# NCERT Solutions Class 12 maths Chapter-4 (Determinants)Exercise 4.3

**NCERT Solutions Class 12 Maths**from class

**12th**Students will get the answers of

**Chapter-4 (Determinants)Exercise 4.3**This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of

**NCERT Board Mathematics Textbook**in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

### 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) **

**Solution:**

**(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. **

**Solution:**

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

For example:- Area of triangle = 0

**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.

**(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.

**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.

**(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.

**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:

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