NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

Mathematics (Maths) Class 12

Created by: Tarun Kaushik

JEE : NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

The document NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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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)

Answer
(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev
(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev
(iii)The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8) is given by the relation, 

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

Hence, the area of the triangle is |-15| = 15 sq units


Question 2: Show that points

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev are collinear

Answer
Area of ΔABC is given by the relation,

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev

= 0

Thus, the area of the triangle formed by points A, B, and C is zero. Hence, the points A, B, and C are collinear.  

Question 3: Find values of k if area of triangle is 4 square units and vertices are
 (i) (k, 0), (4, 0), (0, 2) 

(ii) (−2, 0), (0, 4), (0, k)

Answer
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is the absolute value of the determinant (Δ), where
NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev


(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev
When −k + 4 = − 4, k = 8.
When −k + 4 = 4, k = 0.
Hence, k = 0, 8.


(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev
When k − 4 = − 4, k = 0.
When k − 4 = 4, k = 8.
Hence, k = 0, 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

Answer
(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev
Hence, the equation of the line joining the given points is y = 2x.
(ii) Let P (x, y) be any point on the line joining points A (3, 1) and B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev
Hence, the equation of the line joining the given points is x − 3y = 0.


Question 5: If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is
 A. 12 

B. −2 

C. −12, −2 

D. 12, −2


Answer: D
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,

NCERT Solutions - Determinants, Exercise 4.3 JEE Notes | EduRev
When 5 − k = −7, k = 5 + 7 = 12.
When 5 − k = 7, k = 5 − 7 = −2.
Hence, k = 12, −2.
The correct answer is D.

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