begin{vmatrix}-10&-4\\5&2\end{vmatrix}$
$\displaystyle =-20+20=0$
$\displaystyle \therefore\text{ The points }(5,5),\;(-5,1)\text{ and }(10,7)\text{ are collinear.}$

$\displaystyle \text{(ii)}$
$\displaystyle \Delta=\begin{vmatrix}1&-1&1\\2&1&1\\4&5&1\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}1&-1&1\\1&2&0\\3&6&0\end{vmatrix}$
$\displaystyle =\begin{vmatrix}1&2\\3&6\end{vmatrix}$
$\displaystyle =6-6=0$
$\displaystyle \therefore\text{ The points }(1,-1),\;(2,1)\text{ and }(4,5)\text{ are collinear.}$

$\displaystyle \text{(iii)}$
$\displaystyle \Delta=\begin{vmatrix}3&-2&1\\8&8&1\\5&2&1\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}3&-2&1\\5&10&0\\2&4&0\end{vmatrix}$
$\displaystyle =\begin{vmatrix}5&10\\2&4\end{vmatrix}$
$\displaystyle =20-20=0$
$\displaystyle \therefore\text{ The points }(3,-2),\;(8,8)\text{ and }(5,2)\text{ are collinear.}$

$\displaystyle \text{(iv)}$
$\displaystyle \Delta=\begin{vmatrix}2&3&1\\-1&-2&1\\5&8&1\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}2&3&1\\-3&-5&0\\3&5&0\end{vmatrix}$
$\displaystyle =\begin{vmatrix}-3&-5\\3&5\end{vmatrix}$
$\displaystyle =-15+15=0$
$\displaystyle \therefore\text{ The points }(2,3),\;(-1,-2)\text{ and }(5,8)\text{ are collinear.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{If the points }(a,0),\;(0,b)\text{ and }(1,1)\text{ are collinear, prove that } \\ a+b=ab.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the points }(a,0),\;(0,b)\text{ and }(1,1)\text{ are collinear,}$
$\displaystyle \begin{vmatrix}a&0&1\\0&b&1\\1&1&1\end{vmatrix}=0$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \begin{vmatrix}a&0&1\\-a&b&0\\1-a&1&0\end{vmatrix}=0$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \begin{vmatrix}-a&b\\1-a&1\end{vmatrix}=0$
$\displaystyle \Rightarrow -a-b(1-a)=0$
$\displaystyle \Rightarrow -a-b+ab=0$
$\displaystyle \Rightarrow a+b=ab$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Using determinants, prove that the points }(a,b),\;(a',b')\text{ and } \\ (a-a',b-b')\text{ are collinear if }ab'=a'b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}a&b&1\\a'&b'&1\\a-a'&b-b'&1\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}a&b&1\\a'-a&b'-b&0\\-a'&-b'&0\end{vmatrix}$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \Delta=\begin{vmatrix}a'-a&b'-b\\-a'&-b'\end{vmatrix}$
$\displaystyle =-b'(a'-a)+a'(b'-b)$
$\displaystyle =-a'b'+ab'+a'b'-a'b$
$\displaystyle =ab'-a'b$
$\displaystyle \text{But }ab'=a'b.$
$\displaystyle \therefore \Delta=0$
$\displaystyle \therefore\text{ The points }(a,b),\;(a',b')\text{ and }(a-a',b-b')\text{ are collinear.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the value of }\lambda\text{ so that the points }(1,-5),\;(-4,5)\text{ and }(\lambda,7)\text{ are collinear.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the points }(1,-5),\;(-4,5)\text{ and }(\lambda,7)\text{ to be collinear,}$
$\displaystyle \begin{vmatrix}1&-5&1\\-4&5&1\\\lambda&7&1\end{vmatrix}=0$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \begin{vmatrix}1&-5&1\\-5&10&0\\\lambda-1&12&0\end{vmatrix}=0$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \begin{vmatrix}-5&10\\\lambda-1&12\end{vmatrix}=0$
$\displaystyle \Rightarrow -60-10(\lambda-1)=0$
$\displaystyle \Rightarrow -60-10\lambda+10=0$
$\displaystyle \Rightarrow -10\lambda=50$
$\displaystyle \therefore \lambda=-5$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Find the value of }x\text{ if the area of the triangle with vertices } \\ (x,4),\;(2,-6)\text{ and }(5,4)\text{ is }35\text{ square cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}x&4&1\\2&-6&1\\5&4&1\end{vmatrix}\right|$
$\displaystyle \therefore \frac{1}{2}\left|\begin{vmatrix}x&4&1\\2&-6&1\\5&4&1\end{vmatrix}\right|=35$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \frac{1}{2}\left|\begin{vmatrix}x&4&1\\2-x&-10&0\\5-x&0&0\end{vmatrix}\right|=35$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \frac{1}{2}\left|\begin{vmatrix}2-x&-10\\5-x&0\end{vmatrix}\right|=35$
$\displaystyle \Rightarrow \frac{1}{2}\left|10(5-x)\right|=35$
$\displaystyle \Rightarrow \left|50-10x\right|=70$
$\displaystyle \Rightarrow 50-10x=70\quad\text{or}\quad 50-10x=-70$
$\displaystyle \Rightarrow -10x=20\quad\text{or}\quad -10x=-120$
$\displaystyle \therefore x=-2\quad\text{or}\quad x=12$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Using determinants, find the area of the triangle whose vertices are } \\ (1,4),\;(2,3)\text{ and }(-5,-3).\text{ Are the given points collinear?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}1&4&1\\2&3&1\\-5&-3&1\end{vmatrix}\right|$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}1&4&1\\1&-1&0\\-6&-7&0\end{vmatrix}\right|$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}1&-1\\-6&-7\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|-7-6\right|$
$\displaystyle =\frac{1}{2}\left|-13\right|$
$\displaystyle =\frac{13}{2}\text{ square units}$
$\displaystyle \text{Also, }\begin{vmatrix}1&4&1\\2&3&1\\-5&-3&1\end{vmatrix}=-13\neq0$
$\displaystyle \therefore\text{ The points }(1,4),\;(2,3)\text{ and }(-5,-3)\text{ are not collinear.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Using determinants, find the area of the triangle with vertices } \\ (-3,5),\;(3,-6)\text{ and }(7,2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}-3&5&1\\3&-6&1\\7&2&1\end{vmatrix}\right|$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}-3&5&1\\6&-11&0\\10&-3&0\end{vmatrix}\right|$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}6&-11\\10&-3\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|-18+110\right|$
$\displaystyle =\frac{1}{2}(92)$
$\displaystyle =46\text{ square units}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Using determinants, find the value of }k\text{ so that the points }(k,\,2-2k),\;(-k+1,\,2k)\text{ and }(-4-k,\,6-2k)\text{ may be collinear.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the given points to be collinear,}$
$\displaystyle \begin{vmatrix}k&2-2k&1\\-k+1&2k&1\\-4-k&6-2k&1\end{vmatrix}=0$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \begin{vmatrix}k&2-2k&1\\-2k+1&4k-2&0\\-4-2k&4&0\end{vmatrix}=0$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \begin{vmatrix}-2k+1&4k-2\\-4-2k&4\end{vmatrix}=0$
$\displaystyle \Rightarrow 4(-2k+1)-(4k-2)(-4-2k)=0$
$\displaystyle \Rightarrow -8k+4+16k-8+8k^2-4k=0$
$\displaystyle \Rightarrow 8k^2+4k-4=0$
$\displaystyle \Rightarrow (8k-4)(k+1)=0$
$\displaystyle \therefore k=-1\quad\text{or}\quad k=\frac{1}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If the points }(x,-2),\;(5,2)\text{ and }(8,8)\text{ are collinear, find }x\text{ using determinants.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the points }(x,-2),\;(5,2)\text{ and }(8,8)\text{ are collinear,}$
$\displaystyle \begin{vmatrix}x&-2&1\\5&2&1\\8&8&1\end{vmatrix}=0$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \begin{vmatrix}x&-2&1\\5-x&4&0\\8-x&10&0\end{vmatrix}=0$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \begin{vmatrix}5-x&4\\8-x&10\end{vmatrix}=0$
$\displaystyle \Rightarrow 10(5-x)-4(8-x)=0$
$\displaystyle \Rightarrow 50-10x-32+4x=0$
$\displaystyle \Rightarrow 18-6x=0$
$\displaystyle \therefore x=3$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If the points }(3,-2),\;(x,2)\text{ and }(8,8)\text{ are collinear, find }x\text{ using determinants.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the points }(3,-2),\;(x,2)\text{ and }(8,8)\text{ are collinear,}$
$\displaystyle \begin{vmatrix}3&-2&1\\x&2&1\\8&8&1\end{vmatrix}=0$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \begin{vmatrix}3&-2&1\\x-3&4&0\\5&10&0\end{vmatrix}=0$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \begin{vmatrix}x-3&4\\5&10\end{vmatrix}=0$
$\displaystyle \Rightarrow 10(x-3)-4(5)=0$
$\displaystyle \Rightarrow 10x-30-20=0$
$\displaystyle \Rightarrow 10x-50=0$
$\displaystyle \therefore x=5$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Using determinants, find the equation of the line joining the points.}$
$\displaystyle \text{(i) }(1,2)\text{ and }(3,6)$
$\displaystyle \text{(ii) }(3,1)\text{ and }(9,3)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required line.}$
$\displaystyle \text{Then the area of }\triangle ABP=0.$
$\displaystyle \therefore \begin{vmatrix}1&2&1\\3&6&1\\x&y&1\end{vmatrix}=0$
$\displaystyle \Rightarrow 1(6-y)-2(3-x)+1(3y-6x)=0$
$\displaystyle \Rightarrow 6-y-6+2x+3y-6x=0$
$\displaystyle \Rightarrow 2y-4x=0$
$\displaystyle \therefore y=2x$
$\displaystyle \text{(ii)}$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required line.}$
$\displaystyle \text{Then the area of }\triangle ABP=0.$
$\displaystyle \therefore \begin{vmatrix}3&1&1\\9&3&1\\x&y&1\end{vmatrix}=0$
$\displaystyle \Rightarrow 3(3-y)-1(9-x)+1(9y-3x)=0$
$\displaystyle \Rightarrow 9-3y-9+x+9y-3x=0$
$\displaystyle \Rightarrow -2x+6y=0$
$\displaystyle \therefore x=3y$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Find the values of }k\text{ if the area of the triangle is }4\text{ square units whose vertices are:}$
$\displaystyle \text{(i) }(k,0),\;(4,0),\;(0,2)$
$\displaystyle \text{(ii) }(-2,0),\;(0,4),\;(0,k)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}k&0&1\\4&0&1\\0&2&1\end{vmatrix}\right|=4$
$\displaystyle \text{Expanding along }C_2,$
$\displaystyle \frac{1}{2}\left|2\begin{vmatrix}k&1\\4&1\end{vmatrix}\right|=4$
$\displaystyle \Rightarrow |k-4|=4$
$\displaystyle \Rightarrow k-4=4\quad\text{or}\quad k-4=-4$
$\displaystyle \Rightarrow k=8\quad\text{or}\quad k=0$
$\displaystyle \therefore k=0\text{ or }8$
$\displaystyle \text{(ii)}$
$\displaystyle \text{Area of the triangle}=\frac{1}{2}\left|\begin{vmatrix}-2&0&1\\0&4&1\\0&k&1\end{vmatrix}\right|=4$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \frac{1}{2}\left|-2\begin{vmatrix}4&1\\k&1\end{vmatrix}\right|=4$
$\displaystyle \Rightarrow |-(4-k)|=4$
$\displaystyle \Rightarrow 4-k=4\quad\text{or}\quad 4-k=-4$
$\displaystyle \Rightarrow k=0\quad\text{or}\quad k=8$
$\displaystyle \therefore k=0\text{ or }8$
$\displaystyle \\$