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 $(0, \sqrt{3})$

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) $(8, -15)$

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 $x$ .

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

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 $(11, 8)$ .

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

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 $(11, -8)$ .

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

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.

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

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 $a$ .

$P(2, -1)$ 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 $(-3, 4)$ .

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 $(-4, 3)$ .

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 $PQ$ .

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.

$P(0, 5), Q(5, 10)$ 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 $PQRS$ .

$P (0, -4), Q (6, 2), R (3, 5)$ 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.

$A (1, -3), B (-3, 0)$ 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 $ABCD$.

$A (5, 6), B (1, 5), C (2, 1)$ 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.

$(-3, -2), (-5, -5), (2, -3)$ 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 $AB=CD$ .

$A (-3, -2), B (-6, a), C (-3, -4)$ 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.

$(5, 1), (11, 1)$ 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 $AB=5$ .

$A (3, 1)$ 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 $AB=17$ .

$A = (x+2, -2)$ 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.

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 $10$ .

$PQ=10, P(2, -3)$ 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)   $AB$

$P(2, -7)$; 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]

$P(2, 2)$ 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]

$A(7, 3)$ 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 $9$ .

$A(5, -3)$ 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 $1:2$ .

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: $5x^2+5y^2-34x+70y+58=0$ .

$P(x, y)$ $A(1, -3)$ 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 $a$ .

$A(3, 0), B(a, -2)$ 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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