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

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  • Last Updated : 10 May, 2021
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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

cos B = \frac{a^2 + c^2 - b^2}{2ac}

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 = \sqrt{(9-2)^2 + (2+1)^2} = \sqrt{49 + 9} = \sqrt{58}

b = CA = \sqrt{(0-9)^2 + (0+2)^2} = \sqrt{81 + 4} = \sqrt{85}

and c = AB = \sqrt{(2-0)^2 + (-1 - 0)^2} = \sqrt{4 + 1} = \sqrt{5}

Using the cosine formula,

cos B = \frac{58 + 5 - 85}{2\times\sqrt{58}\times \sqrt{5}}

= \frac{63 - 85}{2\times\sqrt{290}}

= \frac{-22}{\sqrt{290}} = \frac{- 11}{\sqrt{290}}

Hence, cosB = \frac{- 11}{\sqrt{290}}

Question. 3  Four points A(6,3), B(-3,5), C(4,-2) and D(x,3x) are given in such a way that \frac{\triangle DBC}{\triangle ABC} = \frac{1}{2}    , find x.

Solution:

We know the formula that,

area of a triangle = \frac{1}{2}[x_1(y_2-y_1) + x_2(y_3-y_2) + x_3(y_3-y_1)]   , so

area of triangle ABC = \frac{1}{2}[-3(-2-3x) + 4(3x-5) + x(5+2)]

\frac{1}{2}[6+ 9x+ 12x -20 + 5x +2x]

\frac{1}{2}[28x -14] = 14x - 7

Similarly, area of triangle DBC = \frac{1}{2}[6(5+2) - 3(-2-3) + 4(3-5)]

\frac{1}{2}[42+15-8]

\frac{49}{2}

Now, we are given that

\frac{\triangle DBC}{\triangle ABC} = \frac{1}{2}

\Rightarrow \frac{7(2x-1)}{\frac{49}{2}} = \frac{1}{2}

\Rightarrow \frac{14(2x-1)}{49} = \frac{1}{2}

\Rightarrow \frac{28x-14}{49} = \frac{1}{2}

\Rightarrow 56x - 28 = 49

\Rightarrow 56x = 77 \\\Rightarrow x = \frac{11}{8}

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 = \left ( \frac{2 + 11}{2} \right ), \left ( \frac{0 + 6}{2} \right )    = (13/2, 3)

and mid-point of line BD = \left ( \frac{9 + 4}{2} \right ), \left ( \frac{1 + 4}{2} \right )    = \left ( \frac{13}{2}, \frac{5}{2} \right )

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 = \left ( \frac{ax_1 + bx_2 + cx_3}{a+b+c},\frac{ay_1 + by_2 + cy_3}{a+b+c} \right )   , where a, b, c  are length opposite to ∠A, ∠B and ∠C respectively.

Therefore, lets say a = BC = \sqrt{(20-0)^2 + (7-(-8))^2} = \sqrt{20^2 + 15^2} = 25

similarly, b = AC = \sqrt{(-36-0)^2 + (7-(-8))^2} = \sqrt{36^2 + 15^2} = \sqrt{1521} = 39

and c = AB = \sqrt{(-36-20)^2 + (7-7)^2} = 56

Therefore, the coordinates of the incentre will be:

O = \left (\frac{25(-36)+39(20)+56(0)}{25+39+56},\frac{25(7)+39(7)+56(-8)}{25+39+56}\right )

 = \left (\frac{-1}{120},0\right )

 = (-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  = \frac{\sqrt{3}}{4} \times a^2    , where a is the side of the triangle.

Therefore, area of given triangle = \frac{\sqrt{3}}{4} \times (2a)^2 = \sqrt{3}a^2

Also, area of a triangle = \frac{1}{2}\times base \times height

\Rightarrow  \sqrt{3}a^2 = \frac{1}{2}\times BC \times OA

\Rightarrow  \sqrt{3}a^2 = \frac{1}{2}\times 2a \times OA

\Rightarrow  \sqrt{3}a = OA

Thus, the coordinates of point A is \left (\sqrt{3}a,0  \right )

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  \left (-\sqrt{3}a,0  \right )    or (0,a), (0.-a) and \left (\sqrt{3}a,0  \right )

 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 = \sqrt{(x2-x1)^2 + (y2-y1)^2} = \sqrt{(0)^2 + (y2-y1)^2} = |y2-y1|

(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 = \sqrt{(x2-x1)^2 + (y2-y1)^2} = \sqrt{(x2-x1)^2 + (0)^2} = |x2-x1|    

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

\sqrt{(x-7)^2 + (0-6)^2} = \sqrt{(x-3)^2 + (0-4)^2}

\Rightarrow (x-7)^2 + (6)^2 = (x-3)^2 + (4)^2

\Rightarrow x^2 -14x + 49 + 36 = x^2 -6x + 9 + 16

\Rightarrow 8x = 60 \Rightarrow x = \frac{15}{2}

Therefore, coordinate of the point on the x-axis is (\frac{15}{2},0)


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