Note: We are going to use the following formula extensively in solving the following problems. The distance between any two points (x_1, y_1) and (x_2, y_2) is = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

Notes: If a point is on x-axis , its ordinate is 0 , therefore the point on x-axis is taken as (x,0) . Similarly, if the point is on y-axis , its abscissa is 0 , therefore the point on y-axis is taken as (0,y) . For details refer to the following lecture notes.

 

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

i) (-3,6) and (2, -6)

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

iii) (\frac{3}{5}, 2) and (-\frac{1}{5}, 1\frac{2}{5})

iv) (\sqrt{3}+1, 1) and (0, \sqrt{3})

Answer:

i) (-3, 6) and (2, -6)

Distance = \sqrt{(2+3)^2+(-6-6)^2} = \sqrt{25+144} = \sqrt{169} = 13

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

Distance = \sqrt{(a+a)^2+(b+b)^2} = \sqrt{4a^2+4b^2} = 2\sqrt{a^2+b^2}

iii) (\frac{3}{5}, 2) and ( -\frac{1}{5}, 1\frac{2}{5})

Distance = \sqrt{(-\frac{1}{5}-\frac{3}{5})^2+(1\frac{2}{5}-2)^2} = \sqrt{\frac{16}{25}+\frac{9}{25}} = 1

iv) (\sqrt{3}+1, 1) and (0, \sqrt{3})

Distance = \sqrt{(0-\sqrt{3}-1)^2+(\sqrt{3}-1)^2}

= \sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}

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Question 2: Find the distance between the origin and the points:

i) (-8, 6)

ii) (-5, -12)

iii) (8, -15)

Answer:

i) (0,0) \ and \ (-8, 6)

Distance = \sqrt{(-8-0)^2+(6-0)^2} = \sqrt{64+36} = \sqrt{100} = 10

ii) (0,0) \ and \ (-5, -12)

Distance = \sqrt{(-5-0)^2+(-12-0)^2} = \sqrt{25+144} = \sqrt{169} = 13

iii) (0,0) \ and \ (8, -15)

Distance = \sqrt{(8-0)^2+(-15-0)^2} = \sqrt{64+225} = \sqrt{289} = 17

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Question 3: The distance between point (3, 1) and (0, x) is 5 . Find x .

Answer:

(3, 1) \ and \ (0, x)

Distance:  \sqrt{(0-3)^2+(x-1)^2} = 5

 \sqrt{9+(x-1)^2} = 5

 (x-1)^2=25-9=16

 x^2+1-2x=16

 x^2-2x-15=0

 (x-5)(x+3)=0 \Rightarrow x = 5 or -3

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Question 4: Find the coordinate of the point on x-axis which are at a distance of 17 units from the point (11, 8) .

Answer:

(x,0) \ and \ (11, -8)

Distance:  \sqrt{(11-x)^2+(-8-0)^2} =17

 (11-x)^2+64=289

 (11-x)^2=225

 x^2-22x-104=0

 (x-26)(x+4)=0 \Rightarrow x = 26 \ or \ -4

Therefore the points are  (26,0) \ and \ (-4,0)

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Question 5: Find the coordinate of the point on y-axis which are at a distance of 10 units from the point (11, -8) .

Answer:

(0,y) \ and \ (-8, 4)

Distance:  \sqrt{(-8-0)^2+(4-y)^2} =10

64+(4-y)^2=100

(4-y)^2=36

y^2-8y-20=0

(y-10)(y+2)=0 \Rightarrow y = 10 \ or \ -2

Hence the points could be (0, 10) \ and \ (0, -2)

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Question 6: A point A is at a distance of \sqrt{10} units from the point (4, 3) . Find the coordinates of the point A if its ordinate is twice its abscissa.

Answer:

(2a,a) \ and \ (4,3)

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

(4-2a)^2+(3-a)^2=10

5a^2-22a-15=0

(a-5)(5a+3)=0 \Rightarrow a =5 \ or  \ -\frac{3}{5}

Hence the points could be (10,5) \ and \ (-\frac{6}{5}, -\frac{3}{5})

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Question 7: A point P (2, 1) is equidistant from the point (a, 7) and (3, a) . Find a .

Answer:

P(2, -1) is equidistant from (a, 7) and (-3, 9)

Therefore \sqrt{(a-2)^2+(7+1)^2} =  \sqrt{(-3-2)^2+(a+1)^2}

(a-2)^2+64=25+(a+1)^2

a^2+4-4a+64=25+a^2+1+2a

6a=42 \Rightarrow a =7

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

Answer:

Let the point be P(x,0) . Therefore

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

(7-x)^2+36=(-3-x)^2+16

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

20x=60 \Rightarrow x = 3

Therefore the point is (3,0)

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Question 9: What point on y-axis is equidistant from the point (5, 2) and (-4, 3) .

Answer:

Let the point be P (0,y) . Therefore

\sqrt{(5-0)^2+(2-y)^2} =  \sqrt{(-4-0)^2+(3-y)^2}

25+(2-y)^2=16+(3-y)^2

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

2y=-4 \Rightarrow y = -2

Hence the point is (0, -2)

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Question 10: A point P lies on x-axis and another point Q lies on y-axis . Write the ordinate of point P , abscissa of point Q . If the abscissa of point P is -12 and ordinate of point Q is -16 . Calculate the length of the line segment PQ .

Answer:

Let  P(x, 0) and  Q (0, y) . Given  P(-12, 0) and  Q (0, -16)

Distance:  \sqrt{(0+12)^2+(-16-0)^2} = \sqrt{144+256} = \sqrt{400} = 20

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Question 11: Show that the points P(0, 5), Q(5, 10) and R(6, 3) are the vertices of an isosceles triangle.

Answer:

P(0, 5), Q(5, 10) and R(6, 3)

PQ =   \sqrt{(5-0)^2+(10-5)^2} = \sqrt{25+25} = \sqrt{50}  

PR =   \sqrt{(6-0)^2+(3-5)^2} = \sqrt{36+4} = \sqrt{40}

QR =   \sqrt{(6-5)^2+(3-10)^2} = \sqrt{1+49} = \sqrt{50}  

Therefore two sides PQ \ and \ QR are equal which makes it an isosceles triangle.

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Question 12: Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of the rectangle PQRS .

Answer:

P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1)

PQ =   \sqrt{(6-0)^2+(2+4)^2} = \sqrt{36+36} = \sqrt{72}  

QR =   \sqrt{(3-6)^2+(5-2)^2} = \sqrt{9+9} = \sqrt{18}

RS =   \sqrt{(-3-3)^2+(-1-5)^2} = \sqrt{36+36} = \sqrt{72}  

PS =   \sqrt{(-3-0)^2+(-1+4)^2} = \sqrt{9+9} = \sqrt{18}  

Therefore PQ=RS   and QR=PS  .

Hence it is a rectangle.

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Question 13: Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles triangle. Find the area of the triangle.

Answer:

A (1, -3), B (-3, 0) and C (4, 1)

AB =   \sqrt{(-3-1)^2+(0+3)^2} = \sqrt{16+9} = \sqrt{25} = 5  

AC =   \sqrt{(4-1)^2+(1+3)^2} = \sqrt{16+9} = \sqrt{25} = 5

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

For this to be a right angled triangle we should have AB^2+AC^2=BC^2  

AB^2+AC^2 = (5)^2+(5)^2= (50)^2 = BC^2  . Hence proved that it is a right angled triangle.

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

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Question 14: Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of the square ABCD .

Answer:

A (5, 6), B (1, 5), C (2, 1) and D (6, 2)

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

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

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

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

Therefore AB=BC=CD=DA  .

Hence it is a square.

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Question 15: Show that (-3, -2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Answer:

(-3, -2), (-5, -5), (2, -3) and (4, 4)

AB =   \sqrt{(-5+3)^2+(-5+2)^2} = \sqrt{4+9} = \sqrt{13}  

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+36} = \sqrt{83}  

Therefore BC=CD  .

Two sides are equal and the other two sides are of different length. Hence it is a rhombus.

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Question 16: Points  A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of a quadrilateral ABCD . Find a if a is negative and AB=CD .

Answer:

 A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1)

AB =   \sqrt{(-6+3)^2+(a+2)^2} = \sqrt{9+(a+2)^2}  

CD =   \sqrt{(0+3)^2+(-1+4)^2} = \sqrt{9+9} = \sqrt{18}

Therefore  \sqrt{18}=  \sqrt{9+(a+2)^2} 

 a^2+4+4a+9=18

 a^2+4a-5=0

 (a+5)(a-1)=0 \Rightarrow a = -5 \ or \ 1 . Hence a = -5  as it is negative.

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Question 17: The vertices of a triangle are (5, 1), (11, 1) and (11, 9) . Find the coordinates of the circumcenter of the triangle.

Answer:

(5, 1), (11, 1) and (11, 9) are the points

Let the coordinates of the circumcenter = (x, y)

Therefore  \sqrt{(x-5)^2+(y-1)^2} =  \sqrt{(x-11)^2+(y-1)^2}  =  \sqrt{(x-11)^2+(y-9)^2}

Hence  {(x-5)^2+(y-1)^2} =  {(x-11)^2+(y-1)^2}  =  {(x-11)^2+(y-9)^2}

Therefore equation 1:

{(x-5)^2+(y-1)^2} =  {(x-11)^2+(y-1)^2}

x^2+25-10x+y^2+1-2y=x^2+121-22x+y^2+1-2y

26-10x-2y=122-22x-2y

12x=96 \Rightarrow x = 8

Also equation 2:

{(x-11)^2+(y-1)^2}  =  {(x-11)^2+(y-9)^2}

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

16y=80 \Rightarrow y = 5

Hence the coordinates of the circumcenter is (8,5)

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Question 18: Given A (3, 1) and B (0, y-1) . Find y if AB=5 .

Answer:

A (3, 1) and B (0, y-1)

AB =   \sqrt{(0-3)^2+(y-1-1)^2} = \sqrt{9+y^2}

Therefore 9+y^2=25 \Rightarrow y^2=16 \Rightarrow y = 4, -4

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Question 19: Given A = (x+2, -2) and B=(11, 6) . Find x if AB=17 .

Answer:

A = (x+2, -2) and B=(11, 6)

AB =   \sqrt{(11-x-2)^2+(6+2)^2} = \sqrt{289} = 17

(9-x)^2+64=289

(9-x)^2=225

x^2+81-18x-225=0

x^2-18x-144=0

(x+6)(x-24)=0 \Rightarrow x = -6 \ or \ 24

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Question 20: The center of the circle is (2x-1, 3x+1) . Find x if the circle passes through (-3, -1) and the length of the diameter is 20 units.

Answer:

Diameter = 20  units i.e.Radius = 10  units

  \sqrt{(-3-2x+1)^2+(-1-3x-1)^2} = 10

(-2x-2)^2+(-3x-2)^2=100 

4x^2+4+8x+9x^2+4+12x=100 

13x^2+20x-92=0 \Rightarrow x = 2, -3.539 

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Question 21: The length of the line PQ is 10 units and the coordinates of P are (2, -3) , calculate the coordinates of point Q , if its abscissca is 10 .

Answer:

  PQ=10, P(2, -3) and Let   Q(10, y)

  \sqrt{(10-2)^2+(y+3)^2} = 10

  8^2+(y+3)^2=100

  64+y^2+9+6y=100

  y^2+6y-27=0

  (y+9)(y-3)=0 \Rightarrow y = -9, 3

hence the points   Q could be   (10, -9) \ or  \ (10,3)

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Question 22: Point P (2, -7) is the center of the circle with radius 13 units, PT  is perpendicular to chord AB and T=(-2, -4) . Calculate the length of  i) AT ii)   AB

Answer:

P(2, -7)  ; Radius = 13   units; T(-2,-4)  

PT =   \sqrt{(-2-2)^2+(-4+7)^2} = \sqrt{25} = 5  

Therefore AT=\sqrt{13^2-5^2}=\sqrt{169-25} = 12  

AB = 2 \times AT = 2 \times 12 = 24 units.  

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Question 23: Calculate the distance between the two points P(2, 2) and Q (5, 4) , correct to three significant figures.     [1990]

Answer:

P(2, 2) and Q (5, 4)

Distance = \sqrt{(5-2)^2+(4-2)^2} = \sqrt{9+4} = \sqrt{13} = 3.61

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Question 24: Calculate the distance between A(7, 3) and B on the x-axis whose abscissa is 11 .     [1997]

Answer:

A(7, 3) and B(11,0)

Distance = \sqrt{(11-7)^2+(0-3)^2} = \sqrt{16+9} = \sqrt{25} = 5

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Question 25: Calculate the distance between A(5, -3) and B on the y-axis whose ordinate is 9 .

Answer:

A(5, -3) and B (0,9)

Distance = \sqrt{(0-5)^2+(9+3)^2} = \sqrt{25+144} = \sqrt{169} = 13

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Question 26: Find the point on y-axis whose distance from the point A(6, 7) and B(4, -3) are in the ratio of 1:2 .

Answer:

Let point be P(0,y)

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

\frac{AP}{PB} = \frac{1}{2} = \frac{\sqrt{(0-6)^2+(y-7)^2}}{\sqrt{(0-4)^2+(y+3)^2}}

\sqrt{16+(y+3)^2}=2\sqrt{36+(y-7)^2}

16+(y+3)^2=4[36+(y-7)^2]

16+y^2+9+6y=4(36+y^2+49-14y)

y^2+25+6y=340+4y^2-56y

3y^2-62y+315=0 \Rightarrow y = 9 or 11.67

Hence points could be (0,9) \ or \ (0, 11.67)

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Question 27: The distance of point P(x, y) from the points A(1, -3) and B(-2, 2) are in the ratio 2:3 . Show that: 5x^2+5y^2-34x+70y+58=0 .

Answer:

P(x, y)  A(1, -3) and B(-2, 2)

\frac{PA}{PB} = \frac{2}{3}

\frac{\sqrt{(1-x)^2+(-3-y)^2}}{\sqrt{(-2-x)^2+(2-y)^2}} = \frac{2}{3}

9[(1-x)^2+(-3-y)^2] = 4[(-2-x)^2+(2-y)^2]

9[1+x^2-2x+9+y^2+6y]=4[4+x^2+4x+4+y^2-4y]

9[x^2-2x+10+y^2+6y]=4[x^2+4x+8+y^2-4y]

5x^2+5y^2-34x+70y+58=0 . Hence proved.

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Question 28: The point A(3, 0), B(a, -2) and C(4, -1) are vertices of triangle ABC right angles at vertex A . Find the value of a .

Answer:

A(3, 0), B(a, -2) and C(4, -1)

AB = \sqrt{(a-3)^2+(-2-0)^2} = \sqrt{4+(a-3)^2}

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

BC = \sqrt{(4-a)^2+(-1+2)^2} = \sqrt{1+(4-a)^2}  

AB^2+AC^2=BC^2  

4+(a-3)^2+2=1+(4-a)^2  

4+a^2+9-6a+2=1+16+a^2-8a  

2a=2 \Rightarrow a = 1  

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