Class 11 RD Sharma Solutions- Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates- Exercise 22.1

Question 1. If the line segment joining the points P(x₁,y₁) and Q(x₂,y₂) subtends an angle ∅ at the origin O, prove that OP.OQ cos∅ = x₁x₂ + y₁y₂.

Solution:

Since, O is the origin, we can get, OP² = x₁² + y₁²  and OQ² = x₂² + y₂²

Also, by the distance formula we know distance between two points P and Q is:

PQ² = (x₂ – x₁)² + (y₂ – y₁)²

Using the cosine formula, in triangle OPQ, we have

PQ² = OP² + OQ² – 2 (OP). (OQ) cos∅

⇒  (x₂ – x₁)² + (y₂ – y₁)² =  x₁² + y₁² +  x₂² + y₂²  – 2 (OP). (OQ) cos∅

⇒ x₂² + x₁² – 2 x₂x₁ + y₂² + y₁² – 2y₂y₁ = x₁² + y₁² +  x₂² + y₂²  – 2 (OP). (OQ) cos∅

⇒ – 2 (x₁x₂ + y₁y₂) =  – 2 (OP). (OQ) cos∅

⇒ OP.OQ cos∅ = x₁x₂ + y₁y₂

Question 2. The vertices of a triangle ABC are A(0,0), B (2,-1) and C (9,0). Find cos B.

Solution:

Using the cosine formula, we know

Now, lets assume, a = BC, b = CA and c = AB are the sides of the triangle ABC.

Therefore, the distance between two consecutive points can be calculated as:

a = BC = 

b = CA = 

and c = AB = 

Using the cosine formula,

Hence, 

Question. 3  Four points A(6,3), B(-3,5), C(4,-2) and D(x,3x) are given in such a way that  , find x.

Solution:

We know the formula that,

area of a triangle = , so

area of triangle ABC = 

= 

= 

Similarly, area of triangle DBC = 

= 

= 

Now, we are given that

Question. 4 The points A(2,0), B(9,1), C(11,6) and D (4,4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

Solution: 

We know the property of a rhombus that the diagonals bisect each other at the right angles. Thus, both the diagonals must have a common mid-point.

Now mid-point of line AC =  = (13/2, 3)

and mid-point of line BD =  = 

Since, both the lines have different mid-points, we can conclude that the quadrilateral ABCD is not a rhombus.

Question 5. Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are (-36,7), (20,7), and (0,-8).

Solution:

Since, the circle is inscribed in a triangle, the centre of the circle is known as incentre. We know that incentre of a circle (O) inscribed in a triangle is given by the formula:

O = , where a, b, c  are length opposite to ∠A, ∠B and ∠C respectively.

Therefore, lets say a = BC = 

similarly, b = AC = 

and c = AB = 

Therefore, the coordinates of the incentre will be:

O = 

 = (-1,0)

Hence, the coordinate of the centre of the circle is (-1,0)

Question 6. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Solution:

Since, ABC is an equilateral triangle it will have each sides equal, i.e, AB = BC = CA = 2a

Also, area of an equilateral triangle  =  , where a is the side of the triangle.

Therefore, area of given triangle = 

Also, area of a triangle = 

Thus, the coordinates of point A is 

similarly, the coordinates of point B is (0,-a) and the coordinates of point C is (0,a)

Hence, the vertices of a triangle are (0,a), (0,-a) and   or (0,a), (0.-a) and 

 Question 7. Find the distance between P(x1,y1) and Q(x2,y2) when (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Solution: 

We are given two points P(x1,y1) and Q(x2,y2),

(i) when line  PQ is parallel to the y-axis, then we can conclude that the x-coordinate will be constant ⇒ x2 = x1

Thus, by using distance formula:

PQ = 

(ii) when line  PQ is parallel to the x-axis, then we can conclude that the y-coordinate will be constant ⇒ y2 = y1

Thus, by using distance formula:

PQ = 

Question 8. Find a point on the x-axis, which is equidistant from the point (7,6) and (3,4).

Solution:

As given in the question, let the arbitrary point C that lies on the x-axis has coordinate (x,0). Now, this point is equidistant from both the coordinates (7,6) and (3,4), therefore by using distance formula, we get

Therefore, coordinate of the point on the x-axis is