RD Sharma Class 12 Exercise 5.3 Determinants Solutions Maths - Download PDF Free Online

Determinants Excercise: 5.3

Determinants exercise 5.3 question 1 (ii)

$Answer\! : \frac{47}{2}sq.units$
Hints: By putting the values of the co-ordinate in the formulaof area, we will calculate the area of the triangle.
$Given\, \! : \left ( 2,7 \right )\! ,\left ( 1,1 \right )\: and \: \left ( 10,8 \right )\, \! .$
$Explanation\! : V\! ertices\: are \left ( 2,7 \right )\! ,\left ( 1,1 \right )\! , \left ( 10,8 \right )\, \! .$
$Area\; o\! f\; triangles\; is =\! \Delta\! =\frac{1}{2}\begin{vmatrix} X_{1} &Y_{1} & 1\\ X_{2} &Y_{2} & 1\\ X_{3} & Y_{3}& 1 \end{vmatrix}$
$\Rightarrow where, \begin{matrix} X_{1}=2 &Y_{1}=7 & \\ X_{2}=1 &Y_{2}=1 & \\ X_{3}=10 &Y_{3}=8 & \end{matrix}$
$\Rightarrow \! \Delta \, \! =\frac{1}{2}\begin{vmatrix} 2 &7 &1 \\ 1 &1 &1 \\ 10 &8 &1 \end{vmatrix}$
$\Rightarrow \! \Delta \, \! =\frac{1}{2}\left ( 2\! \begin{vmatrix} 1 &1 \\ 8 &1 \end{vmatrix}-7\! \begin{vmatrix} 1 &1 \\ 10 &1 \end{vmatrix}+1\! \begin{vmatrix} 1 &1 \\ 10 &8 \end{vmatrix} \right )$
$=\frac{1}{2}\left [ 2\! \left ( 1-8 \right ) -7\! \left ( 1-10 \right )+1\! \left ( 8-10 \right )\right ]$
$=\frac{1}{2}\left [ 2\! \left ( -7 \right )-7\! \left ( -9 \right )+1\! \left ( -2 \right ) \right ]$
$=\frac{1}{2}\left [ -14+63-2 \right ]$

Determinants exercise 5.3 question 1 (iii)

$Answer\! : 15sq.units.$
Hints: By putting the values of the co-ordinate in the formulaof area, we will calculate the area of the triangle.
$Given\, \! : \left ( -1,-8 \right )\! ,\left ( -2,-3 \right )\: and \: \left ( 3,2 \right )\, \! .$
$Explanation\! : V\! ertices \; are \left ( -1,-8 \right )\! ,\left ( -2,-3 \right )\! , \left ( 3,2 \right )\, \! .$
$Area\; o\! f\; triangles\; is =\! \Delta\! =\frac{1}{2}\begin{vmatrix} X_{1} &Y_{1} & 1\\ X_{2} &Y_{2} & 1\\ X_{3} & Y_{3}& 1 \end{vmatrix}$
$\Rightarrow where, \begin{matrix} X_{1}=-1 &Y_{1}=-8 & \\ X_{2}=-2 &Y_{2}=-3 & \\ X_{3}=3 &Y_{3}=2 & \end{matrix}$
$\Rightarrow \! \Delta \, \! =\frac{1}{2}\begin{vmatrix} -1 &-8 &1 \\ -2 &-3 &1 \\ 3 &2 &1 \end{vmatrix}$
$\Rightarrow \! \Delta \, \! =\frac{1}{2}\left ( -1\! \begin{vmatrix} -3 &1 \\ 2 &1 \end{vmatrix}-\left ( -8 \right )\! \begin{vmatrix} -2 &1 \\ 3 &1 \end{vmatrix}+1\! \begin{vmatrix} -2 &-3 \\ 3 &2 \end{vmatrix} \right )$
$=\frac{1}{2}\left [ -1\! \left ( -3-2 \right ) +8\! \left ( -2-3 \right )+1\! \left ( -4+9 \right )\right ]$
$=\frac{1}{2}\left [ -1\! \left ( -5 \right )+8\! \left ( -5 \right )+1\! \left ( 5\right ) \right ]$
$=\frac{1}{2}\left [ 5-40+5\right ]$
$=\frac{1}{2}\times 30$
$=15sq. units$

Determinants exercise 5.3 question 1 (iv)

$Answer\! : 9sq.units.$
Hints: By putting the values of the co-ordinate in the formulaof area, we will calculate the area of the triangle.
$Given\, \! : \left ( 0,0 \right )\! ,\left ( 6,0 \right )\: and \: \left ( 4,3 \right )\, \! .$
$Explanation\! : V\! ertices \; are \left ( 0,0 \right )\! ,\left (6,0 \right )\! , \left ( 4,3 \right )\, \!.$
$Area\; o\! f\; triangles\; is =\! \Delta\! =\frac{1}{2}\begin{vmatrix} X_{1} &Y_{1} & 1\\ X_{2} &Y_{2} & 1\\ X_{3} & Y_{3}& 1 \end{vmatrix}$
$\Rightarrow where, \begin{matrix} X_{1}=0 &Y_{1}=0 & \\ X_{2}=6 &Y_{2}=0 & \\ X_{3}=4 &Y_{3}=3 & \end{matrix}$
$\Rightarrow \! \Delta \, \! =\frac{1}{2}\begin{vmatrix} 0 &0 &1 \\ 6 &0 &1 \\ 4 &3 &1 \end{vmatrix}$
$\Rightarrow \! \Delta \, \! =\frac{1}{2}\left ( 0\! \begin{vmatrix} 0 &1 \\ 3 &1 \end{vmatrix}-0\! \begin{vmatrix} 6 &1 \\ 4 &1 \end{vmatrix}+1\! \begin{vmatrix} 6 &0 \\ 4 &3 \end{vmatrix} \right )$
$=\frac{1}{2}\left [ 0\! \left ( 0-3 \right ) -0\! \left ( 6-4 \right )+1\! \left ( 18-0 \right )\right ]$
$=\frac{1}{2}\left [ 0-0+18\right ]$
$=\frac{1}{2}\times 18$
$=9sq. units$

Determinants exercise 5.3 question 2 (i)

Answer: Yes, points are collinear.
Hints: First by using the values of vertices find determinant. If value of determinant is zero, then points are collinear.
$Given\! : \left ( 5,5 \right ),\left ( -5,1 \right )\: and \: \left ( 10,7 \right )\! .$
$Explanation\! : V\! ertices\; are \left ( 5,5 \right ),\left ( -5,1 \right )\: and \: \left ( 10,7 \right )\! .$
$Determinant\!= \begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}$
$= \begin{vmatrix} 5 &5 &1 \\ -5 &1 &1 \\ 10 &7 &1 \end{vmatrix}R_{1}\rightarrow R_{1}+R_{2}$
$= \begin{vmatrix} 0 &6 &2 \\ -5 &1 &1 \\ 10 &7 &1 \end{vmatrix}R_{2}\rightarrow R_{2}-R_{3}$
$= \begin{vmatrix} 0 &6 &2 \\ -15 &-6 &0 \\ 10 &7 &1 \end{vmatrix}$
$= 0\begin{vmatrix} -6 &0 \\ 7 &1 \end{vmatrix}-6\begin{vmatrix} -15 &0 \\ 10 &1 \end{vmatrix}+2\begin{vmatrix} -15 &-6 \\ 10 &7 \end{vmatrix}$
$= 0-6\left ( -15-0 \right )+2\left ( -105+60 \right )$
$= 90+2\left ( -45 \right )$
$= 90-90$
$= 0$
Hence, points are collinear.

Determinants exercise 5.3 question 2 (ii)

Answer: Yes, points are collinear.
Hints: First by using the values of vertices find determinant. If value of determinant is zero, then points are collinear.
$Given\! : \left ( 1,-1 \right ),\left ( 2,1 \right )\: and \: \left ( 4,5 \right )\! .$
$Explanation\! : V\! ertices\; are \left ( 1,-1 \right ),\left ( 2,1 \right )\: and \: \left ( 4,5 \right )\! .$
$Determinant\!= \begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}$
$= \begin{vmatrix} 1 &-1 &1 \\ 2 &1 &1 \\ 4 &5 &1 \end{vmatrix}$
$= 1\! \begin{vmatrix} 1 &1 \\ 5 &1 \end{vmatrix}-\left (-1 \right )\! \begin{vmatrix} 2 &1 \\ 4 &1 \end{vmatrix}+1\! \begin{vmatrix} 2 &1 \\ 4 &5 \end{vmatrix}$
$= 1\! \left (1-5 \right )+1\! \left ( 2-4 \right )+1\! \left ( 10-4 \right )$
$= 1\! \left ( -4 \right )+1\! \left ( -2 \right )+1\! \left ( 6 \right )$
$= -4-2+6$
$= -6+6$
$= 0$
Hence, points are collinear.

Determinants exercise 5.3 question 2 (iii)

Answer: Yes, points are collinear.
Hints: First by using the values of vertices find determinant. If value of determinant is zero, then points are collinear.
$Given\! : \left ( 3,-2 \right ),\left ( 8,8 \right )\: and \: \left ( 5,2 \right )\! .$
$Explanation\! : V\! ertices\; are \left ( 3,-2 \right ),\left ( 8,8 \right )\: and \: \left ( 5,2 \right )\! .$
$Determinant\!= \begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}$
$= \begin{vmatrix} 3 &-2 &1 \\ 8 &8 &1 \\ 5 &2 &1 \end{vmatrix}$
$= 3\! \begin{vmatrix} 8 &1 \\ 2 &1 \end{vmatrix}-\left (-2 \right )\! \begin{vmatrix} 8 &1 \\ 5 &1 \end{vmatrix}+1\! \begin{vmatrix} 8 &8 \\ 5 &2 \end{vmatrix}$
$= 3\! \left (8-2 \right )+2\! \left ( 8-5 \right )+1\! \left ( 16-40 \right )$
$= 3\! \left ( 6 \right )+2\! \left ( 3 \right )+1\! \left ( -24 \right )$
$= 18+6-24$
$= 24-24$
$= 0$
Hence, points are collinear.

Determinants exercise 5.3 question 2 (iv)

Answer: Yes, points are collinear.
Hints: First by using the values of vertices find determinant. If value of determinant is zero, then points are collinear.
$Given\! : \left ( 2,3 \right ),\left ( -1,-2 \right )\: and \: \left ( 5,8 \right )\! .$
$Explanation\! : V\! ertices\; are \left ( 2,3 \right ),\left ( -1,-2 \right )\: and \: \left ( 5,8 \right )\! .$
$Determinant\!= \begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}$
$= \begin{vmatrix} 2 &3 &1 \\ -1 &-2 &1 \\ 5 &8 &1 \end{vmatrix}$
$= 2\! \begin{vmatrix} -2 &1 \\ 8 &1 \end{vmatrix}-3\! \begin{vmatrix} -1 &1 \\ 5 &1 \end{vmatrix}+1\! \begin{vmatrix} -1 &-2 \\ 5 &8 \end{vmatrix}$
$= 2\! \left (-2-8 \right )-3\! \left ( -1-5 \right )+1\! \left ( -8-\left ( -10 \right ) \right )$
$= 2\! \left ( -10 \right )-3\! \left ( -6 \right )+1\! \left ( -8+10 \right )$
$= -20+18+2$
$= -20+20$
$= 0$
Hence, points are collinear.

Determinants exercise 5.3 question 3

Answer: Proved.
Hints: First find determinant and the value of a and b, to prove the given equation.
$\mathbf{Given\! :}\left ( a,0 \right )\! , \left ( 0,b \right )\: and\: \left ( 1,1 \right ) are\; collinear.$
Explanation: Determinant=0
$\Rightarrow \begin{vmatrix} a &0 &1 \\ 0 &b &1 \\ 1 &1 &1 \end{vmatrix}=0$
$\Rightarrow a\! \begin{vmatrix} b &1 \\ 1 &1 \end{vmatrix}-0\! \begin{vmatrix} 0 &1 \\ 1 &1 \end{vmatrix}+1\! \begin{vmatrix} 0 &b \\ 1 &1 \end{vmatrix}=0$
$\Rightarrow a\! \left ( b-1 \right )-0 +1\! \left ( 0-b \right )=0$
$\Rightarrow ab-a-b=0$
$\Rightarrow ab=a+b$
$\Rightarrow a+b=ab$
Hence, proved the given equation.

Determinants exercise 5.3 question 4

Answer: Yes, points are collinear.
Hints: Use the values of vertices and find the determinant. If determinant is zero, points will be collinear.
$\mathbf{Given\! :} V\! ertices\; are \left ( a,b \right )\! ,\left ( a',b' \right ) and \left ( a-a',b-b' \right ) and\; ab' = a'b$
Explanation:
Let,
$x_{1}=a, x_{2}=a', x_{3}=a-a'$
$y_{1}=a, y_{2}=a', y_{3}=b-b'$
$Determinant=\begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}$
$=\begin{vmatrix} a &b &1 \\ a' &b' &1 \\ a-a' &b-b' &1 \end{vmatrix}$
$=a\! \begin{vmatrix} b' &1 \\ b-b' &1 \end{vmatrix}-b\! \begin{vmatrix} a' &1 \\ a-a' &1 \end{vmatrix}+1\! \begin{vmatrix} a' &b' \\ a-a' &b-b' \end{vmatrix}$
$=a\! \left ( b'-\! \left ( b-b' \right ) \right )-b\! \left ( a'-\! \left ( a-a' \right ) \right )+1\! \left [ a'\! \left ( b-b' \right )-b'\! \left ( a-a' \right ) \right ]$
$=a\left(b^{\prime}-b+b^{\prime}\right)-b\left(a^{\prime}-a+a^{\prime}\right)+\left(a^{\prime} b-\not a^{\prime} \phi^{\prime}-b^{\prime} a^{\prime}+\not b^{\prime} \alpha^{\prime}\right)$
$=a\left(2 b^{\prime}-b\right)-b\left(2 a^{\prime}-a\right)+1\left(a^{\prime} b-b^{\prime} a\right)$
$=2 a b^{\prime}-a b-2 a^{\prime} b+a b+a^{\prime} b-b^{\prime} a$
$P u t\; a b^{\prime}=a^{\prime} b$
$\Rightarrow 2 a^{\prime} b-a b-2 a^{\prime} b+a b+a^{\prime} b-a^{\prime} b$
$\Rightarrow 0$
Hence, points are collinear

Determinants exercise 5.3 question 5

$\boldsymbol{Answer\! :} \lambda =-5$
$\boldsymbol{H\! int\! :} I\! f \left ( 1,-5 \right )\! , \left ( -4,5 \right )\; and\; \left ( \lambda ,7 \right )$
$\boldsymbol{Explanation\! :} V\! ertices\; are \left ( 1,-5 \right )\! , \left ( -4,5 \right )\; and\; \left ( \lambda ,7 \right )$
$\text { Determinant }=\left|\begin{array}{ccc} 1 & -5 & 1 \\ -4 & 5 & 1 \\ \lambda & 7 & 1 \end{array}\right|=0$
$\Rightarrow 1\left|\begin{array}{ll} 5 & 1 \\ 7 & 1 \end{array}\right|-(-5)\left|\begin{array}{cc} -4 & 1 \\ \lambda & 1 \end{array}\right|+1\left|\begin{array}{cc} -4 & 5 \\ \lambda & 7 \end{array}\right|=0$
$\Rightarrow 1(5-7)+5(-4-\lambda)+1(-28-5 \lambda)=0$
$\Rightarrow -2-20-5\lambda -28-5\lambda =0$
$\Rightarrow -10\lambda -50=0$
$\Rightarrow -10\lambda =50$
$\Rightarrow \lambda =\frac{50}{-10}$
$\Rightarrow \lambda =-5$

Determinants exercise 5.3 question 6

Answer:$x$ = -2
Hints: Using determinant, find area of triangle and find value of $x$.

$\boldsymbol{Given\: \! :} \left ( x,4 \right )\! , \left ( 2,-6 \right )\; and\; \left ( 5,4 \right ).$
area of triangle = 35 sq. Units.
$\boldsymbol{Explanation:} V\! ertices\; are \left ( x,4 \right )\! , \left ( 2,-6 \right )\; and\; \left ( 5,4 \right ).$
$\text { Area of triangle is }=\Delta=\frac{1}{2}\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|$
$\Rightarrow \Delta=\frac{1}{2}\left|\begin{array}{lll} x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \end{array}\right|=35$
$\Delta =\frac{1}{2}\left ( x\! \begin{vmatrix} -6 & 1\\ 4 & 1 \end{vmatrix}-4\! \begin{vmatrix} 2 & 1\\ 5 & 1 \end{vmatrix}+1\! \begin{vmatrix} 2 & -6\\ 5 & 4 \end{vmatrix} \right )=35$
$\Rightarrow \frac{1}{2}[x(-6-4)-4(2-5)+1(8-(-30))]=35$
$\Rightarrow \frac{1}{2}[x(-10)-4(-3)+1(38)]=35$
$\Rightarrow \frac{1}{2}[-10x+12+38]=35$
$\Rightarrow -10x+50=70$
$\Rightarrow -10x=70-50$
$\Rightarrow x=-2$

Determinants exercise 5.3 question 7

$\boldsymbol{Answer\! : }\frac{+13}{2} sq. units.\;\; N\! o, \; points\; are\; not\; collinear.$
Hints: First find the determinant and then area of the triangle. If result found be zero, then given points are collinear.

Given: Vertices are (1,4), (2,3) and (-5, -3).

Explanation: Vertices are (1,4), (2,3) and (-5, -3).

$\text { Area of triangle is }=\Delta=\frac{1}{2}\left|\begin{array}{lll} 1 & 4 & 1 \\ 2 & 3 & 1 \\ -5 & -3 & 1 \end{array}\right|$

$\Rightarrow|\Delta|=\frac{1}{2}\left(1\left|\begin{array}{cc} 3 & 1 \\ -3 & 1 \end{array}\right|-4\left|\begin{array}{cc} 2 & 1 \\ -5 & 1 \end{array}\right|+1\left|\begin{array}{cc} 2 & 3 \\ -5 & -3 \end{array}\right|\right)$

$\Rightarrow \frac{1}{2}[1(3-(-3))-4(2-(-5))+1(-6-(-15))]$

$\Rightarrow \frac{1}{2}[1(6)-4(7)+1(9)]$

$\Rightarrow \frac{1}{2}[6-28+9]$

$\Rightarrow \frac{1}{2}\times (-13)$

$\Rightarrow \left | \frac{-13}{2} \right | sq. units \Rightarrow \frac{13}{2} sq. units$
and determinant is not zero, that is why points are not collinear.

Determinants exercise 5.3 question 8

Answer: 46 sq. units
Hints: To find area of triangle use modulus in the formula of determinant.
Given: (-3,5), (3, -6) and (7,2).
Explanation: Vertices are (-3,5), (3, -6) and (7,2).
$\text { Area of triangle is } \Rightarrow|\Delta|=\left | \frac{1}{2}\begin{vmatrix} -3 &5 &1 \\ 3 &-6 &1 \\ 7 &2 &1 \end{vmatrix} \right |$
$=\left | \frac{1}{2}\left(-3\left|\begin{array}{cc} -6 & 1 \\ 2 & 1 \end{array}\right|-5\left|\begin{array}{cc} 3 & 1 \\ 7 & 1 \end{array}\right|+1\left|\begin{array}{cc} 3 & -6 \\ 7 & 2 \end{array}\right|\right) \right |$
$=\left | \frac{1}{2}[-3(-6-2)-5(3-7)+1(6-(-42))] \right |$
$=\left | \frac{1}{2}[-3(-8)-5(-4)+1(48)] \right |$
$\Rightarrow \frac{1}{2}\left | 92 \right |$
$\Rightarrow 46sq. units$

Determinants exercise 5.3 question 9

$\boldsymbol{Answer\! :} k=-1,\frac{1}{2}$
Hints: If points are collinear then determinant will be zero.
$\boldsymbol{Given\! :} (k,2-2k), (-k+1, 2k), (-4-k, 6-2k)$
$\boldsymbol{Explanation\! :} V\! ertices\; are\; (k,2-2k), (-k+1, 2k), (-4-k, 6-2k)$
$\text { Determinant}=\left|\begin{array}{lll} \; \; \; \; \; \; k & 2-2k & 1 \\ -k+1 & \; \; \; 2k & 1 \\ -4-k & 6-2k & 1 \end{array}\right|=0$
Expanding along row1.
$\Rightarrow k(2k-(6-2k))-(2-2k)(-k+1-(-4-k))+1[(-k+1)(6-2k)-2k(-4-k)]$
$\Rightarrow [2k^{2}-6k+2k^{2}]-(2-2k)(-k+1+4+k)+[-6k+6+2k^{2}-2k]+8k+2k^{2}$
$\Rightarrow [2k^{2}-6k+2k^{2}]-[(2-2k)(5)+(6+4k^{2})]=0$
$\Rightarrow [4k^{2}-6k-10+10k+6+4k^{2}]=0$
$\Rightarrow [8k^{2}-4+4k]=0$
$\Rightarrow 2k^{2}+k-1=0$
$\Rightarrow 2k^{2}+2k-k-1=0$
$\Rightarrow 2k(k+1)-1(k+1)=0$
$\Rightarrow (2k-1)(k+1)=0$
$\begin{bmatrix} k+1 &=0 \\ k &=-1 \end{bmatrix}$ $\begin{bmatrix} 2k-1&=0 \\ k&=\frac{1}{2} \end{bmatrix}$
$k=-1,\frac{1}{2}$

Determinants exercise 5.3 question 10

Answer:$x$=3
Hints: Use vertices in the determinant formula and equate it with zero because if points are collinear then determinant will be zero and then find the value of x.
Given: ($x$, -2), (5,2) and (8,8) are collinear.
Explanation: Vertices are ($x$, -2), (5,2) and (8,8)
$\text { Determinant }=\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|=0$
$\Rightarrow \left|\begin{array}{lll} x & -2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \end{array}\right|=0$
$\Rightarrow x\left|\begin{array}{ll} 2 & 1 \\ 8 & 1 \end{array}\right|-(-2)\left|\begin{array}{cc} 5 & 1 \\ 8 & 1 \end{array}\right|+1\left|\begin{array}{cc} 5 & 2 \\ 8 & 8 \end{array}\right|=0$
$\Rightarrow x(2-8)+2(5-8)+1(40-16)=0$
$\Rightarrow x(-6)+2(-3)+1(24)=0$
$\Rightarrow -6x-6+24=0$
$\Rightarrow -6x+18=0$
$\Rightarrow 6x=18$
$\Rightarrow x=3$

Determinants exercise 5.3 question 11

Answer:$x$=5
Hints: Use vertices in the determinant formula and equate it with zero because if points are collinear then determinant will be zero and then find the value of $x$.
Given: (3,-2), ($x$,2) and (8,8).
Explanation: Vertices are (3,-2), ($x$,2) and (8,8).
$\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|=0$
$\Rightarrow \left|\begin{array}{lll} 3 & -2 & 1 \\ x & 2 & 1 \\ 8 & 8 & 1 \end{array}\right|=0$
$\Rightarrow 3\left|\begin{array}{ll} 2 & 1 \\ 8 & 1 \end{array}\right|-(-2)\left|\begin{array}{cc} x & 1 \\ 8 & 1 \end{array}\right|+1\left|\begin{array}{cc} x & 2 \\ 8 & 8 \end{array}\right|=0$
$\Rightarrow 3(2-8)+2(x-8)+1(8x-16)=0$
$\Rightarrow 3(-6)+2(x-8)+8x-16=0$
$\Rightarrow -18+2x-16+8x-16=0$
$\Rightarrow 10x-50=0$
$\Rightarrow x=\frac{50}{10}$
$\Rightarrow x=5$

Determinants exercise 5.3 question 12 (i)

$\boldsymbol{Answer\! :} 2x=y$
$\boldsymbol{Hints\! :} Take\; the\; third\; point\; as\; (x,y)\; and\;\; use\; the\; f\! ormula\; o\! f\; area\; o\! f\; triangle\; and\; equate\; it\; with\; zero.$
Given: (1,2) and (3,6)
Explanation: Vertices are (1,2) and (3,6)
$Let\; the\; third\; point\; be\; (x,y)$
$Determinant= \Delta =0$
$\Delta =\frac{1}{2}\begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}=0$
$\Rightarrow \frac{1}{2}\begin{vmatrix} x &y &1 \\ 1 &2 &1 \\ 3 &6 &1 \end{vmatrix}=0$
$\Rightarrow x\left|\begin{array}{ll} 2 & 1 \\ 6 & 1 \end{array}\right|-y\left|\begin{array}{cc} 1 & 1 \\ 3 & 1 \end{array}\right|+1\left|\begin{array}{cc} 1 & 2 \\ 3 & 6 \end{array}\right|=0$
$\Rightarrow x(2-6)-y(1-3)+1(6-6)=0$
$\Rightarrow -4x+2y+0=0$
$\Rightarrow -4x+2y=0$
$\Rightarrow -2x+y=0$
$\Rightarrow 2x=y$

Determinants exercise 5.3 question 12 (ii)

$\dpi{100} \boldsymbol{Answer\! :} x=3y$
$\boldsymbol{Hints\! :} Take\; the\; third\; point\; as\; (x,y)\; and\;\; use\; the\; f\! ormula\; o\! f\; area\; o\! f\; triangle\; and\; equate\; it\; with\; zero.$
Given: (3,1) and (9,3)
Explanation: Vertices are (3,1) and (9,3)
$Let\; the\; third\; point\; be\; (x,y)$
$Determinant= \Delta =0$
$\Delta =\frac{1}{2}\begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}=0$
$\Rightarrow \frac{1}{2}\begin{vmatrix} x &y &1 \\ 3 &1 &1 \\ 9 &3 &1 \end{vmatrix}=0$
$\Rightarrow x\left|\begin{array}{ll} 1 & 1 \\ 3 & 1 \end{array}\right|-y\left|\begin{array}{cc} 3 & 1 \\ 9 & 1 \end{array}\right|+1\left|\begin{array}{cc} 3 & 1 \\ 9 & 3 \end{array}\right|=0$
$\Rightarrow x(1-3)-y(3-9)+1(9-9)=0$
$\Rightarrow -2x+6y=0$
$\Rightarrow -x+3y=0$
$\Rightarrow x=3y$

Determinants exercise 5.3 question 13 (i)

Answer:$k$=0
Hints: Putting the values of vertices, in the formula of area of triangle.
Given: ($k$,0), (4,0) and (0,2)
area of triangle= 4 sq. units
Explanation: Vertices are ($k$,0), (4,0) and (0,2)
$\text { Area of triangle is }=\Delta=\frac{1}{2}\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|=4$
$\Rightarrow \Delta=\frac{1}{2}\left|\begin{array}{lll} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array}\right|=4$
$\Delta=\frac{1}{2}\left(k\left|\begin{array}{cc} 0 & 1 \\ 2 & 1 \end{array}\right|-0\left|\begin{array}{cc} 4 & 1 \\ 0 & 1 \end{array}\right|+1\left|\begin{array}{cc} 4 & 0 \\ 0 & 2 \end{array}\right|\right)=4$
$= \frac{1}{2}[k(0-2)-0+1(8-0)]=4$
$\Rightarrow -2k+8=8$
$\Rightarrow -2k=0$
$\Rightarrow k=0$

Determinants exercise 5.3 question 13 (ii)

Answer:$k$=8
Hints: Putting the values of vertices, in the formula of area of triangle.
Given: (-2,0), (0,4) and (0,$k$)
area of triangle= 4 sq. units
Explanation: Vertices are (-2,0), (0,4) and (0,$k$)
$\text { Area of triangle is }=\Delta=\frac{1}{2}\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|=4$
$\Rightarrow \Delta=\frac{1}{2}\left|\begin{array}{lll} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right|=4$
$\Delta=\frac{1}{2}\left(-2\left|\begin{array}{cc} 4 & 1 \\ k & 1 \end{array}\right|-0\left|\begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array}\right|+1\left|\begin{array}{cc} 0 & 4 \\ 0 & k \end{array}\right|\right)=4$
$= \frac{1}{2}[-2(4-k)-0+1(0)]=4$
$\Rightarrow -8+2k=8$
$\Rightarrow 2k=16$
$\Rightarrow k=8$