RD Sharma Class 12 Exercise 5.3 Determinants Solutions Maths

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.

Latest: JEE Main: high scoring chapters | Past 10 year's papers

Don't Miss: Most scoring concepts for NEET | NEET papers with solutions

New: Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now

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)

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

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.

All the solutions in this book are created and verified by the experts according to the latest syllabus. It follows the NCERT pattern too. Hence, the CBSE board students can use it to complete their homework and assignments and prepare for their tests. Scoring better marks in the Determinants chapter will boost up your total score. Therefore, the RD Sharma Chapter 5 Exercise 5.3 Solution will make you understand the concepts effortlessly. Numerous practice questions will make you furthermore clear about exercise 5.3.

You might think that RD Sharma solutions book would cost a lot by looking at its benefits. But that is not the case here. The RD Sharma Class 12 Solutions Determinants Ex 5.3 can be downloaded free from top educational websites like Career 360. RD Sharma Class 12th exercise 5.3 materials are available for free of cost on this website. So, all you need to do is, visit the Career 360 website, look out for the solution book you need and download it to your device. Then, start practicing the sums from day 1 to achieve more marks in your public exam.

Once you commence preparing for your tests using the RD Sharma solutions books, you can witness yourself crossing all your benchmarks. Many students have benefited from the RD Sharma Class 12 Solutions Chapter 5 ex 5.3 to know the concepts of Determinants better. There are high chances that questions in the public exam can be asked from the RD Sharma solution books. Hence, preparing with this material in advance leads you to score better marks in the examinations easily.

Chapter-wise RD Sharma Class 12 Solutions

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

Frequently Asked Question (FAQs)

1. Where can I find the RD Sharma solution books?

The RD Sharma solution books can be found on the Career360 website. In addition, you can download the PDF file for your reference.

2. Where can I clear my doubts regarding the concept of the determinant?

If you have doubts while solving the sums in the determinants chapter, use the RD Sharma Class 12th exercise 5.3 solution books. A clear explanation of each concept in this book and the answer key for every question are given.

3. Can a slow-learner student understand the concepts in the RD Sharma books?

The solutions in this book are provided in various methods, the elaborated ones, and the shortcut methods. Hence, any category of students can use the RD Sharma solution books as their reference guide.

4. How much does the Class 12 RD Sharma Chapter 5 Exercise 5.3 Solution books cost?

The RD Sharma Class 12th exercise 5.3 solution books can be downloaded for free from the Career360 website. So, you do not pay even a single penny to access this best resource.

5. Is it enough to use the RD Sharma boot to score well in the Determinants chapter?

The RD Sharma Class 12 Chapter 5 Exercise 5.3 is the best solutions guide that provides a deeper understanding of the concept of the determinant. Hence it is enough to help you achieve better scores in the public exam.