Note: If the points are (x_1, y_1) and (x_2, y_2) , then the distance between them is equal to \sqrt{(x_2-x_1)^2 + (y_2-x_1)^2 }

Question 1: Find the distance between the following pair of points:

(i) (-6,7) and (-1, -5)

(ii) (a +b, b + c) and (a -b, c -b)

(iii) (a \sin \alpha, - b \cos \alpha) and (-a \cos \alpha, b \sin \alpha)

(iv) (a,0) and (0,b)

Answer:

(i) Distance between the given points (-6,7) and (-1, -5)

= \sqrt{(-1+6)^2 + (-5-7)^2 }

\sqrt{5^2+12^2} = 13

(i) Distance between the given points (a +b, b + c) and (a -b, c -b)

= \sqrt{(a-b-a-b)^2 + (c-b-b-c)^2 }

= \sqrt{(-2b)^2+(-2b)^2}

= \sqrt{4b^2 + 4b^2}

= 2\sqrt{2}b

(i) Distance between the given points (a \sin \alpha, - b \cos \alpha) and (-a \cos \alpha, b \sin \alpha)

= \sqrt{(-a \cos \alpha - a \sin \alpha)^2 + (b \sin \alpha + b \cos \alpha}

= \sqrt{a^2(\cos \alpha + \sin \alpha)^2 + b^2 (\cos \alpha + \sin \alpha)^2}

= \sqrt{a^2 + b^2} (\cos \alpha + \sin \alpha)

(i) Distance between the given points (a,0) and (0,b)

= \sqrt{(0-a)^2 + (b-0)^2 } = \sqrt{a^2+b^2} = 13

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Question 2: Find the value of a when the distance between the points (3, a) and (4,1) is \sqrt{10} .

Answer:

Given: \sqrt{10} = \sqrt{(4-3)^2 + (1-a)^2 }

\Rightarrow \sqrt{10} = \sqrt{1 + 1 + a^2-2a }

Squaring both sides

\Rightarrow 10 = 2 + a^2-2a

\Rightarrow a^2 - 2a -8 = 0

\Rightarrow (a-4)(a+2) = 0

\Rightarrow a = -2, 4

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Question 3: If the points (2, 1) and (1,-2) are equidistant from the point (x, y) , show that x+3y=0 .

Answer:

Given: \sqrt{(2-x)^2 + (1-y)^2 } = \sqrt{(1-x)^2 + (-2-y)^2 }

\Rightarrow 4 + x^2 - 4x + 1 + y^2 - 2y = 1 + x^2 - 2x + 4 + y^2+4y

\Rightarrow 5 - 4x - 2y = 5 - 2x + 4y

\Rightarrow -2x - 6y = 0

\Rightarrow x + 3y = 0 

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Question 4: Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3,0) are 4 .

Answer:

Given: \sqrt{(-3-x)^2 + (0-y)^2 } = 4

9 + x^2 + 6x + y^2 = 16

x^2 + y^2 +6x = 7 … … … … … i)

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

9 + x^2 - 6x + y^2 = 1

x^2 + y^2 -6x = 7 … … … … … ii)

From i) and ii) we get

x^2 + y^2 +6x = x^2 + y^2 -6x

\Rightarrow 12 x = 0 \Rightarrow x = 0

Therefore from i), y^2 = 16-9 \Rightarrow y = \pm \sqrt{7}

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Question 5: The length of a line segment is of 10 units and the coordinates of one end-point are (2,-3) .If the abscissa of the other end is 10 , find the ordinate of the other end.

Answer:

Given: \sqrt{(10-2)^2 + (y+3)^2 } = 10

\Rightarrow 64 + y^2 + 9 + 6y = 100

\Rightarrow y^2 + 6y - 27 = 0

\Rightarrow (y+9)(y-3) = 0

\Rightarrow y = -9, 3

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Question 6: Show that the points (-4,-1), (-2,-4),(4,0) and (2,3) are the vertices points of a rectangle.

Answer:

Let the points be A(-4,-1), B(-2,-4), C(4,0) and D(2,3)

Therefore sides of the figure

AB = \sqrt{(-2+4)^2 + (-4+1)^2 }= \sqrt{4+9} = \sqrt{13}

BC = \sqrt{(4+2)^2 + (0+4)^2 }= \sqrt{36+16} = \sqrt{52}

CD = \sqrt{(4-2)^2 + (0-3)^2 }= \sqrt{4+9} = \sqrt{13}

DA = \sqrt{(2+4)^2 + (3+1)^2 }= \sqrt{36+16} = \sqrt{52}

Diagonals of the figure

AC= \sqrt{(4+4)^2 + (0+1)^2 }= \sqrt{64+1} = \sqrt{65}

BD = \sqrt{(2+2)^2 + (3+4)^2 }= \sqrt{16+49} = \sqrt{65}

Hence, AB = CD, BC = DA, AC=BD .Therefore the figure is a rectangle since the opposite sides are equal and the diagonals are also equal.

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Question 7: Show that the points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.

Answer:

The points are A (1,- 2), B (3, 6), C (5, 10) and D (3, 2)

Therefore sides of the figure

AB = \sqrt{(3-1)^2 + (6+2)^2 }= \sqrt{4+64} = \sqrt{68}

BC = \sqrt{(5-3)^2 + (10-6)^2 }= \sqrt{4+16} = \sqrt{20}

CD = \sqrt{(5-3)^2 + (10-2)^2 }= \sqrt{4+64} = \sqrt{68}

DA = \sqrt{(3-1)^2 + (2+2)^2 }= \sqrt{4+16} = \sqrt{20}

Diagonals of the figure

AC= \sqrt{(5-1)^2 + (10+2)^2 }= \sqrt{16+144} = \sqrt{160}

BD = \sqrt{(3-3)^2 + (2-6)^2 }= \sqrt{0+16} = \sqrt{16}

Hence, AB = CD, BC = DA, AC \neq BD .Therefore the figure is a parallelogram since the opposite sides are equal but the diagonals are unequal.

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Question 8: Prove that the points A(1,7),B(4,2),C(-1,-1) and D (4,4) are the vertices of a square.

Answer:

The points are A(1,7),B(4,2),C(-1,-1) and D (4,4)

Therefore sides of the figure

AB = \sqrt{(4-1)^2 + (2-7)^2 }= \sqrt{9+25} = \sqrt{34}

BC = \sqrt{(-1-4)^2 + (-1-2)^2 }= \sqrt{25+9} = \sqrt{34}

CD = \sqrt{(-1+4)^2 + (-1-4)^2 }= \sqrt{9+25} = \sqrt{34}

DA = \sqrt{(-4-1)^2 + (4-7)^2 }= \sqrt{25+9} = \sqrt{34}

Diagonals of the figure

AC= \sqrt{(-1-1)^2 + (-1-7)^2 }= \sqrt{4+64} = \sqrt{68}

BD = \sqrt{(-4-4)^2 + (4-2)^2 }= \sqrt{64+4} = \sqrt{68}

Hence, AB = CD = BC = DA, AC = BD .Therefore the figure is a square since the opposite sides are equal and also the diagonals are equal.

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Question 9: Prove that the points (3, 0), (6,4) and (- 1, 3) are vertices of a right-angled isosceles triangle.

Answer:

Let the points be A(3, 0), B(6,4) and C(- 1, 3)

Therefore sides of the figure

AB = \sqrt{(6-3)^2 + (4-0)^2 }= \sqrt{9+16} = 5

BC = \sqrt{(-1-6)^2 + (3-4)^2 }= \sqrt{49+1} = 5\sqrt{2}

AC = \sqrt{(-1-3)^2 + (3-0)^2 }= \sqrt{16+9} = 5

We see that AC^2 + AB^2 = BC^2

Therefore \triangle ABC is a right angled triangle at A . Also AB = AC therefore triangle is also an isosceles triangle.

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Question 10: Prove that (2, -2), (-2, 1 ) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.

Answer:

Let the points be A(2, -2), B(-2, 1 ) and C(5, 2)

Therefore sides of the figure

AB = \sqrt{(-2-2)^2 + (1+2)^2 }= \sqrt{16+9} = 5

BC = \sqrt{(5+2)^2 + (2-1)^2 }= \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}

AC = \sqrt{(5-2)^2 + (2+2)^2 }= \sqrt{9+16} = 5

We see that AB^2 + CA^2 = BC^2

Therefore \triangle ABC is a right angled triangle at A .

Area = \frac{1}{2} \times AC \times AB = \frac{1}{2} \times 5 \times 5 = 12.5 sq. units

Hypotenuse = 5\sqrt{2}

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Question 11: Prove that the points (2a,4a),(2a,6a) and (2a+\sqrt{3}a, 5a) are the vertices of an equilateral triangle.

Answer:

Let the points be A(2a,4a), B(2a,6a) and C (2a+\sqrt{3}a, 5a)

Therefore sides of the figure

AB = \sqrt{(2a-2a)^2 + (6a-4a)^2 }= \sqrt{0+4a^2} = 2a

BC = \sqrt{(2a+\sqrt{3}a - 2a)^2 + (5a-6a)^2 }= \sqrt{3a^2+a^2} = 2a

AC = \sqrt{(2a-2a-\sqrt{3}a)^2 + (4a-5a)^2 }= \sqrt{3a^2+a^2} = 2a

We see that AB=BC=CA

Therefore \triangle ABC is an equilateral triangle.

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Question 12: Prove that the points (2, 3), (-4, -6) and (1, 3 / 2) do not form a triangle.

Answer:

Let the points be A(2, 3), B(-4, -6) and C(1, 3 / 2)

Therefore sides of the figure

AB = \sqrt{(-4-2)^2 + (-6-3)^2 }= \sqrt{36+81} = \sqrt{117} = 10.817

BC = \sqrt{(1+4)^2 + (1.5+6)^2 }= \sqrt{25+56.25} =\sqrt{81.25} =  9.014

AC = \sqrt{(2-1)^2 + (3-1.5)^2 }= \sqrt{1+2.25} = \sqrt{3.25} = 1.803

In a triangle, the sum of the length of to sides should be greater then the third side.

AB + BC = 10.817 + 9.014 > 1.803 (CA) \rightarrow TRUE

BC + CA = 9.014 + 1.803 = 10.817 (AB) \rightarrow FALSE

CA + AB = 1.803 + 10.817 > 9.014 (BC) \rightarrow TRUE

Because all three conditions are not TRUE, this is not a triangle.

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Question 13: An equilateral triangle has two vertices at the points (3,4) and (-2,3) , find the coordinates of the third vertex.

Answer:

Given vertices: A(3, 4) and B ( -2, 3)

Let the third vertices be O(x, y)

AB^2 = ( 3 + 2)^2 + ( 4-3)^2 = 25 +1 = 26

OA^2 = ( x+2)^2 + ( y - 3)^2

\Rightarrow OA^2 = x^2 + 4 + 4x + y^2 + 9 - 6y

\Rightarrow OA^2 = x^2 + y^2 + 4x - 6y + 13

OB^2 = ( x - 3)^2 + ( y - 4)^2

\Rightarrow OB^2 = x^2 + 9 - 6x + y^2 + 16 - 8 y

\Rightarrow OB^2 = x^2 + y^2 -6x -8y + 25

OA^2 = OB^2

\Rightarrow x^2 + y^2 + 4x -6y + 13 = x^2 + y^2 -6x -8y + 25

\Rightarrow 4x -6y + 13 = -6x-8y+25

\Rightarrow 10x + 2y = 12

\Rightarrow 5x + y = 16

\Rightarrow y = ( 6-5x)

Also OA^2 = AB^2

\Rightarrow x^2 + y^2 + 4x - 6y + 13 = 26

\Rightarrow x^2 + y^2+ 4x - 6y - 13 = 0

Substituting

x^2 + ( 6-5x)^2+ 4x - 6( 6-5x) - 13 = 0

\Rightarrow x^2 + 36 + 25x^2 - 60x + 4x - 36 + 30x - 13 = 0

\Rightarrow 26c^2 - 26x - 13 = 0

\Rightarrow 2x^2 - 2x - 1= 0

\Rightarrow x = \frac{1 \pm \sqrt{3}}{2}

Therefore x = \frac{1 + \sqrt{3}}{2} or x = \frac{1 - \sqrt{3}}{2}

If x = \frac{1 + \sqrt{3}}{2} \Rightarrow y = 6 - 5 ( \frac{1 + \sqrt{3}}{2} ) = \frac{7 -5\sqrt{3}}{2}

If x = \frac{1 - \sqrt{3}}{2} \Rightarrow y = 6 - 5 ( \frac{1 - \sqrt{3}}{2} ) = \frac{7 + 5\sqrt{3}}{2}

Therefore the coordinate of the third vertices would be \Big( \frac{1 + \sqrt{3}}{2}, \frac{7 -5\sqrt{3}}{2} \Big),  \Big( \frac{1 + \sqrt{3}}{2}, \frac{7 +5\sqrt{3}}{2} \Big)

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Question 14: Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2,3) and (-3, -2) is a rhombus.

Answer:

Let the points be A(2,-1), B(3,4), C(-2,-3) and D(-3,-2)

Therefore sides of the figure

AB = \sqrt{(3-2)^2 + (4+1)^2 }= \sqrt{1+25} = \sqrt{26}

BC = \sqrt{(-2-3)^2 + (3-4)^2 }= \sqrt{25+1} = \sqrt{26}

CD = \sqrt{(-3+2)^2 + (-2+1)^2 }= \sqrt{1+25} = \sqrt{26}

DA = \sqrt{(-3-2)^2 + (-2+1)^2 }= \sqrt{25+1} = \sqrt{26}

Diagonals of the figure

AC= \sqrt{(-2-2)^2 + (3+1)^2 }= \sqrt{16+16} = \sqrt{32}

BD = \sqrt{(-3-3)^2 + (-2-4)^2 }= \sqrt{81+36} = \sqrt{117}

Hence, AB = BC = CD = CA . Therefore we see that the sides are equal but the diagonals are not equal. Hence ABCD   is a Rhombus.

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Question 15: Two vertices of an isosceles triangle are (2,0) and (2,5) . Find the third vertex if the length of the equal sides is 3 .

Answer:

Given A (2, 0), B ( 2, 5)

Length of equal sides = 3

Let the third vertices be ( x, y)

AB = \sqrt{(2-2)^2 + (5-0)^2 }= \sqrt{25} = 5

Therefore equal sides are AC and BC

AC^2 = (x-2)^2 + ( y-0)^2 = x^2 + 4 - 4x + y^2 = 9

\Rightarrow x^2 + y^2 - 4x = 5    … … … … … i)

BC^2 = (x-2)^2 + ( y - 5)^2 = x^2 + 4 - 4x + y^2 + 25 - 10y = 9

\Rightarrow x^2 + y^2 - 4x - 10 y = -20    … … … … … ii)

Subtracting i) from ii)

\Rightarrow -10y = -25

\Rightarrow y = 2.5

Therefore from i)

x^2 + ( 2.5)^2 - 4x = 5

\Rightarrow x^2 - 4x + 6.25 = 5

\Rightarrow x^2 - 4x + 1.25 = 0

\Rightarrow x = \frac{4 \pm \sqrt{11}}{2} 

Therefore the coordinates are \Big( \frac{4 + \sqrt{11}}{2}, \frac{5}{2}   \Big) , \Big( \frac{4 - \sqrt{11}}{2}, \frac{5}{2}   \Big)

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Question 16: Which point on x-axis is equidistant from (5, 9) and (- 4,6) ?

Answer:

Given (5, 9) and ( -4, 6)

Let ( x, 0) be the equidistant point

Therefore (x-5)^2 + ( 0-9)^2 = ( x+4)^2 + ( 0-6)^2

\Rightarrow x^2 + 25 - 10 x + 81 = x^2 + 16 + 8x + 36

\Rightarrow 106 - 10x = 52 + 8x

\Rightarrow 54 = 18x

\Rightarrow x = 3

Therefore the point is (3, 0)

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Question 17: Prove that the points (- 2,5), (0, 1) and (2,-3) are collinear.

Answer:

Given A ( -2, 5), B(0, 1), C ( 2, -3)

AB = \sqrt{(-2-0)^2 + (5-1)^2 }= \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}

BC = \sqrt{(0-2)^2 + (1+3)^2 }= \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}

AC = \sqrt{(2+2)^2 + (-3-5)^2 }= \sqrt{16+64} = \sqrt{80} = 4\sqrt{5}

Therefore AB + BC = AC

Hence the points are collinear.

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Question 18: The coordinates of the point P are (-3,2) . Find the coordinates of the point Q which lies on the line joining P and origin such that OP = OQ .

Answer:

Given P(-3, 2) . Let Q be (x, y)

OP = OQ

\sqrt{(0+3)^2 + (0-2)^2 }= \sqrt{(0-x)^2 + (0-y)^2 }

9+4 = x^2 + y^2

\Rightarrow x^2 + y^2 = 13

Now Slope of OP = Slope of OQ

\Rightarrow \frac{0-2}{0+3} = \frac{0-y}{0-x} 

\Rightarrow - \frac{2}{3} = \frac{y}{x} 

\Rightarrow y = - \frac{2}{3} x

Substituting

x^2 + (- \frac{2}{3} x)^2 = 13

\Rightarrow x^2 + \frac{4}{9} x^2 = 13

\Rightarrow x^2 = 9

\Rightarrow x = \pm 3

If x = 3, y = -2

If x = -3, y = 2

Since Q needs to be in the same line as OP , Q is (3, -2)

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Question 19: Which point on y-axis is equidistant from (2, 3) and (4, 1) ?

Answer:

Given points A(2, 3), B(-4, 1)

Let point C be (0, y)

AC = CB

\Rightarrow ( 0-2)^2 + (y-3)^2 = (0+4)^2 + ( y-1)^2

4 + y^2 + 9 - 6y = 16 + y^2 + 1 - 2y

\Rightarrow 13 - 6y = 17 - 2

\Rightarrow -4-4y = 0

\Rightarrow y = -1

Hence the point is (0, -1)

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Question 20: The three vertices of a parallelogram are (3, 4), (3,8) and (9, 8) . Find the fourth vertex.

Answer:

Given points A( 3, 4), B ( 3, 8), C (9, 8)

Let D be (x, y)

Since ABCD is a parallelogram

AB = CD

\Rightarrow ( 3-3)^2 + (8-4)^2 = ( 9-x)^2 + ( 8-y)^2

\Rightarrow 16 = 81 + x^2 -18x + 64 + y^2 - 16 y

\Rightarrow x^2 + y^2 - 18x - 16y + 129 = 0   … … … … … i)

Also AD = BC

\Rightarrow (3-x)^2 + ( 4 - y)^2 = ( 3-9)^2 + ( 8-8)^2

\Rightarrow 9 + x^2 - 6x + 16 + y^2 -8y = 36+0

\Rightarrow x^2 + y^2 -6x - 8y - 11 = 0   … … … … … ii)

Slope of AB = Slope of CD

\frac{8-4}{3-3} = \frac{8-y}{9-x} 

\Rightarrow (9-x) \times 4 = 0

\Rightarrow x = 9

Hence 9^2 + y^2 -6(9) - 8y- 11=0

\Rightarrow y^2 - 8y + 16 = 0

\Rightarrow (y - 4)^2 = 0

\Rightarrow y = 4

Hence D(9, 4)

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Question 21: Find the circumcenter of the triangle whose vertices are (-2,-3),(-1,0), (7,-6) .

Answer:

Given A(-2, -3), B(-1, 0), C(7, -6)

Let O(x, y) be the circumcenter. Therefore

OA^2 = (x+2)^2 + ( y+3)^2

OB^2 = (x+1)^2 + ( y-0)^2

OC^2 = (x-7)^2 + (y+6)^2

OA^2 = OB^2

\Rightarrow (x+2)^2 + ( y+3)^2 = (x+1)^2 + ( y-0)^2

\Rightarrow x^2 + 4 + 4x + y^2 + 9 + 6y = x^2 + 1 + 2x + y^2

\Rightarrow 4x + 6y + 13 = 2x+1

\Rightarrow 2x+ 6y = -12

\Rightarrow x = -3y -6  … … … … … i)

OB^2 = OC^2

\Rightarrow (x+1)^2 + ( y-0)^2 = (x-7)^2 + (y+6)^2

\Rightarrow x^2 + 1 + 2x + y^2 = x^2 + 49 - 14x + y^2 + 36 + 12y

\Rightarrow 2x+1 = 85 - 14x + 12y

\Rightarrow 16x -12y = 84

\Rightarrow 4x-3y=21    … … … … … ii)

Substituting i) in ii)

4(-3y-6) - 3y = 21

\Rightarrow -12y -24 -3y = 21

\Rightarrow -15 y = 45

\Rightarrow y = -3

Hence x = -3(-3) - 6 =9-6=3

Therefore circumcenter O is ( 3 -3)

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Question 22: Find the angle subtended at the origin by the line segment whose end points are (0, 100) and (10,0) .

Answer:

Given points A(0, 100) and B(10, 0)

Point A is on y axis and point B is on x axis.

Therefore angle subtended by A and B on origin is 90^o

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Question 23: Find the center of the circle passing through (5, -8), (2,- 9) and (2, 1) .

Answer:

Given A(5, -8), B(2, -9), C(2, 1)

Let center be O(x, y) . Therefore

OA^2 = (x-5)^2 + ( y+8)^2

OB^2 = (x-2)^2 + ( y+9)^2

OC^2 = (x-2)^2 + (y-1)^2

OA^2 = OB^2

\Rightarrow (x-5)^2 + ( y+8)^2 = (x-2)^2 + ( y+9)^2

\Rightarrow x^2 + 25 - 10x +y^2 + 64 + 16y = x^2 + 4 - 4x + y^2 + 81 + 18y

\Rightarrow 89-10x+16y = 85-4x+18y

\Rightarrow 6x+2y = 4

\Rightarrow 3x+y= 2    … … … … … i)

OB^2 = OC^2

\Rightarrow (x-2)^2 + ( y+9)^2 = (x-2)^2 + (y-1)^2

\Rightarrow y^2 + 81 + 18y = y^2 + 1 - 2y

\Rightarrow 20y = -80

\Rightarrow y = -4

Substituting in i) we get

3x = 2 - (-4) = 6

\Rightarrow x = 2

Hence the center of the circle is O(2, 4)

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Question 24: Find the value of k , if the point P (0,2) is equidistant from (3, k) and (k,5) .

Answer:

Given P(0, 2), A(3, k), B(k, 5)

Since PA = PB

\Rightarrow (3-0)^2 + (k-2)^2 = (k-0)^2 + (5-2)^2

\Rightarrow 9+k^2 + 4 - 4k = k^2 + 9

\Rightarrow 4-4k=0

\Rightarrow k = 1

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Question 25: If two opposite vertices of a square arc (5, 4) and (1, -6) , find the coordinates of its remaining two vertices.

Answer:

Given A(5,4 ), C(1, -6) .

Let B be (x, y)

Since ABCD is a square, AB = BC

(x-5)^2 + (y-4)^2 = (x-1)^2 + (y+6)^2

\Rightarrow x^2 + 25 -10x + y^2 + 16 -8y = x^2 + 1 - 2x + y^2 +36 + 12y

\Rightarrow 8x+ 20y = 4

\Rightarrow 2x+5y = 1

\Rightarrow y = \frac{1-2x}{5}    … … … … … i)

AB^2 + BC^2 = AC^2

\Rightarrow x-5)^2 + ( y-4)^2 + ( x-1)^2 + (y+6)^2 = (1-5)^2 + ( -6 - 4)^2

\Rightarrow x^2 + 26 - 10x + y^2 + 16 - 8y + x^2 + 1 - 2x + y^2 + 36+ 12y = 16 + 100

\Rightarrow 2x^2 + 2y^2 + 78 - 12x + 4y = 116

\Rightarrow 2x^2 + 2y^2 - 12x + 4y = 38

\Rightarrow x^2 + y^2 -6x + 2y = 19    … … … … … ii)

Substituting i) into ii) we get

x^2 + \Big(  \frac{1-2x}{5} \Big)^2 -6x + 2 \Big(  \frac{1-2x}{5} \Big) = 19

\Rightarrow 25x^2 + 1 + 4x^2 - 4x - 150x + 10 - 20x = 475

\Rightarrow 29x^2 - 174x = 464

\Rightarrow x^2 - 6x = 16

\Rightarrow x^2 - 6x - 16 = 0

\Rightarrow (x-8)(x+2) = 0

\Rightarrow x = 8 \ or \  x = -2

If x = 8   then y = -3

If x = -2 , then y = 1

Hence the other two corners of the square are (8, -3) and ( -2, 1)

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Question 26: Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.

Answer:

Let the points be A(-3, 2), B(-5, -5), C(2,-3) and D(4, 4)

Therefore sides of the figure

AB = \sqrt{(-5+3)^2 + (-5-2)^2 }= \sqrt{4+49} = \sqrt{53}

BC = \sqrt{(2+5)^2 + (-3+5)^2 }= \sqrt{49+4} = \sqrt{53}

CD = \sqrt{(4-2)^2 + (4+3)^2 }= \sqrt{4+49} = \sqrt{53}

DA = \sqrt{(4+3)^2 + (4-2)^2 }= \sqrt{49+4} = \sqrt{53}

Diagonals of the figure

AC= \sqrt{(2+3)^2 + (-3-2)^2 }= \sqrt{25+25} = \sqrt{50} = 5 \sqrt{2}

BD = \sqrt{(4+5)^2 + (4+5)^2 }= \sqrt{81+81} = \sqrt{162} = 9 \sqrt{9}

Hence, AB = BC = CD = CA . Therefore we see that the sides are equal but the diagonals are not equal. Hence ABCD   is a Rhombus.

Area = \frac{1}{2} ( Product of lengths of diagonal)

= \frac{1}{2} 5 \sqrt{2} \times \sqrt{9} = 45 sq. units

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Question 27: Find the coordinates of the circumcenter of the triangle whose vertices are (3, 0), (-1,-6) and (4,-1) . Also, find its circumradius.

Answer:

Given: A ( 3, 0), B ( -1, -6) , C ( 4, -1)

Let O(x, y) be the center

OA^2 = ( x-3)^2 + (y-0)^2

OB^2 = (x+1)^2 +( y + 6)^2

OC^2 = ( x-4)^2 + ( y+1)^2

OA^2 = OB^2

\Rightarrow ( x-3)^2 + (y-0)^2 = (x+1)^2 +( y + 6)^2

\Rightarrow x^2 + 9 - 6x + y^2 = x^2 + 1 + 2x + y^2 + 36 + 12 y

\Rightarrow 9-6x = 1 + 2x + 36 + 12y

\Rightarrow 8x + 12y + 28 = 0

\Rightarrow 2x + 3y + 7 = 0    … … … … … i)

OB^2 = OC^2

\Rightarrow (x+1)^2 +( y + 6)^2 = ( x-4)^2 + ( y+1)^2

\Rightarrow x^2 + 1 + 2x + y^2 + 36 12 y = x^2 + 16 -8x + y + 1 + 2y

\Rightarrow 37 + 2x + 12y = 17 - 8x + 2y

\Rightarrow 10x + 10 y + 20 = 0

\Rightarrow x + y + 2 = 0  … … … … … ii)

Solving i) and ii) we get y = -3

Therefore x = 1

Hence the center is (1, -3)

Therefore Circumradius = \sqrt{ (1-3)^2 + ( -3-0)^2} = \sqrt{13} units

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Question 28: Find a point on the x-axis which is equidistant from the points (7 6) and (-3,4) .

Answer:

Given A ( 7, 6) , B ( -3, 4)

Let the point be O (x, 0)

Given OA = OB \Rightarrow OA^2 = OB^2

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

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

85 - 14x = 6x + 25

20x = 60

x = 3

Therefore point is (3, 0)

Link to the rest of Exercise 19.4

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