Class 11: Cartesian System of Rectangular Co-ordinates – Exercise 22.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: If the line segment joining the points $P (x_1 , y_1)$ and $Q (x_2 , y_2)$ subtends an angle $\alpha$ at the origin $O$ , prove that: $OP \cdot OQ \cos \alpha = x_1x_2 + y_1y_2$ .

Answer:

Consider the figure given.

$OP^2 = {x_1}^2 + {y_1}^2$

$OQ^2 = {x_2}^2 + {y_2}^2$

$\therefore PQ^2 = ( x_2-x_1)^2 + ( y_2-y_1)^2$

Using cosine formula in $\triangle OPQ$ we get

$PQ^2 = OP^2 + OQ^2 - 2 \cdot OP \cdot OQ. \cos \alpha$

$\Rightarrow ( x_2-x_1)^2 + ( y_2-y_1)^2 = ( {x_1}^2+{y_1}^2) + ( {x_2}^2+{y_2}^2) - 2 \cdot OP \cdot OQ. \cos \alpha$

$\Rightarrow {x_2}^2 + {x_1}^2 - 2 x_1 x_2 + {y_2}^2 + {y_1}^2 - 2y_1y_2 = {x_2}^2 + {x_1}^2 + {y_2}^2 + {y_1}^2 - 2 \cdot OP \cdot OQ. \cos \alpha$

$\Rightarrow 2 \cdot OP \cdot OQ. \cos \alpha = 2 x_1 x_2 + 2y_1y_2$

$\Rightarrow \cdot OP \cdot OQ. \cos \alpha = x_1 x_2 +y_1y_2$

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

Answer:

Given $\triangle ABC, A ( 0,0), B( 2, -1) , C( 9, 2)$

We know $\cos B =$ $\frac{a^2 + c^2 - b^2}{2ac}$ where $a = BC, b = CA, c = AB$

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

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

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

$\therefore \cos B =$ $\frac{58+5-85}{2 \cdot \sqrt{58} \cdot \sqrt{5}}$ $=$ $\frac{-11}{\sqrt{290}}$

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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 $x$ .

Answer:

Given $A(6,3), B(-3, 5) , C(4, -2$ ) and $D(x, 3x)$

We know that the area of a Triangle with vertices $(x_1, y_1), (x_2, y_2),(x_3, y_3)$

Area $=$ $\frac{1}{2}$ $\Big[ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3( y_1-y_2) \Big]$

$\therefore \ Ar. \ \triangle DBA =$ $\frac{1}{2}$ $\Big[ -3( -2-3x) + 4 ( 3x-5) + x( 5+2) \Big] = 7(2x-1)$

Also Area of $\triangle ABC =$ $\frac{1}{2}$ $\Big[ 6( 5+2( - 3 ( -2-3) + 4( 3 - 5) \Big] =$ $\frac{49}{2}$

Given $\frac{\triangle DBA}{\triangle ABC}$ $=$ $\frac{7(2x-1)}{\frac{49}{2}}$

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

$\Rightarrow 8x - 4 = 7$

$\Rightarrow x =$ $\frac{11}{8}$

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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.

Answer:

Given $A (2, 0), B (9,1), C (11, 6)$ and $D (4, 4)$

$AB = \sqrt{(2-9)^2 + ( 0-1)^2} = \sqrt{49 + 1} = \sqrt{50}$

$BC= \sqrt{(9-11)^2 + ( 1-6)^2} = \sqrt{4 + 25} = \sqrt{29}$

$CD = \sqrt{(11-4)^2 + ( 6-4)^2} = \sqrt{49 + 4} = \sqrt{53}$

$AD = \sqrt{(2-4)^2 + ( 0-4)^2} = \sqrt{4 + 16} = \sqrt{20}$

Since $AB \neq BC \neq CD \neq AD$ , therefore $ABCD$ is not a rhombus.

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

Answer:

Given the vertices of the triangle $(-36, 7), (20, 7)$ and $(0, - 8)$ .

We know that if the vertices of a triangle are $(x_1, y_1), (x_2, y_2),(x_3, y_3)$ , then the in-center of the circle is

$\Big($ $\frac{ax_1+bx_2+cx_3}{a+b+c}$ $,$ $\frac{ay_1+by_2+cy_3}{a+b+c}$ $\Big)$ where $a = BC, b = AC, c= AB$

Now

$a = BC = \sqrt{(20-0)^2 + (7+8)^2} = \sqrt{400 + 225} = 25$

$b = CA= \sqrt{(0+36)^2 + ( =8-7)^2} = \sqrt{1296 + 225} = 39$

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

Therefore the coordinates of the in-center are

$\Big($ $\frac{25(-36)+39(20)+56(0)}{25+39+56}$ $,$ $\frac{25(7)+39(7)+56(-8)}{25+39+56}$ $\Big) = \Big($ $\frac{-120}{120}$ $, 0 \Big) = (-1, 0)$

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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