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**NCERT mathematics class 12 chapter Determinants Exercise 4.3 Solutions**. You can find the numerical questions solutions for the**chapter 4/Exercise 4.3**of**NCERT class 12 mathematics**in this page. So is the case if you are looking for**NCERT class 12 Maths**related topic**Determinants Exercise 4.3**solutions. This page contains Exercise 4.3 solutions. If you’re looking for summary of the chapter or other exercise solutions, they’re available at●

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Exercise 4.3 Solutions

1. Find area of the triangle with vertices at the poing 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) To find the area of the triangle with vertices at the points (1, 0), (6, 0), (4, 3)

We know that the area of a triangle whose vertices are {(x_1, y_1),} {(x_2, y_2)} and {(x_3, y_3)} can be expressed as a determinant as

{Δ = \dfrac12 × \left|\begin{array}{ccc} x_1 & y_1 & 1 \\[5pt] x_2 & y_2 & 1 \\[5pt] x_3 & y_3 & 1 \end{array}\right|}

So, the area of the given triangle can be expressed as

{Δ = \dfrac12 × \left|\begin{array}{ccc} 1 & 0 & 1 \\[5pt] 6 & 0 & 1 \\[5pt] 4 & 3 & 1 \end{array}\right|}

Expanding along {\text{C}_2,} we have,

Δ

{= \dfrac12 × \left(-0 \left|\begin{array}{cc} 6 & 1 \\[5pt] 4 & 1 \end{array}\right| + 0 \left|\begin{array}{cc} 1 & 1 \\[5pt] 4 & 1 \end{array}\right| - 1 \left|\begin{array}{cc} 1 & 1 \\[5pt] 6 & 1 \end{array}\right|\right)}

{= \dfrac12 × \left[-0 + 0 - 3\left[1(1) - 1(6)\right]\right]}

{= \dfrac12 × \left[-3(1 - 6)\right]}

{= \dfrac12 × (-3) × (-5)}

{= \dfrac12 × 15}

{= \dfrac{15}{2}} sq. units.

∴ The area of the triangle with vertices at the points (1, 0), (6, 0), (4, 3) is \dfrac{15}{2} sq. units.

(ii) To find the area of the triangle with vertices at the points (2, 7), (1, 1), (10, 8)

We know that the area of a triangle whose vertices are {(x_1, y_1),} {(x_2, y_2)} and {(x_3, y_3)} can be expressed as a determinant as

{Δ = \dfrac12 × \left|\begin{array}{ccc} x_1 & y_1 & 1 \\[5pt] x_2 & y_2 & 1 \\[5pt] x_3 & y_3 & 1 \end{array}\right|}

So, the area of the given triangle can be expressed as

{Δ = \dfrac12 × \left|\begin{array}{ccc} 2 & 7 & 1 \\[5pt] 1 & 1 & 1 \\[5pt] 10 & 8 & 1 \end{array}\right|}

Expanding along {\text{R}_2,} we have,

Δ

{= \dfrac12 × \left(-1 \left|\begin{array}{cc} 7 & 1 \\[5pt] 8 & 1 \end{array}\right| + 1 \left|\begin{array}{cc} 2 & 1 \\[5pt] 10 & 1 \end{array}\right| - 1 \left|\begin{array}{cc} 2 & 7 \\[5pt] 10 & 8 \end{array}\right|\right)}

{= \dfrac12 × \left[(-1)\left(7(1) - 1(8)\right) + 1\left(2(1) - 1(10)\right) - 1\left(2(8) - 7(10)\right)\right]}

{= \dfrac12 × \left[(-1)(7 - 8) + 1(2 - 10) - 1(16 - 70)\right]}

{= \dfrac12 × \left[(-1)(-1) + 1(-8) - 1(-54)\right]}

{= \dfrac12 × (1 - 8 + 54)}

{= \dfrac12 × 47}

= \dfrac{47}{2} sq. units.

∴ The area of the triangle with vertices at the points (2, 7), (1, 1), (10, 8) is \dfrac{47}{2} sq. units.

(iii) To find the area of the triangle with vertices at the points (-2, -3), (3, 2), (-1, -8)

We know that the area of a triangle whose vertices are {(x_1, y_1),} {(x_2, y_2)} and {(x_3, y_3)} can be expressed as a determinant as

{Δ = \dfrac12 × \left|\begin{array}{ccc} x_1 & y_1 & 1 \\[5pt] x_2 & y_2 & 1 \\[5pt] x_3 & y_3 & 1 \end{array}\right|}

So, the area of the given triangle can be expressed as

{Δ = \dfrac12 × \left|\begin{array}{ccc} -2 & -3 & 1 \\[5pt] 3 & 2 & 1 \\[5pt] -1 & -8 & 1 \end{array}\right|}

Expanding along {\text{R}_1,} we have,

Δ

{= \dfrac12 × \left(-2 \left|\begin{array}{cc} 2 & 1 \\[5pt] -8 & 1 \end{array}\right| - (-3) \left|\begin{array}{cc} 3 & 1 \\[5pt] -1 & 1 \end{array}\right| + 1 \left|\begin{array}{cc} 3 & 2 \\[5pt] -1 & -8 \end{array}\right|\right)}

{= \dfrac12 × \big[(-2)\big(2(1) - 1(-8)\big) + 3\big(3(1) - 1(-1)\big) + 1\big(3(-8) - 2(-1)\big)\big]}

{= \dfrac12 × \big[(-2)(2 + 8) + 3(3 + 1) + 1(-24 + 2)\big]}

{= \dfrac12 × \big[(-2)10 + 3(4) + 1(-22)\big]}

{= \dfrac12 × (-20 + 12 - 22)}

{= \dfrac12 × -30}

= |-15| (∵ area of a triangle is always positive)

= 15 sq. units.

∴ The area of the triangle with vertices at the points (-2, -3), (3, 2), (-1, -8) is 15 sq. units.

2. Show that points {\text{A} (a, b + c),} {\text{B} (b, c + a),} {\text{C} (c, a + b)} are collinear.

The given points are {\text{A} (a, b + c),} {\text{B} (b, c + a),} {\text{C} (c, a + b)}

To show that the given points are collinear, we show that the are of the triable formed by these 3 points is zero.

So, the area of the triangle formed by the given three points can be expressed as

{Δ = \dfrac12 × \left|\begin{array}{ccc} a & b + c & 1 \\[5pt] b & c + a & 1 \\[5pt] c & a + b & 1 \end{array}\right|}

Applying {\text{C}_1 → \text{C}_1 + \text{C}_2,} we have,

{Δ = \dfrac12 × \left|\begin{array}{ccc} a + b + c & b + c & 1 \\[5pt] b + c + a & c + a & 1 \\[5pt] c + a + b & a + b & 1 \end{array}\right|}

Taking common factor {(a + b + c)} from {\text{C}_1,} we have,

{Δ = \dfrac12 × (a + b + c) \left|\begin{array}{ccc} 1 & b + c & 1 \\[5pt] 1 & c + a & 1 \\[5pt] 1 & a + b & 1 \end{array}\right|}

As {\text{C}_1} and {\text{C}_3} are same, the value of the given determinant is zero. So,

Δ

{= \dfrac12 × (a + b + c) \left|\begin{array}{ccc} 1 & b + c & 1 \\[5pt] 1 & c + a & 1 \\[5pt] 1 & a + b & 1 \end{array}\right|}

{= \dfrac12 × (a + b + c) × 0}

= 0

As the area of the triangle formed by the points {\text{A} (a, b + c),} {\text{B} (b, c + a),} {\text{C} (c, a + b)} is 0, it can deduced that the points {\text{A} (a, b + c),} {\text{B} (b, c + a),} {\text{C} (c, a + b)} are collinear.

3. Find 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)

(i) To find the value of k if area of triangle is 4 sq. units and vertices are (k, 0), (4, 0) and (0, 2)

So, the area of the triangle formed by the given three points can be expressed as

{Δ = \dfrac12 × \left|\begin{array}{ccc} k & 0 & 1 \\[5pt] 4 & 0 & 1 \\[5pt] 0 & 2 & 1 \end{array}\right|}

Expanding along {\text{C}_2,} we have,

Δ

{= \dfrac12 × \left[-0 \left|\begin{array}{cc} 4 & 1 \\[5pt] 0 & 1 \end{array}\right| + 0 \left|\begin{array}{cc} k & 1 \\[5pt] 0 & 1 \end{array}\right| - 2 \left|\begin{array}{cc} k & 1 \\[5pt] 4 & 1 \end{array}\right| \right]}

{= \dfrac12 × \big[-0 + 0 - 2\big(k(1) - 1(4)\big)\big]}

{= \dfrac12 × (-2)(k - 4)}

= -(k – 4)

It is given in the problem that the area of the triangle is 4 sq. units.

{⇒\left|- (k - 4)\right|}

=

4 (∵ area of a triangle is always positive)

{⇒ -(k - 4)}

=

±4

{⇒ k - 4}

=

±4

{⇒ k}

=

±4 + 4

=

4 + 4 or -4 + 4

=

8 or 0

∴ The value of k for the area of the triangle whose vertices are (k, 0), (4, 0) and (0, 2) to be equal to 4 sq. units is 0 or 8

(ii) To find the value of k if area of triangle is 4 sq. units and vertices are (-2, 0), (0, 4) and (0, k)

So, the area of the triangle formed by the given three points can be expressed as

{Δ = \dfrac12 × \left|\begin{array}{ccc} -2 & 0 & 1 \\[5pt] 0 & 4 & 1 \\[5pt] 0 & k & 1 \end{array}\right|}

Expanding along {\text{C}_1,} we have,

Δ

{= \dfrac12 × \left[-2 × \left|\begin{array}{cc} 4 & 1 \\[5pt] k & 1 \end{array}\right| - 0 × \left|\begin{array}{cc} 0 & 1 \\[5pt] k & 1 \end{array}\right| + 0 × \left|\begin{array}{cc} 0 & 1 \\[5pt] 4 & 1 \end{array}\right| \right]}

{= \dfrac12 × \big[(-2)\big(4(1) - 1(k)\big) - 0 + 0\big]}

{= \dfrac12 × (-2)(4 - k)}

{= -(4 - k)}

{= k - 4}

It is given in the problem that the area of the triangle is 4 sq. units.

{⇒\left|(k - 4)\right|}

=

4 (∵ area of a triangle is always positive)

{⇒ (k - 4)}

=

±4

{⇒ k}

=

±4 + 4

=

4 + 4 or -4 + 4

=

8 or 0

∴ The value of k for the area of the triangle whose vertices are (-2, 0), (0, 4) and (0, k) to be equal to 4 sq. units is 0 or 8

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.

(i) To find the equation of line joining the points (1, 2) and (3, 6) using determinants.

Consider any point {\text{P} (x , y)} on the line joining the points (1, 2) and (3, 6).

Now, the area of the triangle formed by the three points {(x, y),} (1, 2), and (3, 6) in the determinant form is given as

{Δ = \dfrac12 × \left|\begin{array}{ccc} 1 & 2 & 1 \\[5pt] 3 & 6 & 1 \\[5pt] x & y & 1 \end{array}\right|}

As the points {(x, y),} (1, 2), and (3, 6) are collinear, the area of the triangle formed by these three poings will be 0. Thus,

{\dfrac12 × \left|\begin{array}{ccc} 1 & 2 & 1 \\[5pt] 3 & 6 & 1 \\[5pt] x & y & 1 \end{array}\right|= 0}

Expanding the determinant in L.H.S. along {\text{R}_3,} we have,

{x \left|\begin{array}{cc} 2 & 1 \\[5pt] 6 & 1 \end{array}\right| - y \left|\begin{array}{cc} 1 & 1 \\[5pt] 3 & 1 \end{array}\right| + 1 \left|\begin{array}{cc} 1 & 2 \\[5pt] 3 & 6 \end{array}\right| = 0}

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

{⇒ -4x + 2y + 0 = 0}

{⇒ 2x - y = 0}

∴ The equation of line joining (1, 2) and (3, 6) using determinants is {2x - y = 0}

(ii) To find the equation of line joining the points (3, 1) and (9, 3) using determinants.

Consider any point {\text{P} (x , y)} on the line joining the points (3, 1) and (9, 3).

Now, the area of the triangle formed by the three points {(x, y),} (3, 1), and (9, 3) in the determinant form is given as

{Δ = \dfrac12 × \left|\begin{array}{ccc} 3 & 1 & 1 \\[5pt] 9 & 3 & 1 \\[5pt] x & y & 1 \end{array}\right|}

As the points {(x, y),} (3, 1), and (9, 3) are collinear, the area of the triangle formed by these three poings will be 0. Thus,

{\dfrac12 × \left|\begin{array}{ccc} 3 & 1 & 1 \\[5pt] 9 & 3 & 1 \\[5pt] x & y & 1 \end{array}\right|= 0}

Expanding the determinant in L.H.S. along {\text{R}_3,} we have,

{x \left|\begin{array}{cc} 1 & 1 \\[5pt] 3 & 1 \end{array}\right| - y \left|\begin{array}{cc} 3 & 1 \\[5pt] 9 & 1 \end{array}\right| + 1 \left|\begin{array}{cc} 3 & 1 \\[5pt] 9 & 3 \end{array}\right| = 0}

{⇒ x[1(1) - 1(3)] - y[3(1) - 1(9)] + 1[3(3) - 1(9)] = 0}

{⇒ -2x + 6y + 0 = 0}

{⇒ x - 3y = 0}

∴ The equation of line joining (3, 1) and (9, 3) using determinants is {x - 3y = 0}

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 ✔

The area of the triangle formed by the given three points (2, -6), (5, 4) and {(k, 4)} can be expressed as

{Δ = \dfrac12 × \left|\begin{array}{ccc} 2 & -6 & 1 \\[5pt] 5 & 4 & 1 \\[5pt] k & 4 & 1 \end{array}\right|}

Expanding along {\text{R}_1,} we have,

Δ

{= \dfrac12 × \left(2 \left|\begin{array}{cc} 4 & 1 \\[5pt] 4 & 1 \end{array}\right| - (-6) \left|\begin{array}{cc} 5 & 1 \\[5pt] k & 1 \end{array}\right| + 1 \left|\begin{array}{cc} 5 & 4 \\[5pt] k & 4 \end{array}\right| \right)}

{= \dfrac12 × \big[2(0) + 6\big(5(1) - 1(k)\big) + 1\big(5(4) - 4(k)\big)\big]}

{= \dfrac12 × [6(5 - k) + 1(20 - 4k)]}

{= \dfrac12 × (30 - 6k + 20 - 4k)}

{= \dfrac12 × (50 - 10k)}

= 25 – 5k

It is given in the problem that the area of the triangle is 35 sq. units.

{⇒\left|25 - 5k\right|}

=

35 (∵ area of a triangle is always positive)

{⇒ 25 - 5k}

=

±35

{⇒ 5(k - 5)}

=

±35

{⇒ k - 5}

=

{\dfrac{\pm35}{5}}

=

±7

{⇒ k}

=

±7 + 5

=

7 + 5 or -7 + 5

=

12 or -2

∴ option D is the correct answer.