Class 9: Co-ordinate Geometry – Exercise 19.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

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

$\displaystyle \text{(i) } (-6,7) \text{ and } (-1, -5)$

$\displaystyle \text{(ii) } (a +b, b + c) \text{ and } (a -b, c -b)$

$\displaystyle \text{(iii) } (a \sin \alpha, - b \cos \alpha) \text{ and } (-a \cos \alpha, b \sin \alpha)$

$\displaystyle \text{(iv) } (a,0) \text{ and } (0,b)$

Answer:

(i) Distance between the given points $\displaystyle (-6,7) \text{ and } (-1, -5)$

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

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

(ii) Distance between the given points $\displaystyle (a +b, b + c) \text{ and } (a -b, c -b)$

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

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

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

$\displaystyle = 2\sqrt{2}b$

(iii) Distance between the given points $\displaystyle (a \sin \alpha, - b \cos \alpha) \text{ and } (-a \cos \alpha, b \sin \alpha)$

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

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

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

(iv) Distance between the given points $\displaystyle (a,0) \text{ and } (0,b)$

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

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

Answer:

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

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

Squaring both sides

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

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

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

$\displaystyle \Rightarrow a = -2, 4$

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

Answer:

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

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

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

$\displaystyle \Rightarrow -2x - 6y = 0$

$\displaystyle \Rightarrow x + 3y = 0$

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

Answer:

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

$\displaystyle 9 + x^2 + 6x + y^2 = 16$

$\displaystyle x^2 + y^2 +6x = 7$ … … … … … i)

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

$\displaystyle 9 + x^2 - 6x + y^2 = 1$

$\displaystyle x^2 + y^2 -6x = 7$ … … … … … ii)

From i) and ii) we get

$\displaystyle x^2 + y^2 +6x = x^2 + y^2 -6x$

$\displaystyle \Rightarrow 12 x = 0 \Rightarrow x = 0$

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

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

Answer:

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

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

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

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

$\displaystyle \Rightarrow y = -9, 3$

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Question 6: Show that the points $\displaystyle (-4,-1), (-2,-4),(4,0) \text{ and } (2,3)$ are the vertices points of a rectangle.

Answer:

Let the points be $\displaystyle A(-4,-1), B(-2,-4), C(4,0) \text{ and } D(2,3)$

Therefore sides of the figure

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

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

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

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

Diagonals of the figure

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

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

Hence, $\displaystyle AB = CD, BC = DA, AC=BD$ .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 $\displaystyle A (1,- 2), B (3, 6), C (5, 10) \text{ and } D (3, 2)$ are the vertices of a parallelogram.

Answer:

The points are $\displaystyle A (1,- 2), B (3, 6), C (5, 10) \text{ and } D (3, 2)$

Therefore sides of the figure

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

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

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

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

Diagonals of the figure

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

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

Hence, $\displaystyle AB = CD, BC = DA, AC \neq BD$ .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 $\displaystyle A(1,7),B(4,2),C(-1,-1) \text{ and } D (4,4)$ are the vertices of a square.

Answer:

The points are $\displaystyle A(1,7),B(4,2),C(-1,-1) \text{ and } D (4,4)$

Therefore sides of the figure

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

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

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

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

Diagonals of the figure

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

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

Hence, $\displaystyle AB = CD = BC = DA, AC = BD$ .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 $\displaystyle (3, 0), (6,4) \text{ and } (- 1, 3)$ are vertices of a right-angled isosceles triangle.

Answer:

Let the points be $\displaystyle A(3, 0), B(6,4) \text{ and } C(- 1, 3)$

Therefore sides of the figure

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

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

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

We see that $\displaystyle AC^2 + AB^2 = BC^2$

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

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Question 10: Prove that $\displaystyle (2, -2), (-2, 1 ) \text{ and } (5, 2)$ 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 $\displaystyle A(2, -2), B(-2, 1 ) \text{ and } C(5, 2)$

Therefore sides of the figure

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

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

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

We see that $\displaystyle AB^2 + CA^2 = BC^2$

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

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

Hypotenuse $\displaystyle = 5\sqrt{2}$

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

Answer:

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

Therefore sides of the figure

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

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

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

We see that $\displaystyle AB=BC=CA$

Therefore $\displaystyle \triangle ABC$ is an equilateral triangle.

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

Answer:

Let the points be $\displaystyle A(2, 3), B(-4, -6) \text{ and } C(1, 3 / 2)$

Therefore sides of the figure

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

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

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

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

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

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

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

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 $\displaystyle (3,4) \text{ and } (-2,3)$ , find the coordinates of the third vertex.

Answer:

Given vertices: $\displaystyle A(3, 4) \text{ and } B ( -2, 3)$

Let the third vertices be $\displaystyle O(x, y)$

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

$\displaystyle OA^2 = ( x+2)^2 + ( y - 3)^2$

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

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

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

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

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

$\displaystyle OA^2 = OB^2$

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

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

$\displaystyle \Rightarrow 10x + 2y = 12$

$\displaystyle \Rightarrow 5x + y = 16$

$\displaystyle \Rightarrow y = ( 6-5x)$

Also $\displaystyle OA^2 = AB^2$

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

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

Substituting

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

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

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

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

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

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

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

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

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

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Question 14: Show that the quadrilateral whose vertices are $\displaystyle (2, -1), (3, 4), (-2,3) \text{ and } (-3, -2)$ is a rhombus.

Answer:

Let the points be $\displaystyle A(2,-1), B(3,4), C(-2,-3) \text{ and } D(-3,-2)$

Therefore sides of the figure

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

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

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

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

Diagonals of the figure

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

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

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

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

Answer:

$\displaystyle \text{Given } A (2, 0), B ( 2, 5)$

Length of equal sides $\displaystyle = 3$

Let the third vertices be $\displaystyle ( x, y)$

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

Therefore equal sides are $\displaystyle AC \text{ and } BC$

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

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

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

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

Subtracting i) from ii)

$\displaystyle \Rightarrow -10y = -25$

$\displaystyle \Rightarrow y = 2.5$

Therefore from i)

$\displaystyle x^2 + ( 2.5)^2 - 4x = 5$

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

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

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

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

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Question 16: Which point on x-axis is equidistant from $\displaystyle (5, 9) \text{ and } (- 4,6)$ ?

Answer:

$\displaystyle \text{Given } (5, 9) \text{ and } ( -4, 6)$

Let $\displaystyle ( x, 0)$ be the equidistant point

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

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

$\displaystyle \Rightarrow 106 - 10x = 52 + 8x$

$\displaystyle \Rightarrow 54 = 18x$

$\displaystyle \Rightarrow x = 3$

Therefore the point is $\displaystyle (3, 0)$

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Question 17: Prove that the points $\displaystyle (- 2,5), (0, 1) \text{ and } (2,-3)$ are collinear.

Answer:

$\displaystyle \text{Given } A ( -2, 5), B(0, 1), C ( 2, -3)$

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

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

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

Therefore $\displaystyle AB + BC = AC$

Hence the points are collinear.

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

Answer:

$\displaystyle \text{Given } P(-3, 2)$ . Let $\displaystyle Q$ be $\displaystyle (x, y)$

$\displaystyle OP = OQ$

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

$\displaystyle 9+4 = x^2 + y^2$

$\displaystyle \Rightarrow x^2 + y^2 = 13$

Now Slope of $\displaystyle OP =$ Slope of $\displaystyle OQ$

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

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

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

Substituting

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

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

$\displaystyle \Rightarrow x^2 = 9$

$\displaystyle \Rightarrow x = \pm 3$

If $\displaystyle x = 3, y = -2$

If $\displaystyle x = -3, y = 2$

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

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Question 19: Which point on y-axis is equidistant from $\displaystyle (2, 3) \text{ and } (4, 1)$ ?

Answer:

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

Let point $\displaystyle C$ be $\displaystyle (0, y)$

$\displaystyle AC = CB$

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

$\displaystyle 4 + y^2 + 9 - 6y = 16 + y^2 + 1 - 2y$

$\displaystyle \Rightarrow 13 - 6y = 17 - 2$

$\displaystyle \Rightarrow -4-4y = 0$

$\displaystyle \Rightarrow y = -1$

Hence the point is $\displaystyle (0, -1)$

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Question 20: The three vertices of a parallelogram are $\displaystyle (3, 4), (3,8) \text{ and } (9, 8)$ . Find the fourth vertex.

Answer:

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

Let $\displaystyle D$ be $\displaystyle (x, y)$

Since $\displaystyle ABCD$ is a parallelogram

$\displaystyle AB = CD$

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

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

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

Also $\displaystyle AD = BC$

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

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

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

Slope of $\displaystyle AB =$ Slope of $\displaystyle CD$

$\displaystyle \frac{8-4}{3-3} = \frac{8-y}{9-x}$

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

$\displaystyle \Rightarrow x = 9$

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

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

$\displaystyle \Rightarrow (y - 4)^2 = 0$

$\displaystyle \Rightarrow y = 4$

Hence $\displaystyle D(9, 4)$

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

Answer:

$\displaystyle \text{Given } A(-2, -3), B(-1, 0), C(7, -6)$

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

$\displaystyle OA^2 = (x+2)^2 + ( y+3)^2$

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

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

$\displaystyle OA^2 = OB^2$

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

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

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

$\displaystyle \Rightarrow 2x+ 6y = -12$

$\displaystyle \Rightarrow x = -3y -6$ … … … … … i)

$\displaystyle OB^2 = OC^2$

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

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

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

$\displaystyle \Rightarrow 16x -12y = 84$

$\displaystyle \Rightarrow 4x-3y=21$ … … … … … ii)

Substituting i) in ii)

$\displaystyle 4(-3y-6) - 3y = 21$

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

$\displaystyle \Rightarrow -15 y = 45$

$\displaystyle \Rightarrow y = -3$

Hence $\displaystyle x = -3(-3) - 6 =9-6=3$

Therefore circumcenter $\displaystyle O$ is $\displaystyle ( 3 -3)$

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

Answer:

Given points $\displaystyle A(0, 100) \text{ and } B(10, 0)$

Point $\displaystyle A$ is on y axis and point $\displaystyle B$ is on x axis.

Therefore angle subtended by $\displaystyle A \text{ and } B$ on origin is $\displaystyle 90^o$

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Question 23: Find the center of the circle passing through $\displaystyle (5, -8), (2,- 9) \text{ and } (2, 1)$ .

Answer:

$\displaystyle \text{Given } A(5, -8), B(2, -9), C(2, 1)$

Let center be $\displaystyle O(x, y)$ . Therefore

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

$\displaystyle OB^2 = (x-2)^2 + ( y+9)^2$

$\displaystyle OC^2 = (x-2)^2 + (y-1)^2$

$\displaystyle OA^2 = OB^2$

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

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

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

$\displaystyle \Rightarrow 6x+2y = 4$

$\displaystyle \Rightarrow 3x+y= 2$ … … … … … i)

$\displaystyle OB^2 = OC^2$

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

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

$\displaystyle \Rightarrow 20y = -80$

$\displaystyle \Rightarrow y = -4$

Substituting in i) we get

$\displaystyle 3x = 2 - (-4) = 6$

$\displaystyle \Rightarrow x = 2$

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

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Question 24: Find the value of $\displaystyle k$ , if the point $\displaystyle P (0,2)$ is equidistant from $\displaystyle (3, k) \text{ and } (k,5)$ .

Answer:

$\displaystyle \text{Given } P(0, 2), A(3, k), B(k, 5)$

Since $\displaystyle PA = PB$

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

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

$\displaystyle \Rightarrow 4-4k=0$

$\displaystyle \Rightarrow k = 1$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } A(5,4 ), C(1, -6)$ .

Let $\displaystyle B$ be $\displaystyle (x, y)$

Since $\displaystyle ABCD$ is a square, $\displaystyle AB = BC$

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

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

$\displaystyle \Rightarrow 8x+ 20y = 4$

$\displaystyle \Rightarrow 2x+5y = 1$

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

$\displaystyle AB^2 + BC^2 = AC^2$

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

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

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

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

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

Substituting i) into ii) we get

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

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

$\displaystyle \Rightarrow 29x^2 - 174x = 464$

$\displaystyle \Rightarrow x^2 - 6x = 16$

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

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

$\displaystyle \Rightarrow x = 8 \ or \ x = -2$

If $\displaystyle x = 8$ then $\displaystyle y = -3$

If $\displaystyle x = -2$ , then $\displaystyle y = 1$

Hence the other two corners of the square are $\displaystyle (8, -3) \text{ and } ( -2, 1)$

$\displaystyle \\$

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

Answer:

Let the points be $\displaystyle A(-3, 2), B(-5, -5), C(2,-3) \text{ and } D(4, 4)$

Therefore sides of the figure

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

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

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

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

Diagonals of the figure

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

Let $\displaystyle O(x, y)$ be the center

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

$\displaystyle OB^2 = (x+1)^2 +( y + 6)^2$

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

$\displaystyle OA^2 = OB^2$

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

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

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

$\displaystyle \Rightarrow 8x + 12y + 28 = 0$

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

$\displaystyle OB^2 = OC^2$

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

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

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

$\displaystyle \Rightarrow 10x + 10 y + 20 = 0$

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

Solving i) and ii) we get $\displaystyle y = -3$

Therefore $\displaystyle x = 1$

Hence the center is $\displaystyle (1, -3)$

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } A ( 7, 6) , B ( -3, 4)$

Let the point be $\displaystyle O (x, 0)$

$\displaystyle \text{Given } OA = OB \Rightarrow OA^2 = OB^2$

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

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

$\displaystyle 85 - 14x = 6x + 25$

$\displaystyle 20x = 60$

$\displaystyle x = 3$

Therefore point is $\displaystyle (3, 0)$

$\displaystyle \\$

Question 29: (i) show that the points $\displaystyle A(5,6), 8(1, 5), C(2,1) \text{ and } D(6,2)$ are the vertices of a square.

(ii) Prove that the points $\displaystyle A (2,3), B(-2,2),C(-1,-2)$ , and $\displaystyle D(3,-1)$ are the vertices of a square $\displaystyle ABCD$ .

(iii) Name the type of $\displaystyle \triangle PQR$ formed by the points $\displaystyle P(\sqrt{2}, \sqrt{2}), Q( -\sqrt{2}, -\sqrt{2})$ , and $\displaystyle R(-\sqrt{6}, \sqrt{6})$

Answer:

i) Let the points be $\displaystyle A(5,6), B(1, 5), C(2, 1) \text{ and } D(6, 2)$

Therefore sides of the figure

$\displaystyle AB = \sqrt{(5-1)^2 + (6-5)^2 }= \sqrt{16+1} = \sqrt{17}$

$\displaystyle BC = \sqrt{(1-2)^2 + (5-1)^2 }= \sqrt{1+16} = \sqrt{17}$

$\displaystyle CD = \sqrt{(2-6)^2 + (1-2)^2 }= \sqrt{16+1} = \sqrt{17}$

$\displaystyle DA = \sqrt{(6-5)^2 + (2-6)^2 }= \sqrt{1+16} = \sqrt{17}$

Diagonals of the figure

$\displaystyle AC= \sqrt{(5-2)^2 + (6-1)^2 }= \sqrt{9+25} = \sqrt{34}$

$\displaystyle BD = \sqrt{(1-6)^2 + (5-2)^2 }= \sqrt{25+9} = \sqrt{34}$

Hence, $\displaystyle AB = BC=CD=DA$ . Also diagonals are equal. Therefore the figure is a square.

ii) Let the points be $\displaystyle A(2,3), B(-2,2), C(-1,-2) \text{ and } D(3, -1)$

Therefore sides of the figure

$\displaystyle AB = \sqrt{(2+2)^2 + (3-2)^2 }= \sqrt{16+1} = \sqrt{17}$

$\displaystyle BC = \sqrt{(-2+1)^2 + (2+2)^2 }= \sqrt{1+16} = \sqrt{17}$

$\displaystyle CD = \sqrt{(-1-3)^2 + (-2+1)^2 }= \sqrt{16+1} = \sqrt{17}$

$\displaystyle DA = \sqrt{(3-2)^2 + (-1-3)^2 }= \sqrt{1+16} = \sqrt{17}$

Diagonals of the figure

$\displaystyle AC= \sqrt{(2+1)^2 + (3+2)^2 }= \sqrt{9+25} = \sqrt{34}$

$\displaystyle BD = \sqrt{(-2-3)^2 + (2+1)^2 }= \sqrt{25+9} = \sqrt{34}$

Hence, $\displaystyle AB = BC=CD=DA$ . Also diagonals are equal. Therefore the figure is a square.

iii) Let the points be $\displaystyle P(\sqrt{2}, \sqrt{2}), Q( -\sqrt{2}, -\sqrt{2}), R(-\sqrt{6}, \sqrt{6})$

Therefore sides of the figure

$\displaystyle PQ = \sqrt{(\sqrt{2} + \sqrt{2})^2 + (\sqrt{2} + \sqrt{2})^2 }= \sqrt{8+8} = \sqrt{16}$

$\displaystyle QR = \sqrt{(-\sqrt{2} + \sqrt{6})^2 + (-\sqrt{2} - \sqrt{6})^2 }= \sqrt{2+ 6 - 2 \sqrt{12}+ 2 + 6 + 2\sqrt{12}} = \sqrt{16}$

$\displaystyle RP = \sqrt{(-\sqrt{6} - \sqrt{2})^2 + (\sqrt{6} - \sqrt{2})^2 }= \sqrt{6+ 2 + 2 \sqrt{12}+ 6 + 2 - 2\sqrt{12}} = \sqrt{16}$

Hence, $\displaystyle PQ = QR = RP$ . Therefore the figure is an equilateral triangle.

$\displaystyle \\$

Question 30: Find the point on x-axis which is equidistant from the points $\displaystyle (-2, 5) \text{ and } (2,-3)$ .

Answer:

$\displaystyle \text{Given } A ( -2, 5), B ( 2, -3)$

Let the equidistant point be $\displaystyle O(x, 0)$

$\displaystyle OA = OB \Rightarrow OA^2 = OB^2$

$\displaystyle \Rightarrow (x+2)^2 + ( 0-5)^2 = (x-2)^2 + (0+3)^2$

$\displaystyle \Rightarrow x^2 + 4 + 4x + 25 = x^2 + 4 - 4x + 9$

$\displaystyle \Rightarrow 8x = -29 + 13$

$\displaystyle \Rightarrow 8x = -16$

$\displaystyle \Rightarrow x = -2$

Therefore point is $\displaystyle ( -2, 0)$

$\displaystyle \\$

Question 31: Find the value of $\displaystyle x$ such that $\displaystyle PQ = QR$ where the coordinates of $\displaystyle P, Q \text{ and } R$ are $\displaystyle (6,-1), (1,3) \text{ and } (x,8)$ respectively.

Answer:

Given: $\displaystyle P(6, -1), Q(1, 3), R(x, -8)$

$\displaystyle PQ = QR \Rightarrow PQ^2 = QR^2$

$\displaystyle \Rightarrow (1-6)^2 + ( 3+1)^2 = ( x-1)^2 + ( 8-3)^2$

$\displaystyle \Rightarrow 25 + 16 = x^2 + 1 - 2x + 25$

$\displaystyle \Rightarrow x^2 - 2x - 15 = 0$

$\displaystyle \Rightarrow (x-5)(x+3) = 0$

$\displaystyle \Rightarrow x = 5 \ or \ -3$

$\displaystyle \\$

Question 32: Prove, that the points $\displaystyle (0, 0), (5,5) \text{ and } (-5, 5)$ are the vertices of a right isosceles triangle.

Answer:

Let the points be $\displaystyle A(0,0), B(5,5), C(-5, 5)$

Therefore sides of the figure

$\displaystyle AB = \sqrt{(5-0)^2 + (5-0)^2 }= \sqrt{25+25} = \sqrt{50}$

$\displaystyle BC = \sqrt{(-5-5)^2 + (5-5)^2 }= \sqrt{100}$

$\displaystyle CA = \sqrt{(-5-0)^2 + (5-0)^2 }= \sqrt{25+25} = \sqrt{50}$

Since $\displaystyle AB = CA, \triangle ABC$ is an Isosceles triangle.

$\displaystyle \\$

Question 33: If the point $\displaystyle P(x, y)$ is equidistant from the points $\displaystyle A(5,1) \text{ and } B(1,5)$ , prove that $\displaystyle x=y$ .

Answer:

Given: $\displaystyle A(5, 1), B(1, 5) , P(x, y)$

$\displaystyle AP = BP \Rightarrow AP^2 = BP^2$

$\displaystyle \Rightarrow ( x-5)^2 + ( y - 1)^2 = (x-1)^2 + ( y - 5)^2$

$\displaystyle \Rightarrow x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 1 - 2x + y^2 + 25 - 10y$

$\displaystyle \Rightarrow 26 - 10x - 2y = 26 - 2x - 10y$

$\displaystyle \Rightarrow -8x = - 8y$

$\displaystyle \Rightarrow x = y$

$\displaystyle \\$

Question 34: If $\displaystyle Q (0, 1 )$ is equidistant from $\displaystyle P (5, -3) \text{ and } R (x, 6)$ , find the values of $\displaystyle x$ . Also, find the distances $\displaystyle QR \text{ and } PR$ .

Answer:

Given: $\displaystyle P(5, -3), R(x, 6) , Q (0, 1)$

$\displaystyle QP = QR \Rightarrow QP^2 = QR^2$

$\displaystyle \Rightarrow (0-5)^2 + ( 1+3)^2 = (0-x)^2 + ( 1-6)^2$

$\displaystyle \Rightarrow 25 + 16 = x^2 + 25$

$\displaystyle \Rightarrow x = \pm 4$

When $\displaystyle x = 4, QR = \sqrt{4^2 + 25} = \sqrt{16 + 25} = \sqrt{41}$ units

When $\displaystyle x = -4, QR = \sqrt{(-4)^2 + 25} = \sqrt{16 + 25} = \sqrt{41}$ units

$\displaystyle \\$

Question 35: Find the values of $\displaystyle y$ for which the distance between the points $\displaystyle P (2, -3) \text{ and } Q (10, y)$ is $\displaystyle 10$ units.

Answer:

Given: $\displaystyle P(2, -3), Q(10, y) , PQ =10$

$\displaystyle (10-2)^2 + (y+3)^2 = 10^2$

$\displaystyle \Rightarrow 64 + ( y + 3)^2 = 100$

$\displaystyle \Rightarrow ( y + 3) ^2 = 36$

$\displaystyle \Rightarrow y + 3 = \pm 6$

$\displaystyle \Rightarrow y = 3, -9$

$\displaystyle \\$

Question 36: Find the center of the circle passing through $\displaystyle (6, -6), (3,-7) \text{ and } (3, 3)$ .

Answer:

Given: $\displaystyle A ( 6, -6), B(3, -7), C ( 3,3)$

Let center be $\displaystyle O(x, y)$

$\displaystyle OA^2 = ( x-6)^2 + ( y+6)^2$

$\displaystyle OB^2 = ( x-3)^2 + ( y+7)^2$

$\displaystyle OC^2 = ( x-3)^2 + ( y-3)^2$

$\displaystyle OA^2 = OB^2$

$\displaystyle \Rightarrow ( x-6)^2 + ( y+6)^2 = ( x-3)^2 + ( y+7)^2$

$\displaystyle \Rightarrow x^2 + 36 - 12 x + y^2 + 36 + 12 y = x^2 + 9 - 6x + y^2 + 49 + 14 y$

$\displaystyle \Rightarrow 72-12x+12y = 58 - 6x +14y$

$\displaystyle \Rightarrow 14 = 6x + 2y$

$\displaystyle \Rightarrow 3x+y = 7$ … … … … … i)

$\displaystyle OB^2 = OC^2$

$\displaystyle \Rightarrow ( x-3)^2 + ( y+7)^2 = ( x-3)^2 + ( y-3)^2$

$\displaystyle \Rightarrow y^2 + 49 + 14y = y^2 +9 - 16y$

$\displaystyle \Rightarrow 20y = - 40$

$\displaystyle \Rightarrow y = -2$ … … … … … ii)

Substituting in i)

$\displaystyle 3x + ( -2) = 7$

$\displaystyle \Rightarrow 3x = 9$

$\displaystyle \Rightarrow x = 3$

Therefore center is $\displaystyle ( 3, -2)$

$\displaystyle \\$

Question 37: Two opposite vertices of a square are $\displaystyle (-1,2) \text{ and } (3,2)$ . Find the coordinates of other two vertices.

Answer:

$\displaystyle \text{Given } A ( -1, 2), C(3, 2) , B(x, y)$

$\displaystyle AB^2 = BC^2$

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

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

$\displaystyle \Rightarrow 1+2x = 9-6x$

$\displaystyle \Rightarrow 8x = 8$

$\displaystyle \Rightarrow x= 1$ … … … … … i)

Also

$\displaystyle AB^2 + BC^2 = AC^2$

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

$\displaystyle \Rightarrow x^2 + 1 + 2x + y^2 + 4 - 4y + x^2 + 9 - 6x + y^2 + 4 - 4y = 16$

$\displaystyle \Rightarrow 2x^2 + 2y^2 - 4x - 8y + 2 = 0$

From i), $\displaystyle x = 1$

$\displaystyle \Rightarrow 2 + 2y^2 - 4 - 8y + 2 = 0$

$\displaystyle \Rightarrow 2y^2 - 8y= 0$

$\displaystyle \Rightarrow y^2 - 4y=0$

$\displaystyle \Rightarrow y( y - 4) = 0$

$\displaystyle \Rightarrow y = 0 \ or \ y = 4$

Hence the coordinates are $\displaystyle B( 1, 4), D(1, 0)$

$\displaystyle \\$

Question 38: Name the quadrilateral formed, if any, by the following points, and give reasons for your answers:

$\displaystyle \text{(i) } A (-1, - 2), B(1, 0), C (-1, 2),D (-3, 0)$

$\displaystyle \text{(ii) } A (-3, 5), B (3, 1),C (0, 3), D (-1, -4)$

$\displaystyle \text{(iii) } A (4,5), B (7 ,6), C (4,3), D (1,2)$

Answer:

i) Let the points be $\displaystyle A(-1, -2), B(1,0), C(-1,2) \text{ and } D(-3,0)$

Therefore sides of the figure

$\displaystyle AB = \sqrt{(1+1)^2 + (0+2)^2 }= \sqrt{4+4} = \sqrt{8}$

$\displaystyle BC = \sqrt{(-1-1)^2 + (2-0)^2 }= \sqrt{4+4} = \sqrt{8}$

$\displaystyle CD = \sqrt{(-3+1)^2 + (0-2)^2 }= \sqrt{4+4} = \sqrt{8}$

$\displaystyle DA = \sqrt{(-3+1)^2 + (0+2)^2 }= \sqrt{4+4} = \sqrt{8}$

Diagonals of the figure

$\displaystyle AC= \sqrt{(-1+1)^2 + (2+2)^2 }= \sqrt{16} = 4$

$\displaystyle BD = \sqrt{(-3-1)^2 + (0-0)^2 }= \sqrt{16} = 4$

All four sides are equal and also the diagonals are equal. Hence this is a square.

ii) Let the points be $\displaystyle A(-3, 5), B(3, 1), C(0, 3) \text{ and } D(-1,-4)$

Therefore sides of the figure

$\displaystyle AB = \sqrt{(3+3)^2 + (1-5)^2 }= \sqrt{81+16} = \sqrt{97}$

$\displaystyle BC = \sqrt{(0-3)^2 + (3-1)^2 }= \sqrt{9+4} = \sqrt{13}$

$\displaystyle CD = \sqrt{(-1-0)^2 + (-4-3)^2 }= \sqrt{1+49} = \sqrt{50}$

$\displaystyle DA = \sqrt{(-1+3)^2 + (-4-5)^2 }= \sqrt{4+81} = \sqrt{85}$

All four sides are unequal . Hence this is a quadrilateral.

iii) Let the points be $\displaystyle A(4,5), B(7,6), C(4,3) \text{ and } D(1,2)$

Therefore sides of the figure

$\displaystyle AB = \sqrt{(7-4)^2 + (6-5)^2 }= \sqrt{9+1} = \sqrt{10}$

$\displaystyle BC = \sqrt{(4-7)^2 + (3-6)^2 }= \sqrt{9+9} = \sqrt{18}$

$\displaystyle CD = \sqrt{(1-4)^2 + (2-3)^2 }= \sqrt{9+1} = \sqrt{10}$

$\displaystyle DA = \sqrt{(1-4)^2 + (2-5)^2 }= \sqrt{9+9} = \sqrt{18}$

Diagonals of the figure

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

$\displaystyle BD = \sqrt{(1-7)^2 + (2-6)^2 }= \sqrt{36+16} = \sqrt{52}$

All four sides are equal and but the diagonals are unequal, this is a Rhombus.

$\displaystyle \\$

Question 39: Find the equation of the perpendicular bisector of the line segment joining points $\displaystyle (7,1) \text{ and } (3,5)$ .

Answer:

Given: $\displaystyle A ( 7, 1), B ( 3, 5)$

Mid point of $\displaystyle AB = \Big( \frac{7+3}{2} , \frac{1+5}{2} \Big) = (5, 3)$

Slope of $\displaystyle AB = \frac{5-1}{3-7} = \frac{4}{-4} = -1$

Therefore slope of perpendicular bisector $\displaystyle = 1$

Therefore equation of perpendicular bisector is

$\displaystyle y - 3 = 1 ( x - 5)$

$\displaystyle \Rightarrow y = x - 2$

$\displaystyle \\$

Question 40: Prove that the points $\displaystyle (3, 0), (4, 5), (-1, 4) \text{ and } (-2, -1)$ , taken in order, form a rhombus. Also, find its area.

Answer:

Let the points be $\displaystyle A(3,0), B(4,5), C(-1,4) \text{ and } D(-2,-1)$

Therefore sides of the figure

$\displaystyle AB = \sqrt{(4-3)^2 + (5-0)^2 }= \sqrt{1+25} = \sqrt{26}$

$\displaystyle BC = \sqrt{(-1-4)^2 + (4-5)^2 }= \sqrt{25+1} = \sqrt{26}$

$\displaystyle CD = \sqrt{(-2+1)^2 + (-1-4)^2 }= \sqrt{1+25} = \sqrt{26}$

$\displaystyle DA = \sqrt{(-2-3)^2 + (-1-0)^2 }= \sqrt{25+1} = \sqrt{26}$

Diagonals of the figure

$\displaystyle AC= \sqrt{(-1-3)^2 + (4-0)^2 }= \sqrt{16+16} = 4\sqrt{2}$

$\displaystyle BD = \sqrt{(-2-4)^2 + (-1-5)^2 }= \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$

All four sides are equal and but the diagonals are unequal, this is a Rhombus.

Area of the Rhombus $\displaystyle = \frac{1}{2}$ (product of diagonals)

$\displaystyle = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = 24$ sq. units.

$\displaystyle \\$

Question 41: In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at $\displaystyle A (3, 1), B (6,4) \text{ and } C (8, 6)$ . Do you think they are seated in a line?

Answer:

Given: $\displaystyle A ( 3, 1), B ( 6, 4) , C(8, 6)$

$\displaystyle AB = \sqrt{(6-3)^2 + (4-1)^2 }= \sqrt{9+9} = 3\sqrt{2}$

$\displaystyle BC = \sqrt{(8-6)^2 + (6-4)^2 }= \sqrt{4+4} = 2\sqrt{2}$

$\displaystyle CA = \sqrt{(3-8)^2 + (1-6)^2 }= \sqrt{25+25} = 5\sqrt{2}$

$\displaystyle \Rightarrow AB + BC = CA$

Hence they are all in a straight line.

$\displaystyle \\$

Question 42: Find a point on y-axis which is equidistant from the points $\displaystyle (5 , - 2) \text{ and } (- 3, 2)$ .

Answer:

Given: $\displaystyle A ( 5, -2), B ( -3, 2)$

Let the equidistant point be $\displaystyle O(0, y)$

$\displaystyle \text{Given } OA = OB \Rightarrow OA^2 = OB^2$

Therefore $\displaystyle (0-5)^2 + ( y + 2)^2 = ( 0+3)^2 + ( y - 2)^2$

$\displaystyle \Rightarrow 25 + y^2 + 4 + 4y = 9 + y^2 + 4 - 4y$

$\displaystyle \Rightarrow 29 + 4y = 13 - 4y$

$\displaystyle \Rightarrow 8y = - 16$

$\displaystyle \Rightarrow y = -2$

Therefore the point is $\displaystyle (0, -2)$

$\displaystyle \\$

Question 43: Find a relation between $\displaystyle x \text{ and } y$ such that the point $\displaystyle (x, y)$ is equidistant from the points $\displaystyle (3,6) \text{ and } (-3,4)$ .

Answer:

$\displaystyle \text{Given } A ( 3, 6), B ( -3, 4)$

Let equidistant point $\displaystyle O(x, y)$

$\displaystyle \text{Given } OA = OB \Rightarrow OA^2 = OB^2$

$\displaystyle \Rightarrow (x-3)^2 + ( y - 6)^2 = (x+3)^2 + ( y -4)^2$

$\displaystyle \Rightarrow x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 - 4y$

$\displaystyle \Rightarrow 45 - 6x - 12y = 25 + 6x - 8y$

$\displaystyle \Rightarrow 12x + 4y = 20$

$\displaystyle \Rightarrow 3x + y = 5$

$\displaystyle \\$

Question 44: If a point $\displaystyle A(0,2)$ is equidistant from the points $\displaystyle B (3,p) \text{ and } C(p,5)$ , then find the value of $\displaystyle p$ .

Answer:

Given: $\displaystyle A(0, 2), B(3, p), C(p, 5)$

Since $\displaystyle AB = AC \Rightarrow AB^2 = AC^2$

$\displaystyle \Rightarrow (3-0)^2 + ( p-2)^2 = ( p - 0)^2 + ( 5-2)^2$

$\displaystyle \Rightarrow 9+p^2 +4 - 4p = p^2 + 9$

$\displaystyle \Rightarrow 13-4p = 9$

$\displaystyle \Rightarrow 4p = 4$

$\displaystyle \Rightarrow p = 1$

$\displaystyle \\$

Question 45: Prove that the points $\displaystyle (7, 10), (-2,5) \text{ and } (3, -4)$ are the vertices of an isosceles right triangle.

Answer:

Given: $\displaystyle A ( 7, 10), B ( -2, 5) , C(3, -4)$

$\displaystyle AB = \sqrt{(-2-7)^2 + (5-10)^2 }= \sqrt{81+25} = \sqrt{106}$

$\displaystyle BC = \sqrt{(3+2)^2 + (-4-5)^2 }= \sqrt{25+81} = \sqrt{106}$

$\displaystyle CA = \sqrt{(7-3)^2 + (10+4)^2 }= \sqrt{16+196} = \sqrt{212}$

Therefore $\displaystyle AB = AC$ ( i.e. two sides of the triangle are equal)

Also $\displaystyle AB^2 + BC^2 = AC^2$

$\displaystyle \Rightarrow 106 + 106 = 212$

Therefore $\displaystyle \triangle ABC$ is a right angled isosceles triangle.

$\displaystyle \\$

Question 46: If the point $\displaystyle P(x, 3)$ is equidistant from the points $\displaystyle A(7,-1) \text{ and } B(6,8)$ , find the value of $\displaystyle x$ and find the distance $\displaystyle AP$ .

Answer:

Give: $\displaystyle P(x, 3), A ( 7, -1), B(6, 8)$

Since $\displaystyle PA = PB \Rightarrow PA^2 = PB^2$

$\displaystyle \Rightarrow (7-x)^2 + ( -1 -3)^2 = (6-x)^2 + (8-3)^2$

$\displaystyle \Rightarrow 49 + x^2 - 14x + 16 = 36 + x^2 - 12x + 25$

$\displaystyle 65 - 14x = 61 - 12x$

$\displaystyle 4 = 2x \ or \ x = 2$

Hence the point is $\displaystyle (2, 3)$

$\displaystyle AP = \sqrt{(7-2)^2 + ( -1-3)^2} = \sqrt{25+16} = \sqrt{41}$

$\displaystyle \\$

Question 47: It $\displaystyle A(3,y)$ is equidistant from points $\displaystyle P (8,-3) \text{ and } Q(7,6)$ , find the value of $\displaystyle y$ and find the distance $\displaystyle AP$ .

Answer:

Given: $\displaystyle A ( 3, y), P(8, -3) , Q(7,6)$

Since $\displaystyle AP = AQ \Rightarrow AP^2 = AQ^2$

$\displaystyle \Rightarrow (8-3)^2 + ( -3-y)^2 = ( 7-3)^2 + ( 6-y)^2$

$\displaystyle \Rightarrow 25 + 9 + y^2 + 6y = 16 + 36 + y^2 - 12y$

$\displaystyle \Rightarrow 34 + 6y = 52 - 12y$

$\displaystyle \Rightarrow 18 y = 18$

$\displaystyle \Rightarrow y = 1$

Therefore $\displaystyle A ( 3, 1)$

$\displaystyle AP = \sqrt{(8-3)^2 + ( -3-1)^2} = \sqrt{25+16} = \sqrt{41}$

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Question 48: If $\displaystyle (0, - 3) \text{ and } (0, 3)$ are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

Answer:

Given: $\displaystyle A (0, -3), B(0, 3)$

Let third vertices be $\displaystyle C(x, y)$

$\displaystyle AC^2 = ( x-0)^2 + (y+3)^2$

$\displaystyle AB^2 = ( 0-0)^2 + ( 3+3)^2 = 36$

$\displaystyle BC^2 = (x-0)^2 + (y-3)^2$

Since $\displaystyle AC^2 = AB^2$

$\displaystyle \Rightarrow x^2 + (y+3)^2 = 36$

$\displaystyle \Rightarrow x^2 + y^2 + 9 + 6y = 36$

or $\displaystyle \Rightarrow x^2 + y^2 + 6y = 27$ … … … … … i)

$\displaystyle AC^2 = BC^2$

$\displaystyle \Rightarrow x^2 + (y+3)^2 = x^2 + ( y-3)^2$

$\displaystyle \Rightarrow x^2 + y^2 +9 + 6y = x^2 + y^2 + 9 - 6y$

$\displaystyle \Rightarrow 9 + 6y = 9 - 6y$

$\displaystyle \Rightarrow 12y = 0$

$\displaystyle \Rightarrow y = 0$

Substituting in i)

$\displaystyle x^2 = 27 \Rightarrow x = \pm 3\sqrt{3}$

Therefore $\displaystyle C$ could be $\displaystyle ( 3\sqrt{3}, 0)$ or $\displaystyle (- 3\sqrt{3}, 0)$

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Question 49: If the point $\displaystyle P (2, 2)$ is equidistant from the point s $\displaystyle A (-2, k) \text{ and } B (-2k, 3)$ , find $\displaystyle k$ . Also, find the length of $\displaystyle AP$ .

Answer:

Given: $\displaystyle P(2, 2), A ( -2, k), B ( -2k, -3)$

$\displaystyle AP^2 = ( -2-2)^2 + ( k-2)^2 = 16 + ( k-2)^2$

$\displaystyle BP^2 = (-2k-2)^2 + ( -3-2)^2 = ( -2k-2)^2 + 25$

$\displaystyle AP^2 = BP^2$

$\displaystyle \Rightarrow 16 + ( k-2)^2 = ( -2k-2)^2 + 25$

$\displaystyle \Rightarrow k^2 + 4 - 4k = 9 + 4k^2 + 4 + 8k$

$\displaystyle \Rightarrow 3k^2 + 12k + 9 = 0$

$\displaystyle \Rightarrow k^2 + 4k + 3 = 0$

$\displaystyle \Rightarrow (k+3)(k+1)=0$

$\displaystyle \Rightarrow k = -3 \ or \ k = -1$

If $\displaystyle k = -3 \Rightarrow AP = \sqrt{(-2-2)^2+(-3-2)^2} = \sqrt{16+25} = \sqrt{41}$

If $\displaystyle k = -1 \Rightarrow AP = \sqrt{(-2-2)^2+(-1-2)^2} = \sqrt{16+9} = \sqrt{25} = 5$

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Question 50: If the point $\displaystyle A(0,2)$ is equidistant from the points $\displaystyle B(3,p) \text{ and } C(p,5)$ , find $\displaystyle p$ . Also, find the length of $\displaystyle AB$ .

Answer:

Given: $\displaystyle A(0, 2), B(3, p), C(p, 5)$

$\displaystyle AB^2 = ( 3-0)^2 + ( p-2)^2$

$\displaystyle AC^2 = (p-0)^2 + ( 5-2)^2$

$\displaystyle AB^2 = AC^2$

$\displaystyle (3-0)^2 + ( p - 2)^2 = ( p-0)^2 + (5-2)^2$

$\displaystyle 9+p^2+4 - 4p = p^2 + 9$

$\displaystyle 4-4p = 0$

$\displaystyle p=1$

Therefore $\displaystyle AB = \sqrt{9 + ( 1-2)^2 } = \sqrt{10}$

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Question 51: If the point $\displaystyle P (k -1,2)$ is equidistant from the points $\displaystyle A(3,k) \text{ and } B(k, 5)$ , find the values of $\displaystyle k$ .

Answer:

Given: $\displaystyle P(k-1, 2), A(3, k), B(k, 5)$

$\displaystyle AP^2 = ( 3-k+1)^2 + ( k-2)^2 = (4-k)^2 + ( k-2)^2$

$\displaystyle BP^2 = (k-k+1)^2 + ( 5-2)^2 = 1+9 = 10$

$\displaystyle AP^2 = BP^2$

$\displaystyle \Rightarrow (4-k)^2 + ( k-2)^2 =10$

$\displaystyle \Rightarrow 16 +k^2 -8k + k^2 + 4 - 4k = 10$

$\displaystyle \Rightarrow 2k^2 - 12k + 20 = 0$

$\displaystyle \Rightarrow k^2 -6k + 5 = 0$

$\displaystyle \Rightarrow (k-1)(k-5) = 0$

$\displaystyle \Rightarrow k = 1 \ or \ k = 5$

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Question 52: If $\displaystyle (-4,3) \text{ and } (4,3)$ are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the (i) interior, (ii) exterior of the triangle.

Answer:

Given: $\displaystyle A (-4, 3), B(4, 3)$

Let third vertices be $\displaystyle C(x, y)$

$\displaystyle AB^2 = (4+4)^2 + ( 3- 3)^2 = 64$

$\displaystyle BC^2 = ( x-4)^2 + ( y - 3)^2$

$\displaystyle AC^2 = ( x+ 4)^2 + ( y - 3)^2$

$\displaystyle BC^2 = AC^2$

$\displaystyle \Rightarrow ( x-4)^2 + ( y - 3)^2 = ( x+ 4)^2 + ( y - 3)^2$

$\displaystyle \Rightarrow x^2 + 16 - 8x = x^2 + 16 + 8x$

$\displaystyle \Rightarrow 16x = 0$

$\displaystyle \Rightarrow x = 0$

$\displaystyle BC^2 = AB^2$

$\displaystyle \Rightarrow ( x-4)^2 + ( y - 3)^2 = 64$

$\displaystyle \Rightarrow 16 + (y-3)^2 = 64$

$\displaystyle \Rightarrow (y-3)^2 = 48$

$\displaystyle \Rightarrow y - 3 = \pm 4\sqrt{3}$

$\displaystyle \Rightarrow y = 3 \pm 4\sqrt{3}$

Hence $\displaystyle C$ is i) $\displaystyle (0, 3 - 4\sqrt{3})$ and ii) $\displaystyle ( 0, 3+4\sqrt{3})$

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Question 53: Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance traveled by Ayush in reaching the office? (Assume that all distances covered are in straight hires). If the house is situated at $\displaystyle (2,4)$ , bank at $\displaystyle (5,8),$ school at $\displaystyle (13,14)$ and office at $\displaystyle (13, 26)$ and coordinates are in kilometers.

Answer:

Given: House $\displaystyle ( 2, 4)$ , Bank $\displaystyle ( 5, 8)$ , School $\displaystyle ( 13, 14)$ , Office $\displaystyle (13, 26)$

Distance from House to Office $\displaystyle = \sqrt{(13-2)^2 + (26-4)^2} = \sqrt{121+ 484} = \sqrt{605} = 24.6$ km

Distance of House to Bank $\displaystyle = \sqrt{(5-2)^2 + ( 8-4)^2} = \sqrt{9+16} = 5$ km

Distance of Bank to School $\displaystyle = \sqrt{(13-5)^2+ (14-8)^2} = \sqrt{64+36} = 10$ km

Distance from school to office $\displaystyle = \sqrt{(13-13)^2 + (26-14)^2} = \sqrt{144} = 12$ km

Therefore total distance traveled $\displaystyle = 5 + 10 + 12 = 27$ km

Extra distance $\displaystyle = 27 - 24.6 = 2.4$ km

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Question 54: The center of a circle is $\displaystyle (2a, 1)$ . Find the values of $\displaystyle a$ if the circle passes through the point $\displaystyle (11,-a)$ and has diameter $\displaystyle 10\sqrt{2}$ units.

Answer:

Given: $\displaystyle O(2a, 1), A(11, -a)$

Radius $\displaystyle = 5\sqrt{2}$

Therefore $\displaystyle (5\sqrt{2})^2 = (11-2a)^2 + ( -a -1)^2$

$\displaystyle 50 = 121 + 4a^2 - 44 a + a^2 +1 + 2a$

$\displaystyle 5a^2 - 42a + 72 = 0$

$\displaystyle a = 6 \ or \ 2.4$

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Question 55: Find a point which is equidistant from the points $\displaystyle A(-5,4) \text{ and } B (-1,6)$ . How many such points are there?

Answer:

Given: $\displaystyle A(-5, 4), B(-1, 6)$

Let $\displaystyle C(x, y)$ be equidistant

$\displaystyle AC^2 = BC^2$

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

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

$\displaystyle \Rightarrow 41+ 10x - 8y = 37 + 2x - 12y$

$\displaystyle \Rightarrow 8x + 4y = -4$

$\displaystyle \Rightarrow 2x + y + 1 = 0$ … … … … … i)

Therefore all points satisfying the equation i) will be equidistant from the two given points. Hence we have infinite such points.

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Question 56: The points $\displaystyle A (2,9), B (a,5) \text{ and } C (5, 5)$ are the vertices of a $\displaystyle \triangle ABC$ right angled at $\displaystyle B$ . Find the values of $\displaystyle a$ and hence the area of $\displaystyle \triangle ABC$ .