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

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

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 07:12 PM IST

The CBSE board students are recommended to use the RD Sharma solutions books for their references. This is due to the high standard answers provided in the book. It makes the students less dependent on the tutors to clear their silly doubts. When it comes to mathematics, there are high chances that students get more doubts while solving it. Significantly, the students who find it hard to cope with the Determinants portions can use the RD Sharma Class 12th exercise 5.3 solution books.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 5 Determinants - Other Exercise

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)

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)

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)

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)

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

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

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.

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)

$\dpi{100} \boldsymbol{Answer\! :} 2x=y$
$\dpi{100} \boldsymbol{Hints\! :} Take\; the\; third\; point\; as\; (x,y)\; and\;\; use\; the\; f\! ormula\; o\! f\; area\; o\! f\; triangle\; and\; equate\; it\; with\; zero.$
Given: (1,2) and (3,6)
Explanation: Vertices are (1,2) and (3,6)
$\dpi{100} Let\; the\; third\; point\; be\; (x,y)$
$\dpi{100} Determinant= \Delta =0$
$\dpi{100} \Delta =\frac{1}{2}\begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}=0$
$\dpi{100} \Rightarrow \frac{1}{2}\begin{vmatrix} x &y &1 \\ 1 &2 &1 \\ 3 &6 &1 \end{vmatrix}=0$
$\dpi{100} \Rightarrow x\left|\begin{array}{ll} 2 & 1 \\ 6 & 1 \end{array}\right|-y\left|\begin{array}{cc} 1 & 1 \\ 3 & 1 \end{array}\right|+1\left|\begin{array}{cc} 1 & 2 \\ 3 & 6 \end{array}\right|=0$
$\dpi{100} \Rightarrow x(2-6)-y(1-3)+1(6-6)=0$
$\dpi{100} \Rightarrow -4x+2y+0=0$
$\dpi{100} \Rightarrow -4x+2y=0$
$\dpi{100} \Rightarrow -2x+y=0$
$\dpi{100} \Rightarrow 2x=y$

Determinants exercise 5.3 question 12 (ii)

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

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,$k$)
$\text { Area of triangle is }=\Delta=\frac{1}{2}\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|=4$
$\Rightarrow \Delta=\frac{1}{2}\left|\begin{array}{lll} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right|=4$
$\Delta=\frac{1}{2}\left(-2\left|\begin{array}{cc} 4 & 1 \\ k & 1 \end{array}\right|-0\left|\begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array}\right|+1\left|\begin{array}{cc} 0 & 4 \\ 0 & k \end{array}\right|\right)=4$
$= \frac{1}{2}[-2(4-k)-0+1(0)]=4$
$\Rightarrow -8+2k=8$
$\Rightarrow 2k=16$
$\Rightarrow k=8$

The 5th chapter for Class 12 mathematics, Determinants, is a simple chapter where many students aim to score their best. RD Sharma class 12th exercise 5.3, has around 21 questions to be solved, including its subparts. The concepts in this exercise are Area of the Triangle with vertices at given points, Collinear points, and the Equation of lines using Determinants. All these can be referred to at the RD Sharma Class 12 Chapter 5 Exercise 5.3.

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