Class 12 RD Sharma Solutions Chapter 6 Determinants Exercise 6.3 (original) (raw)

Last Updated : 21 Aug, 2024

**Question 1. Find the area of the triangle with vertices at the points:

****(i) (3, 8), (−4, 2) and (5, −1)**

**Solution:

Given (3, 8), (−4, 2) and (5, −1) are the vertices of the triangle.

We know that, if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

A = \frac{1}{2} \left| \begin{array}{cc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{array} \right|

= \frac{1}{2} \left| \begin{array}{cc} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & -1 & 1 \\ \end{array} \right|

= \frac{1}{2}[3(2+1)-8(-4-5)+1(4-10)]

= \frac{1}{2}[9+72-6]

= \frac{75}{2}

**Therefore, area of the triangle is \frac{75}{2} sq. units.

****(ii) (2, 7), (1, 1) and (10, 8)**

**Solution:

Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle.

The area of the triangle is given by,

A = \frac{1}{2} \left| \begin{array}{cc} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \\ \end{array} \right|

= \frac{1}{2}[2(1-8)-7(1-10)+1(8-10)]

= \frac{1}{2}[-14+63-2]

= \frac{47}{2}

**Therefore, area of the triangle is \frac{47}{2} sq. units.

****(iii) (−1, −8), (−2, −3) and (3, 2)**

**Solution:

Given (−1, −8), (−2, −3) and (3, 2) are the vertices of the triangle.

The area of the triangle is given by,

A = \frac{1}{2} \left| \begin{array}{cc} -1 & -8 & 1 \\ -2 & -3 & 1 \\ 3 & 2 & 1 \\ \end{array} \right|

= \frac{1}{2}|-1(-3-2)+8(-2-3)+1(-4+9)|

= \frac{1}{2}|5-40+5|

= \frac{30}{2}

= 15

**Therefore, area of the triangle is 15 sq. units.

****(iv) (0, 0), (6, 0), (4, 3)**

**Solution:

Given (0, 0), (6, 0), (4, 3) are the vertices of the triangle.

The area of the triangle is given by,

A = \frac{1}{2} \left| \begin{array}{cc} 0 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \\ \end{array} \right|

= \frac{1}{2}[0-0+1(18-0)]

= \frac{18}{2}

= 9

**Therefore, area of the triangle is 9 sq. units.

**Question 2. Using the determinants show that the following points are collinear:

****(i) (5, 5), (−5, 1) and (10, 7)**

**Solution:

Given points are (5, 5), (−5, 1) and (10, 7).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

A = \frac{1}{2} \left| \begin{array}{cc} 5 & 5 & 1 \\ -5 & 1 & 1 \\ 10 & 7 & 1 \\ \end{array} \right|

= \frac{1}{2}[5(1-7)-5(-5-10)+1(-35-10)]

= \frac{1}{2}[-30+75-45]

= 0

As the area of the triangle is 0, the points are collinear.

**Hence proved.

****(ii) (1, −1), (2, 1) and (4, 5)**

**Solution:

Given points are (1, −1), (2, 1) and (10, 8).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

A = \frac{1}{2} \left| \begin{array}{cc} 1 & -1 & 1 \\ 2 & 1 & 1 \\ 4 & 5 & 1 \\ \end{array} \right|

= \frac{1}{2}[1(1-5)+1(2-4)+1(10-4)]

= \frac{1}{2}[-4-2+6]

= 0

As the area of the triangle is 0, the points are collinear.

**Hence proved.

****(iii) (3, −2), (8, 8) and (5, 2)**

**Solution:

Given points are (3, −2), (8, 8) and (5, 2).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

A = \frac{1}{2} \left| \begin{array}{cc} 3 & -2 & 1 \\ 8 & 8 & 1 \\ 5 & 2 & 1 \\ \end{array} \right|

= \frac{1}{2}[3(8-2)+2(8-5)+1(16-40)]

= \frac{1}{2}[18+6-24]

= 0

As the area of the triangle is 0, the points are collinear.

**Hence proved.

****(iv) (2, 3), (−1, −2) and (5, 8)**

**Solution:

Given points are (2, 3), (−1, −2) and (5, 8).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

A = \frac{1}{2} \left| \begin{array}{cc} 2 & 3 & 1 \\ -1 & -2 & 1 \\ 5 & 8 & 1 \\ \end{array} \right|

= \frac{1}{2}[2(-2-8)-3(-1-5)+1(-8+10)]

= \frac{1}{2}[-20+18+2]

= 0

As the area of the triangle is 0, the points are collinear.

**Hence proved.

**Question 3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.

**Solution:

Given points (a, 0), (0, b) and (1, 1) are collinear.

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

=> \frac{1}{2} \left| \begin{array}{cc} a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1 \\ \end{array} \right|=0

=> \frac{1}{2}[a(b-1)-0+1(0-b)]=0

=> ab − a − b = 0

=> a + b = ab

**Hence proved.

**Question 4. Using the determinants prove that the points (a, b), (a’, b’), and (a – a’, b – b) are collinear if a b’ = a’ b.

**Solution:

Given points are (a, b), (a’, b’) and (a – a’, b – b) and a b’ = a’ b.

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

=> \frac{1}{2} \left| \begin{array}{cc} a & b & 1 \\ a’ & b’ & 1 \\ a – a’ & b – b’ & 1 \\ \end{array} \right|=0

=> \frac{1}{2}[a(b’-b+b’)-b(a’-a+a’)+1(a’b-a’b’-ab’+a’b’)]=0

=> \frac{1}{2}[ab’-ab+ab’-a’b+ab-a’b+a’b-ab’)]=0

=> a b’ − a’ b = 0

=> a b’ = a’ b

**Hence proved.

**Question 5. Find the value of λ so that the points (1, −5), (−4, 5), and (λ, 7) are collinear.

**Solution:

Given points are (1, −5), (−4, 5) and (λ, 7).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

=> \frac{1}{2} \left| \begin{array}{cc} 1 & -5 & 1 \\ -4 & 5 & 1 \\ λ & 7 & 1 \\ \end{array} \right|=0

=> \frac{1}{2}[1(5-7)+5(-4-λ)+1(-28-5λ)]=0

=> −2 − 20 − 5λ − 28 − 5λ = 0

=> 10λ = 50

=> λ = 5

**Therefore, the value of λ is 5.

**Question 6. Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6), and (5, 4).

**Solution:

Given points are (x, 4), (2, −6) and (5, 4).

We know that, if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,

A = \frac{1}{2} \left| \begin{array}{cc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{array} \right|

=> \frac{1}{2} \left| \begin{array}{cc} x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \\ \end{array} \right| = 35

=> \left| \begin{array}{cc} x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \\ \end{array} \right| = \pm70

=> − 10x + 12 + 38 = ±70

=> – 10x + 50 = ±70

Taking positive sign, we get

=> – 10x + 50 = 70

=> 10x = – 20

=> x = – 2

Taking negative sign, we get

=> – 10x + 50 = – 70

=> 10x = 120

=> x = 12

**Therefore, the value of x is 12 or –2.

**Question 7. Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3), (–5, –3). Are the given points collinear?

**Solution:

Given points are (1, 4), (2, 3), (–5, –3).

So, area = \frac{1}{2} \left| \begin{array}{cc} 1 & 4 & 1 \\ 2 & 3 & 1 \\ -5 & -3 & 1 \\ \end{array} \right|

= \frac{1}{2}|1(3+3)-4(2+5)+1(-6+15)|

= \frac{1}{2}|6-28+9|

= \frac{13}{2}

**As the area of the triangle formed by these three points is not zero, the points are not collinear.

**Question 8. Using determinants, find the area of the triangle with vertices (–3, 5), (3, –6), (7, 2).

Given points are (–3, 5), (3, –6), (7, 2).

So, area = \frac{1}{2} \left| \begin{array}{cc} -3 & 5 & 1 \\ 3 & -6 & 1 \\ 7 & 2 & 1 \\ \end{array} \right|

= \frac{1}{2}|-3(-6-2)-5(3-7)+1(6+42)|

= \frac{1}{2}|24+20+48|

= \frac{92}{2}

= 46

**Therefore, the area of the triangle is 46 sq. units.

**Question 9. Using determinants, find the value of k so that the points (k, 2–2k), (–k+1, 2k), (–4–k, 6–2k) may be collinear.

**Solution:

Given points are (k, 2–2k), (–k+1, 2k), (–4–k, 6–2k). As the points are collinear, the area must be 0.

=> \frac{1}{2} \left| \begin{array}{cc} k & 2–2k & 1 \\ –k+1 & 2k & 1 \\ –4–k & 6–2k & 1 \\ \end{array} \right|=0

=> k(2k – 6 + 2k) – (2–2k) (–k + 1 + 4 + k) + [(1–k) (6–2k) – 2k (–4–k)] = 0

=> 8k2 + 4k – 4 = 0

=> 8k2 + 8k – 4k – 4 = 0

=> 8k (k+1) – 4 (k+1) = 0

=> 8k = 4 or k = –1

=> k = 1/2 or k = –1

**Therefore, the value of k is 1/2 or –1.

**Question 10. If the points (x, –2), (5, 2), (8, 8) are collinear, find x using determinants.

**Solution:

Given points are (x, –2), (5, 2), (8, 8). As the points are collinear, the area must be 0.

=> \frac{1}{2} \left| \begin{array}{cc} x & -2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \\ \end{array} \right|=0

=> x(2 – 8) + 2(5 – 8) + 1(40–16) = 0

=> –6x – 6 + 24 = 0

=> 6x = 18

=> x = 3

**Therefore, the value of x is 3.

**Question 11. If the points (3, –2), (x, 2), (8, 8) are collinear, find x using determinants.

**Solution:

Given points are (3, –2), (x, 2), (8, 8). As the points are collinear, the area must be 0.

=> \frac{1}{2} \left| \begin{array}{cc} 3 & -2 & 1 \\ x & 2 & 1 \\ 8 & 8 & 1 \\ \end{array} \right|=0

=> 3(2 – 8) + 2(x – 8) + 1(8x–16) = 0

=> –18 + 2x – 16 + 8x – 16 = 0

=> 10x = 50

=> x = 5

**Therefore, the value of x is 5.

**Question 12. Using determinants, find the equation of

****(i) the line joining the points (1, 2) and (3, 6).**

**Solution:

Let (x, y), (1, 2), (3, 6) be the points on the line. As these points are collinear, we get,

=> \frac{1}{2} \left| \begin{array}{cc} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \\ \end{array} \right|=0

=> \frac{1}{2}[x(2-6)-y(1-3)+1(6-6)]=0

=> −4x + 2y = 0

=> 2x − y = 0

**Therefore, the required equation is 2x − y = 0.

****(ii) the line joining the points (3, 1) and (9, 3).**

**Solution:

Let (x, y), (3, 1), (9, 3) be the points on the line. As these points are collinear, we get,

=> \frac{1}{2} \left| \begin{array}{cc} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \\ \end{array} \right|=0

=> \frac{1}{2}[x(1-3)-y(3-6)+1(9-9)]=0

=> −2x + 6y = 0

=> x − 3y = 0

**Therefore, the required equation is x − 3y = 0.

**Question 13. Find the values of k

****(i) if area of triangle whose vertices are (k, 0) (4, 0) (0, 2) is 4 square units.**

**Solution:

Given points are (k, 0) (4, 0) (0, 2). According to the question, we have,

=> \frac{1}{2} \left| \begin{array}{cc} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \\ \end{array} \right|=4

=> |k(0−2) − 0 + 1(8−0)| = 8

=> −2k + 8 = ±8

=> −2k + 8 = 8 or −2k + 8 = −8

=> k = 0 or k = 8

**Therefore, the value of k is 0 or 8.

****(ii) if area of triangle whose vertices are (−2, 0) (0, 4) (0, k) is 4 square units.**

**Solution:

Given points are (−2, 0) (0, 4) (0, k). According to the question, we have,

=> \frac{1}{2} \left| \begin{array}{cc} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \\ \end{array} \right|=4

=> |−2(4−k) − 0 + 1(0)| = 8

=> −8 + 2k = ±8

=> −8 + 2k = 8 or −8 + 2k = −8

=> k = 8 or k = 0

**Therefore, the value of k is 0 or 8.

Summary

Chapter 6, Exercise 6.3 typically focuses on the application of determinants in solving systems of linear equations. The main topics covered usually include:

This exercise emphasizes the practical application of determinants in solving real-world problems and understanding the nature of solutions to systems of equations.