## Hello Students! In this post, you will find the complete** ****NCERT Solutions for Maths Class 12 Exercise 4.2**.

You can download the **PDF of NCERT Books Maths Chapter 4** for your easy reference while studying **NCERT Solutions for Maths Class 12 Exercise 4.2**.

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**NCERT Solutions for Maths Class 12 Exercise 4.1**

**NCERT Solutions for Maths Class 12 Exercise 4.3**

**NCERT Solutions for Maths Class 12 Exercise 4.4**

**NCERT Solutions for Maths Class 12 Exercise 4.5**

**NCERT Solutions for Maths Class 12 Exercise 4.2**

**Maths Class 12 Ex 4.2 Question 1.**

**Find the 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)** Area of a triangle =

= 1/2 [1(0 – 3) + 1(18 – 0)]

= 1/2 [1 × (–3) + 1 × 18]

= 1/2 [–3 + 18]

= 1/2 [15]

= 7.5 sq. units

**Maths Class 12 Ex 4.2 Question 2.**

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

**Solution:**

Let us assume that the vertices of ∆ABC are A (a, b + c), B (b, c + a) and C (c, a + b).

**Maths Class 12 Ex 4.2 Question 3.**

**Find the value 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)** Area of a triangle = 4 sq. units (given)

Area of the triangle =

= 1/2 [–2k + 8]

= –k + 4**Case (a):** –k + 4 = 4 ⇒ k = 0**Case (b):** –k + 4 = –4 ⇒ k = 8

Hence, k = 0, 8

**(ii)** The area of the triangle whose vertices are (–2, 0), (0, 4), (0, k).

**Maths Class 12 Ex 4.2 Question 4.**

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

**Solution:****(i)** The given points are (1, 2) and (3, 6)

Equation of the line is

**Maths Class 12 Ex 4.2 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:**

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

Area of a triangle =

= 1/2 [50 – 10k] = 25 – 5k

∴ 25 – 5k = 35 or 25 – 5k = –35

–5k = 10 or 5k = 60

⇒ k = –2 or k = 12

Hence, option (D) is correct.

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