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:

2020-12-21_15-44-00Consider 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

Answer:

Given ABC is an equilateral triangle with side 2a . Let A be ( x, 0) .

2020-12-21_15-49-43\therefore AB = BC = CA

\Rightarrow AB^2 = BC^2 = CA^2

\Rightarrow a^2 + a^2  = (2a)^2

\Rightarrow x^2 = 3a^2

\Rightarrow x = \pm \sqrt{3} a

Therefore the vertices are (0, a), ( 0, -a) , ( \sqrt{3} a, 0) or (0, a), ( 0, -a) , ( -\sqrt{3} a, 0)

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Question 7: Find the distance between P (x_1, y1) and Q (x_2, y_2) when (i) PQ is parallel to the y-axis (ii) PQ is parallel to the x-axis.

Answer:

Given P (x_1, y1) and Q (x_2, y_2)

We know PQ = \sqrt{ (x_1-x_2)^2 + ( y_1-y_2)^2}

i) When PQ is parallel to x-axis, then x_1 = x_ 2

\therefore PQ = \sqrt{( y_1-y_2)^2 } = | y_1-y_2|

i) When PQ is parallel to y-axis, then y_1 = y_ 2

\therefore PQ = \sqrt{( x_1-x_2)^2 } = | x_1-x_2|

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

Answer:

Given points A(7,6)  and B(3,4)

Let C(x, 0) be the equidistant point from A and B

\therefore AC = BC  \Rightarrow AC^2 = BC^2

\Rightarrow ( 7-x)^2 + ( 6-0)^2 = ( 3-x)^2 + ( 4-0)^2

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

\Rightarrow 85 - 14 x = 25 - 6x

\Rightarrow 60 = 8x

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

\therefore C = \Big( \frac{15}{2} , 0 \Big)