OP Malhotra Class 11 Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Ex 15(c)

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S Chand Class 11 ICSE Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Ex 15(c)

Question 1.
Find the area of the triangle whose vertices are
(i) (4, 2),(4, 5) and (-2, 2);
(ii) (0, 0),(-2, 3) and (10, 7);
(iii) (a, 0),(0, b) and (x, y).
Solution:
(i) Required area = \(\frac { 1 }{ 2 }\) | (20 – 8) + (8 + 10) + (- 4 – 8)|
= \(\frac { 1 }{ 2 }\) | 12 + 18 – 12| = 9 sq. units

(ii) Required area = \(\frac { 1 }{ 2 }\) | (0 – 0) + (- 14 – 30) + (0 + 0)|
= 22 sq. units

(iii) Required area = \(\frac { 1 }{ 2 }\) | (ab – 0) + (0 + bx) + (0 – ay)|
= \(\frac { 1 }{ 2 }\) | ab – bx – ay|
= \(\frac { 1 }{ 2 }\) (bx + ay – ab) sq. units

Question 2.
Find the area of the quadrilateral whose vertices are
(i) (1, 1),(7,- 3),(12, 2) and (7, 21);
(ii) (1, 1),(3, 4),(5, – 2) and (4, – 7).
Solution:
(i) Required area = \(\frac { 1 }{ 2 }\) | (- 3 – 7) + (14 + 36) + (252 – 14) + (7 – 21)|
= \(\frac { 1 }{ 2 }\) | – 10 + 50 + 238 – 14 | = 132 sq. units

(ii) Required area = \(\frac { 1 }{ 2 }\) | (4 – 3) + (- 6 – 20) + (- 35 + 8) + (4 + 7) |
= \(\frac { 1 }{ 2 }\) | 1 – 26 – 27 + 11 |
= \(\frac { 41 }{ 2 }\) sq. units

Question 3.
Show that the following points are collinear :
(i) (1, 4),(3, – 2) and (- 3, 16);
(ii) (- 5, 1), (5, 4) and (10, 7).
Solution:
(i) Given vertices of △ABC are A(1, 4); B(3, – 2) and C(- 3, 16)
∴ area of △ABC = \(\frac { 1 }{ 2 }\) | (- 2 – 12) + (48 – 6) + (- 12 – 16)|
= \(\frac { 1 }{ 2 }\) | – 14 + 42 – 28 | = 0 sq. units

Hence the points A, B and C are lies on same straight line.
∴ given points A, B and C are collinear.

(ii) Let the given points are A(- 5, 1); B(5, 4) and C(10, 7).
Here area of △ABC = \(\frac { 1 }{ 2 }\) |(- 20 – 5) + (35 – 40) + (10 + 35) |
= \(\frac { 1 }{ 2 }\) | – 25 – 5 + 45 | = \(\frac { 15 }{ 2 }\) ≠ o
Thus given points are not collinear.

Question 4.
If (7, a),(- 5, 2) and (3, 6) are collinear, find a.
Solution:
Since given points are A(7, a), B(- 5, 2) and C(3, 6) are collinear.
So area of △ABC having vertices A, B and C be equal to 0.
area of △ABC =0

⇒ \(\frac { 1 }{ 2 }\) | (14 + 5a) + (- 30 – 6) + (3a – 42) | = 0
⇒ | 14 + 5a – 36 + 3a – 42 | = 0 ⇒ | 8a – 64 | = 0 ⇒ 8a = 64 ⇒ a = 8

Question 5.
If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (- 5, 6), (7, – 4) and (-2, k) be zero, find the value of k.
Solution:
area of quadrilateral ABCD = \(\frac { 1 }{ 2 }\) | (6 + 10) + (20 – 42) + (7k – 8) + (- 4 – k) |
= \(\frac { 1 }{ 2 }\) | 16 – 22 + 7k – 8 – 4 – k | = \(\frac { 1 }{ 2 }\) | – 18 + 6k |

given area of quadrilateral ABCD = 0 ⇒ \(\frac { 1 }{ 2 }\) | – 18 + 6k | = 0
⇒ 6k = 18
⇒ k = + 3

Question 6.
A, B, C are the points (-1, 5), (3, 1) and (5, 7) respectively. D, E, F, are the mid-points of BC, CA, AB respectively. Prove that △ABC = 4 ΔDEF.
Solution:
Since D be the mid-point of BC.

∴ Coordinates of D are \(\left(\frac{3+5}{2}, \frac{1+7}{2}\right)\) i.e. D(4, 4)
Also E be the mid-point of AC.
∴ Coordinates of E are \(\left(\frac{-1+5}{2}, \frac{7+5}{2}\right)\) i.e. (2, 6)
Further F be the mid-point of AB.
Thus, coordinates of F are \(\left(\frac{3-1}{2}, \frac{1+5}{2}\right)\) i.e. (1, 3)

∴ area of △ABC = \(\frac { 1 }{ 2 }\) | (- 1 – 15) + (21 – 5) + (25 + 7) |
= \(\frac { 1 }{ 2 }\) | 16 + 0 – 8 |
= 4 sq. units
∴ area of △ABC = 4 area of △DEF

Question 7.
The straight lines y = m₁x + c₁, y = m₂x + c₂, and x = 0 intersect in the three points P, Q, and R. Find the area of the triangle PQR. What is the value of the area if c₁ = c₁?
Solution:
Given eqns. of straight lines are ;
y = m₁x + c₁ …(1)
y = m₂x + c₂ …(2)
x = 0 …(3)
Let eqn. (1) and eqn. (2) intersect at point P