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

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

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

\displaystyle \text{i) } (-3,6) \text{ and } (2, -6)  

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

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

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

Answer:

\displaystyle \text{i) } (-3, 6) \text{ and } (2, -6)  

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

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

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

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

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

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

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

\displaystyle = \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:

\displaystyle \text{i) } (-8, 6)  

\displaystyle \text{ii) } (-5, -12)  

\displaystyle \text{iii) } (8, -15)  

Answer:

\displaystyle \text{i) } (0,0) \text{ and } (-8, 6)  

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

\displaystyle \text{ii) } (0,0) \text{ and } (-5, -12)  

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

\displaystyle \text{iii) } (0,0) \text{ and } (8, -15)  

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

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Question 3: The distance between point \displaystyle (3, 1) \text{ and } (0, x) is \displaystyle 5 . Find \displaystyle x

Answer:

\displaystyle (3, 1) \text{ and } (0, x)  

\displaystyle \text{Distance } \sqrt{(0-3)^2+(x-1)^2} = 5  

\displaystyle \sqrt{9+(x-1)^2} = 5  

\displaystyle (x-1)^2=25-9=16  

\displaystyle x^2+1-2x=16  

\displaystyle x^2-2x-15=0  

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

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

Answer:

\displaystyle (x,0) \text{ and } (11, -8)  

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

\displaystyle (11-x)^2+64=289  

\displaystyle (11-x)^2=225  

\displaystyle x^2-22x-104=0  

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

Therefore the points are \displaystyle (26,0) \text{ and } (-4,0)  

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

Answer:

\displaystyle (0,y) \text{ and } (-8, 4)  

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

\displaystyle 64+(4-y)^2=100  

\displaystyle (4-y)^2=36  

\displaystyle y^2-8y-20=0  

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

Hence the points could be \displaystyle (0, 10) \text{ and } (0, -2)  

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

Answer:

\displaystyle (2a,a) \text{ and } (4,3)  

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

\displaystyle (4-2a)^2+(3-a)^2=10  

\displaystyle 5a^2-22a-15=0  

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

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

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

Answer:

\displaystyle P(2, -1) is equidistant from \displaystyle (a, 7) \text{ and } (-3, 9)  

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

\displaystyle (a-2)^2+64=25+(a+1)^2  

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

\displaystyle 6a=42 \Rightarrow a =7  

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Question 8: What point on \displaystyle x-axis is equidistant from the point \displaystyle (7, 6) \text{ and } (-3, 4)

Answer:

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

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

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

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

\displaystyle 20x=60 \Rightarrow x = 3  

Therefore the point is \displaystyle (3,0)  

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Question 9: What point on \displaystyle y-axis is equidistant from the point \displaystyle (5, 2) \text{ and } (-4, 3)

Answer:

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

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

\displaystyle 25+(2-y)^2=16+(3-y)^2  

\displaystyle 9+4+y^2-4y=9+y^2-6y  

\displaystyle 2y=-4 \Rightarrow y = -2  

Hence the point is \displaystyle (0, -2)  

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

Answer:

Let \displaystyle P(x, 0) \text{ and } Q (0, y) . Given \displaystyle P(-12, 0) \text{ and } Q (0, -16)  

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

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

Answer:

\displaystyle P(0, 5), Q(5, 10) \text{ and } R(6, 3)  

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

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

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

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

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

Answer:

\displaystyle P (0, -4), Q (6, 2), R (3, 5) \text{ and } S (-3, -1)  

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

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

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

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

Therefore \displaystyle PQ=RS \text{ and } QR=PS .

Hence it is a rectangle.

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

Answer:

\displaystyle A (1, -3), B (-3, 0) \text{ and } C (4, 1)  

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

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

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

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

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

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

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

Answer:

\displaystyle A (5, 6), B (1, 5), C (2, 1) \text{ and } D (6, 2)  

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

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

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

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

Therefore \displaystyle AB=BC=CD=DA .

Hence it is a square.

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

Answer:

\displaystyle (-3, -2), (-5, -5), (2, -3) \text{ and } (4, 4)  

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

\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+36} = \sqrt{83}  

Therefore \displaystyle 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 \displaystyle A (-3, -2), B (-6, a), C (-3, -4) \text{ and } D (0, -1) are the vertices of a quadrilateral \displaystyle ABCD . Find a if a is negative and \displaystyle AB=CD

Answer:

\displaystyle A (-3, -2), B (-6, a), C (-3, -4) \text{ and } D (0, -1)  

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

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

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

\displaystyle a^2+4+4a+9=18  

\displaystyle a^2+4a-5=0  

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

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

Answer:

\displaystyle (5, 1), (11, 1) \text{ and } (11, 9) are the points

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

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

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

Therefore equation 1:

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

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

\displaystyle 26-10x-2y=122-22x-2y  

\displaystyle 12x=96 \Rightarrow x = 8  

Also equation 2:

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

\displaystyle y^2+1-2y=y^2+81-18y  

\displaystyle 16y=80 \Rightarrow y = 5  

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

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

Answer:

\displaystyle A (3, 1) \text{ and } B (0, y-1)  

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

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

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

Answer:

\displaystyle A = (x+2, -2) \text{ and } B=(11, 6)  

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

\displaystyle (9-x)^2+64=289  

\displaystyle (9-x)^2=225  

\displaystyle x^2+81-18x-225=0  

\displaystyle x^2-18x-144=0  

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

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

Answer:

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

\displaystyle \sqrt{(-3-2x+1)^2+(-1-3x-1)^2} = 10  

\displaystyle (-2x-2)^2+(-3x-2)^2=100  

\displaystyle 4x^2+4+8x+9x^2+4+12x=100  

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

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

Answer:

\displaystyle PQ=10, P(2, -3) and Let \displaystyle Q(10, y)  

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

\displaystyle 8^2+(y+3)^2=100  

\displaystyle 64+y^2+9+6y=100  

\displaystyle y^2+6y-27=0  

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

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

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

Answer:

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

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

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

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

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

Answer:

\displaystyle P(2, 2) \text{ and } Q (5, 4)  

\displaystyle \text{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 \displaystyle A(7, 3) \text{ and } B on the \displaystyle x-axis whose abscissa is \displaystyle 11 [1997]

Answer:

\displaystyle A(7, 3) \text{ and } B(11,0)  

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

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Question 25: Calculate the distance between \displaystyle A(5, -3) \text{ and } B on the \displaystyle y-axis whose ordinate is \displaystyle 9

Answer:

\displaystyle A(5, -3) \text{ and } B (0,9)  

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

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

Answer:

Let point be \displaystyle P(0,y)  

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

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

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

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

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

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

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

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

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

Answer:

\displaystyle P(x, y) A(1, -3) \text{ and } B(-2, 2)  

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

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

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

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

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

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

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

Answer:

\displaystyle A(3, 0), B(a, -2) \text{ and } C(4, -1)  

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

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

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

\displaystyle AB^2+AC^2=BC^2  

\displaystyle 4+(a-3)^2+2=1+(4-a)^2  

\displaystyle 4+a^2+9-6a+2=1+16+a^2-8a  

\displaystyle 2a=2 \Rightarrow a = 1