Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.3

Last Updated : 01 May, 2024

Question 1. Find area of the triangle with vertices at the point given in each of the following :

(i) (1, 0), (6, 0), (4, 3)

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

(iii) (–2, –3), (3, 2), (–1, –8)

Solution:

(i) (1, 0), (6, 0), (4, 3)

Area\ of\ triangle=\frac{1}{2}\begin{vmatrix}1 & 0 & 1\\6&0&1\\4&3&1\end{vmatrix}\\=\frac{1}{2}[1(0-3)-0(6-4)+1(18-0)]\\=\frac{15}{2}square\ units

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

Area\ of\ triangle=\frac{1}{2}\begin{vmatrix}2 & 7 & 1\\1&1&1\\10&8&1\end{vmatrix}\\=\frac{1}{2}[2(1-8)-7(1-10)+1(8-10)]\\=\frac{47}{2}square\ units

(iii) (–2, –3), (3, 2), (–1, –8)

Area\ of\ triangle=\frac{1}{2}\begin{vmatrix}-2 & -3 & 1\\3&2&1\\-1&-8&1\end{vmatrix}\\=\frac{1}{2}[-2(10)+3(4)-22]\\=15\ square\ units

Question 2. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Solution:

If the area of the triangle is equal to zero then the points are collinear.

For example:- Area of triangle = 0

Area\ of\ triangle=\frac{1}{2}\begin{vmatrix}a&b+c&1\\b&c+a&1\\c&a+b&1\end{vmatrix}\\ =\frac{1}{2}[a(c+a-a-b)-(b+c)(b-c)+1\{b(a+b)-c(c+a)\}]\\=\frac{1}{2}(ac-ab-b^2+c^2+ab+b^2-c^2-ac)\\0\\Hence,\ the\ points\ are\ collinear\ as\ the\ area\ of\ triangle\ is\ equal\ to\ zero

Question 3. Find the values of k if area of triangle is 4 sq. units and vertices are

(i) (k, 0), (4, 0), (0, 2)
(ii)(-2, 0), (0, 4), (0, k)

Solution:

(i) (k, 0), (4, 0), (0, 2)

Given: Area of triangle = ±4 sq. units

i.e. \frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=4\\\frac{1}{2}[k(0-2)-0+1(8-0)]=4\\\frac{1}{2}(-2k+4)=4\\-k+4=4\\Here, -k+4=\pm4\\if,\\-k+4=4\ and-k+4=-4\\k=0\ and\ k=8

(ii) (-2, 0), (0, 4), (0, k)

\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=4\\\frac{1}{2}\begin{vmatrix}-2&0&1\\0&4&1\\0&k&1\end{vmatrix}\\\frac{1}{2}(-8+2k)=4\\-k+4=4\\Here, -k+4=\pm4\\if,\\-k+4=4\ and-k+4=-4\\k=0\ and\ k=8

Question 4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Solution:

(i) Find equation of line joining (1, 2) and (3, 6) using determinants.

Let us assume,

A(x, y) be any vertex of a triangle

All points are on one line (collinear) if the area of triangle is zero.

\frac{1}{2}\begin{vmatrix}1&2&1\\3&6&1\\x&y&1\end{vmatrix}=0\\\frac{1}{2}[x(2-6)-y(1-3)+1(6-6)]=0\\-4x+2y=0\\y=2x\\Which\ is\ equation\ of\ a\ line.

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Let us assume,

A(x, y) be any vertex of a triangle

All points are on one line (collinear) if the area of triangle is zero.

\frac{1}{2}\begin{vmatrix}x&y&1\\3&1&1\\9&3&1\end{vmatrix}\\\frac{1}{2}[x(1-3)-y(3-9)+1(9-9)]=0\\-2x+6y=0\\x-3y=0\\It\ is\ a\ equation\ of\ line.

Question 5. If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2

Solution:

(D) is the correct option.

As:

\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=35\\\frac{1}{2}[2(4-4)-(-6)(5-k)+1(20-4k)=35\\By\ solving\ we\ will\ get\\25-5k=\pm35\\25-5k=35\ and\ 25-5k=-35\\k=-2\ and\ k=12.

Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.4

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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.3

Question 1. Find area of the triangle with vertices at the point given in each of the following : (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (–2, –3), (3, 2), (–1, –8)

Question 2. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Question 3. Find the values of k if area of triangle is 4 sq. units and vertices are

(i) (k, 0), (4, 0), (0, 2) (ii)(-2, 0), (0, 4), (0, k)

Question 4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants. (ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Question 5. If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is (A) 12 (B) –2 (C) –12, –2 (D) 12, –2

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Question 1. Find area of the triangle with vertices at the point given in each of the following :

(i) (1, 0), (6, 0), (4, 3)

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

(iii) (–2, –3), (3, 2), (–1, –8)

(i) (k, 0), (4, 0), (0, 2)
(ii)(-2, 0), (0, 4), (0, k)

Question 4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Question 5. If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2