OP Malhotra Class 11 Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Chapter Test

Interactive Class 11 OP Malhotra Solutions Chapter 15 Basic Concepts of Points and their Coordinates Chapter Test engage students in active learning and exploration.

S Chand Class 11 ICSE Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Chapter Test

Question 1.
Show that the points (4, 4), (3, 5) and (- 1, – 1) are the vertices of a right triangle.
Solution:
Let the given points are A(4, 4), B(3, 5) and C (- 1,- 1)
∴ | AB | = \(\sqrt{(3-4)^2+(5-4)^2}\) = \(\sqrt{1+1}\) = \(\sqrt{2}\)
| BC | = \(\sqrt{(-1-3)^2+(-1-5)^2}\) =\(\sqrt{16+36}\) = \(\sqrt{52}\) = \(2 \sqrt{13}\)
| CA | = \(\sqrt{(-1-4)^2+(-1-4)^2}\) = \(\sqrt{25+25}\) = \(\sqrt{50}\) = \(5 \sqrt{2}\)
AB² + CA² = 2 + 50 = 52 = BC²
Thus the points A, B and C are the vertices of right angled △ABC, right angle d at A.

Question 2.
A quadrilateral has the vertices at the points (-4, 2), (2, 6), (8, 5) and (9, – 7). Show that the mid-points of the sides of this quadrilateral are the vertices of a parallelogram.
Solution:
Let the given points are the vertices of quadrilateral ABCD.
Let E, F, G and H are the mid-points of side AB, BC, CD and AD of quadrilateral ABCD.
Then by mid-point formula,
The coordinates of E, F, G and H are

Hence both diagonals of quad. EFGH bisect each other. Thus the quadrilateral EFGH forms a parallelogram.

Question 3.
Show that the points (a, b + c),(b, c + a),(c, a + b) are collinear.
Solution:
Let the given points are (a, b + c),(b, c + a) and (c, a + b).

∴ area of △ABC = \(\frac { 1 }{ 2 }\) [{(a(c + a) – b (b + c)} + {b (a + b) – c (c + a)} + {c (b + c) – a (a + b)}]
= \(\frac { 1 }{ 2 }\) [ac + a² – b² – bc + ab + b² – c² – ac + bc + c² – a² – ab]
= \(\frac { 1 }{ 2 }\) × 0 = 0
Hence the given points A, B and C lies on same straight line and hence collinear.

Question 4.
Centroid of a triangle is (1, 4) and two of its vertices are (4, – 3) and (- 9, 7). Find the area of the triangle.
Solution:
Let the coordinates of third vertex of triangle be (α, β)
we know that, centroid of Δ having vertices are (x₁, y₁) : (x₂, y₂) and (x₃, y₃) be given by
\(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
∴ Centroid of △ABC having vertices are A(4, – 3); B(- 9, 7) and C(α, β) be
\(\left(\frac{4-9+\alpha}{3}, \frac{-3+7+\beta}{3}\right)\) i.e. \(\left(\frac{\alpha-5}{3}, \frac{\beta+4}{3}\right) \text {. }\)
Also given coordinates of centroid are G(1, 4).
∴ 1 = \(\frac{\alpha-5}{3}\) ⇒ α = 8 and \(\frac{\beta+4}{3}\) = 4 ⇒ β = 8
Thus coordinates of third vertex are (8, 8).
∴ area of △ABC = \(\frac { 1 }{ 2 }\) | (28 – 27) + (- 72 – 56) + (- 24 – 32)|
= \(\frac { 1 }{ 2 }\) | 1 – 128 – 56 | = \(\frac { 183 }{ 2 }\) sq. units

Question 5.
Find the third vertex of a triangle if two of its vertices are (- 1, 4) and (5, 2) and the medians through these vertices meet at (0, – 3).
Solution:
Let the third vertex of △ABC be C(α, β).
We know that, centroid of Δ is the point of intersection of all the three medians of a triangle.

∴ Coordinates of centroid are \(\left(\frac{-1+5+\alpha}{3}, \frac{4+2+\beta}{3}\right)\)
Also, given coordinates of centroid G are (0, – 3).
∴ 0 = \(\frac{-1+5+\alpha}{3}\) ⇒ α + 4 = 0 ⇒ α = – 4
and -3 = \(\frac{6+\beta}{3}\) ⇒ β = – 6 – 9 = – 15
Hence the centroid of third vertex of △ABC are (- 4, – 15).