NCERT Solutions for Class 12 Maths Chapter 4 – Determinants Ex 4.3
Page No 122:
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,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6480/Chapter%204_html_27997392.gif)
(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6480/Chapter%204_html_1ae5503b.gif)
(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)
is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6480/Chapter%204_html_m477135a4.gif)
Hence, the area of the triangle is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6480/Chapter%204_html_mbd4265a.gif)
Page No 123:
Question 2:
Show that points
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6481/Chapter%204_html_m6ee10de5.gif)
Answer:
Area of ΔABC is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6481/Chapter%204_html_4574753c.gif)
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
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_m7d88ee23.gif)
It is given that the area of triangle is 4 square units.
∴Δ = ± 4.
(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
Δ =![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_5c597026.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_5c597026.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_5271de10.gif)
∴−k + 4 = ± 4
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,
Δ =![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_48f576c9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_48f576c9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_19a19c67.gif)
∴k − 4 = ± 4
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.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6490/Chapter%204_html_m4c94f002.gif)
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.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6490/Chapter%204_html_2932c449.gif)
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:
Answer: D
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6501/Chapter%204_html_m1ffb3930.gif)
It is given that the area of the triangle is ±35.
Therefore, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6501/Chapter%204_html_m20edecc0.gif)
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.