\displaystyle \textbf{Note: } \text{We shall use the following distance formula extensively in solving the problems in this chapter.}
\displaystyle \text{The distance between two points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is }d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
\displaystyle \textbf{Notes:}
\displaystyle \text{(i) If a point lies on the }x\text{-axis, its ordinate is }0.\text{ Hence, a point on the }x\text{-axis is represented as }(x,0).
\displaystyle \text{(ii) If a point lies on the }y\text{-axis, its abscissa is }0.\text{ Hence, a point on the }y\text{-axis is represented as }(0,y).
\displaystyle \text{For a detailed explanation, refer to the lecture notes.}

\displaystyle \textbf{Question 1: } \text{Find the distance between the following pairs of points:}
\displaystyle \text{(i) }(-3,6)\text{ and }(2,-6)\qquad\text{(ii) }(-a,-b)\text{ and }(a,b)
\displaystyle \text{(iii) }\left(\frac{3}{5},2\right)\text{ and }\left(-\frac{1}{5},1\frac{2}{5}\right)\qquad\text{(iv) }(\sqrt{3}+1,1)\text{ and }(0,\sqrt{3})
\displaystyle \text{Answer:}
\displaystyle \text{(i) Distance }=\sqrt{(2+3)^2+(-6-6)^2}=\sqrt{25+144}=\sqrt{169}=13
\displaystyle \text{(ii) Distance }=\sqrt{(a+a)^2+(b+b)^2}=\sqrt{4a^2+4b^2}=2\sqrt{a^2+b^2}
\displaystyle \text{(iii) Distance }=\sqrt{\left(-\frac{1}{5}-\frac{3}{5}\right)^2+\left(1\frac{2}{5}-2\right)^2}=\sqrt{\frac{16}{25}+\frac{9}{25}}=1
\displaystyle \text{(iv) 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}
\\

\displaystyle \textbf{Question 2: } \text{Find the distance between the origin and the following points:}
\displaystyle \text{(i) }(-8,6)\qquad\text{(ii) }(-5,-12)\qquad\text{(iii) }(8,-15)
\displaystyle \text{Answer:}
\displaystyle \text{(i) The points are }(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) The points are }(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) The points are }(0,0)\text{ and }(8,-15).
\displaystyle \text{Distance }=\sqrt{(8-0)^2+(-15-0)^2}=\sqrt{64+225}=\sqrt{289}=17
\\

\displaystyle \textbf{Question 3: } \text{The distance between the points }(3,1)\text{ and }(0,x)\text{ is }5.\text{ Find }x.
\displaystyle \text{Answer:}
\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
\displaystyle \therefore x=5\text{ or }x=-3.
\\

\displaystyle \textbf{Question 4: } \text{Find the coordinates of the points on the }x\text{-axis which are at a} \\ \text{distance of }17\text{ units from the point }(11,8).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point be }(x,0).
\displaystyle \text{Distance between }(x,0)\text{ and }(11,8)\text{ is }17.
\displaystyle \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
\displaystyle \therefore x=26\text{ or }-4.
\displaystyle \therefore \text{The required points are }(26,0)\text{ and }(-4,0).
\\

\displaystyle \textbf{Question 5: } \text{Find the coordinates of the points on the }y\text{-axis which are at a} \\ \text{distance of }10\text{ units from the point }(-8,4).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point be }(0,y).
\displaystyle \text{Distance between }(0,y)\text{ and }(-8,4)\text{ is }10.
\displaystyle \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
\displaystyle \therefore y=10\text{ or }y=-2.
\displaystyle \therefore \text{The required points are }(0,10)\text{ and }(0,-2).
\\

\displaystyle \textbf{Question 6: } \text{A point }A\text{ is at a distance of }\sqrt{10}\text{ units from the point }(4,3). \\ \text{ Find the coordinates of }A\text{ if its ordinate is twice its abscissa.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the point }A\text{ be }(a,2a).
\displaystyle \text{Distance between }(a,2a)\text{ and }(4,3)\text{ is }\sqrt{10}.
\displaystyle \sqrt{(4-a)^2+(3-2a)^2}=\sqrt{10}
\displaystyle (4-a)^2+(3-2a)^2=10
\displaystyle 16+a^2-8a+9+4a^2-12a=10
\displaystyle 5a^2-20a+15=0
\displaystyle a^2-4a+3=0
\displaystyle (a-1)(a-3)=0
\displaystyle \therefore a=1\text{ or }a=3.
\displaystyle \therefore \text{The required points are }(1,2)\text{ and }(3,6).
\\

\displaystyle \textbf{Question 7: } \text{A point }P(2,1)\text{ is equidistant from the points }(a,7)\text{ and } \\ (3,a).\text{ Find }a.
\displaystyle \text{Answer:}
\displaystyle \text{Since }P(2,1)\text{ is equidistant from }(a,7)\text{ and }(3,a),
\displaystyle \sqrt{(a-2)^2+(7-1)^2}=\sqrt{(3-2)^2+(a-1)^2}
\displaystyle (a-2)^2+36=1+(a-1)^2
\displaystyle a^2-4a+4+36=1+a^2-2a+1
\displaystyle -4a+40=-2a+2
\displaystyle 2a=38
\displaystyle \therefore a=19.
\\

\displaystyle \textbf{Question 8: } \text{What point on the }x\text{-axis is equidistant from the points }(7,6) \\ \text{ and } (-3,4)?
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point be }P(x,0).
\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
\displaystyle \therefore x=3.
\displaystyle \therefore \text{The required point is }(3,0).
\\

\displaystyle \textbf{Question 9: } \text{What point on the }y\text{-axis is equidistant from the points }(5,2) \\ \text{ and }(-4,3)?
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point be }P(0,y).
\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 25+4+y^2-4y=16+9+y^2-6y
\displaystyle 2y=-4
\displaystyle \therefore y=-2.
\displaystyle \therefore \text{The required point is }(0,-2).
\\

\displaystyle \textbf{Question 10: } \text{A point }P\text{ lies on the }x\text{-axis and another point }Q\text{ lies} \\ \text{on the }y\text{-axis. Write the ordinate of }P\text{ and abscissa of }Q.\text{ If the abscissa of }P\text{ is }-12 \\ \text{ and ordinate of }Q\text{ is }-16,\text{ calculate }PQ.
\displaystyle \text{Answer:}
\displaystyle \text{Since }P\text{ lies on the }x\text{-axis, its ordinate is }0.
\displaystyle \text{Since }Q\text{ lies on the }y\text{-axis, its abscissa is }0.
\displaystyle \therefore P=(-12,0)\text{ and }Q=(0,-16).
\displaystyle PQ=\sqrt{(0+12)^2+(-16-0)^2}=\sqrt{144+256}=\sqrt{400}=20
\\

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

\displaystyle \textbf{Question 12: } \text{Prove that the points }P(0,-4),\ Q(6,2),\ R(3,5)\text{ and }S(-3,-1)\text{ are} \\ \text{the vertices of the rectangle }PQRS.
\displaystyle \text{Answer:}
\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}
\displaystyle \therefore PQ=RS\text{ and }QR=PS.
\displaystyle \text{Thus, }PQRS\text{ is a parallelogram.}
\displaystyle PR=\sqrt{(3-0)^2+(5+4)^2}=\sqrt{9+81}=\sqrt{90}
\displaystyle QS=\sqrt{(-3-6)^2+(-1-2)^2}=\sqrt{81+9}=\sqrt{90}
\displaystyle \therefore PR=QS.
\displaystyle \text{Since the diagonals of the parallelogram are equal, }PQRS\text{ is a rectangle.}
\\

\displaystyle \textbf{Question 13: } \text{Prove that the points }A(1,-3),\ B(-3,0)\text{ and }C(4,1)\text{ are the vertices} \\ \text{of an isosceles triangle. Find the area of the triangle.}
\displaystyle \text{Answer:}
\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{9+16}=\sqrt{25}=5
\displaystyle BC=\sqrt{(4+3)^2+(1-0)^2}=\sqrt{49+1}=\sqrt{50}
\displaystyle \therefore AB=AC.
\displaystyle \text{Hence, }A(1,-3),\ B(-3,0)\text{ and }C(4,1)\text{ form an isosceles triangle.}
\displaystyle \text{Also, }AB^2+AC^2=5^2+5^2=25+25=50=BC^2.
\displaystyle \therefore \triangle ABC\text{ is right-angled at }A.
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times AB\times AC=\frac{1}{2}\times5\times5=\frac{25}{2}\text{ sq. units.}
\\

\displaystyle \textbf{Question 14: } \text{Show that the points }A(5,6),\ B(1,5),\ C(2,1)\text{ and }D(6,2)\text{ are the} \\ \text{vertices of the square }ABCD.
\displaystyle \text{Answer:}
\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}
\displaystyle \therefore AB=BC=CD=DA.
\displaystyle AC=\sqrt{(2-5)^2+(1-6)^2}=\sqrt{9+25}=\sqrt{34}
\displaystyle BD=\sqrt{(6-1)^2+(2-5)^2}=\sqrt{25+9}=\sqrt{34}
\displaystyle \therefore AC=BD.
\displaystyle \text{Since all four sides are equal and the diagonals are equal, }ABCD\text{ is a square.}
\\

\displaystyle \textbf{Question 15: } \text{The points }A(3,0),\ B(a,-2)\text{ and }C(4,-1)\text{ are the vertices of } \\ \triangle ABC,\text{ right-angled at }A.\text{ Find the value of }a.
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(a-3)^2+(-2-0)^2}=\sqrt{(a-3)^2+4}
\displaystyle AC=\sqrt{(4-3)^2+(-1-0)^2}=\sqrt{1+1}=\sqrt{2}
\displaystyle BC=\sqrt{(4-a)^2+(-1+2)^2}=\sqrt{(4-a)^2+1}
\displaystyle \text{Since }\angle A=90^\circ,\text{ by Pythagoras theorem, }AB^2+AC^2=BC^2.
\displaystyle (a-3)^2+4+2=(4-a)^2+1
\displaystyle a^2-6a+9+6=a^2-8a+16+1
\displaystyle 2a=2
\displaystyle a=1
\displaystyle \therefore \text{The required value of }a\text{ is }1.
\\

\displaystyle \textbf{Question 16: } \text{Points }A(-3,-2),\ B(-6,a),\ C(-3,-4)\text{ and }D(0,-1)\text{ are the vertices} \\ \text{of a quadrilateral }ABCD.\text{ Find }a\text{ if }a\text{ is negative and }AB=CD.
\displaystyle \text{Answer:}
\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}
\displaystyle \text{Since }AB=CD,
\displaystyle \sqrt{9+(a+2)^2}=\sqrt{18}
\displaystyle 9+(a+2)^2=18
\displaystyle (a+2)^2=9
\displaystyle a+2=\pm3
\displaystyle a=1\text{ or }a=-5
\displaystyle \text{Since }a\text{ is negative, }a=-5.
\\

\displaystyle \textbf{Question 17: } \text{The vertices of a triangle are }(5,1),\ (11,1)\text{ and }(11,9).\text{ Find the} \\ \text{coordinates of the circumcenter of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the coordinates of the circumcenter be }(x,y).
\displaystyle \text{Since the circumcenter is equidistant from all the vertices,}
\displaystyle \sqrt{(x-5)^2+(y-1)^2}=\sqrt{(x-11)^2+(y-1)^2}=\sqrt{(x-11)^2+(y-9)^2}
\displaystyle \therefore (x-5)^2+(y-1)^2=(x-11)^2+(y-1)^2
\displaystyle x^2-10x+25+y^2-2y+1=x^2-22x+121+y^2-2y+1
\displaystyle 12x=96
\displaystyle x=8
\displaystyle \text{Also, }(x-11)^2+(y-1)^2=(x-11)^2+(y-9)^2
\displaystyle y^2-2y+1=y^2-18y+81
\displaystyle 16y=80
\displaystyle y=5
\displaystyle \therefore \text{The coordinates of the circumcenter are }(8,5).
\\

\displaystyle \textbf{Question 18: Given }A(3,1)\text{ and }B(0,y-1).\text{ Find }y\text{ if }AB=5.
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(0-3)^2+\{(y-1)-1\}^2}=5
\displaystyle \sqrt{9+(y-2)^2}=5
\displaystyle 9+(y-2)^2=25
\displaystyle (y-2)^2=16
\displaystyle y^2-4y+4=16
\displaystyle y^2-4y-12=0
\displaystyle (y-6)(y+2)=0
\displaystyle \therefore y=6\text{ or }y=-2.
\\

\displaystyle \textbf{Question 19: } \text{Given }A(x+2,-2)\text{ and }B(11,6).\text{ Find }x\text{ if }AB=17.
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(11-x-2)^2+(6+2)^2}=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
\displaystyle \therefore x=-6\text{ or }x=24.
\\

\displaystyle \textbf{Question 20: } \text{The center of the circle is }(2x-1,3x+1).\text{ Find }x\text{ if the circle} \\ \text{passes through }(-3,-1)\text{ and the diameter is }20\text{ units.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter}=20\text{ units}\Rightarrow \text{Radius}=10\text{ units.}
\displaystyle \sqrt{\{-3-(2x-1)\}^2+\{-1-(3x+1)\}^2}=10
\displaystyle (-2x-2)^2+(-3x-2)^2=100
\displaystyle 4x^2+8x+4+9x^2+12x+4=100
\displaystyle 13x^2+20x-92=0
\displaystyle x=\frac{-20\pm\sqrt{20^2-4(13)(-92)}}{2(13)}
\displaystyle x=\frac{-20\pm\sqrt{5184}}{26}=\frac{-20\pm72}{26}
\displaystyle \therefore x=2\text{ or }x=-\frac{46}{13}.
\\

\displaystyle \textbf{Question 21: } \text{The length of }PQ\text{ is }10\text{ units and the coordinates of }P\text{ are } \\ (2,-3).\text{ Calculate the coordinates of }Q,\text{ if its abscissa is }10.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }PQ=10,\ P(2,-3)\text{ and }Q(10,y).
\displaystyle \sqrt{(10-2)^2+(y+3)^2}=10
\displaystyle 8^2+(y+3)^2=100
\displaystyle 64+y^2+6y+9=100
\displaystyle y^2+6y-27=0
\displaystyle (y+9)(y-3)=0
\displaystyle \therefore y=-9\text{ or }y=3.
\displaystyle \therefore Q=(10,-9)\text{ or }(10,3).
\\

\displaystyle \textbf{Question 22: } \text{Point }P(2,-7)\text{ is the center of a circle with radius }13\text{ units. } \\ PT\text{ is perpendicular to chord }AB\text{ and }T=(-2,-4).\text{ Calculate:}
\displaystyle \text{(i) }AT\qquad\text{(ii) }AB
\displaystyle \text{Answer:}
\displaystyle \text{Given, }P(2,-7),\ T(-2,-4)\text{ and radius }PA=13\text{ units.}
\displaystyle PT=\sqrt{(-2-2)^2+(-4+7)^2}=\sqrt{16+9}=\sqrt{25}=5
\displaystyle \text{Since the perpendicular from the center to a chord bisects the chord, }AT=\frac{AB}{2}.
\displaystyle \text{In right }\triangle PTA,\ PA^2=PT^2+AT^2
\displaystyle AT=\sqrt{PA^2-PT^2}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\text{ units}
\displaystyle AB=2\times AT=2\times12=24\text{ units}
\\

\displaystyle \textbf{Question 23: } \text{Calculate the distance between the points }P(2,2)\text{ and } \\ Q(5,4),\text{ correct to three significant figures.}\hfill\text{[1990]}
\displaystyle \text{Answer:}
\displaystyle \text{Distance }=\sqrt{(5-2)^2+(4-2)^2}=\sqrt{9+4}=\sqrt{13}
\displaystyle \therefore \text{Distance }=\sqrt{13}\approx3.61\text{ units (correct to three significant figures).}
\\

\displaystyle \textbf{Question 24: } \text{Calculate the distance between }A(7,3)\text{ and }B\text{ on the } \\ x\text{-axis whose abscissa is }11.\hfill\text{[1997]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }B\text{ lies on the }x\text{-axis and its abscissa is }11,\ B=(11,0).
\displaystyle \text{Distance }=\sqrt{(11-7)^2+(0-3)^2}=\sqrt{16+9}=\sqrt{25}=5\text{ units}
\\

\displaystyle \textbf{Question 25: } \text{Calculate the distance between }A(5,-3)\text{ and }B\text{ on the } \\ y\text{-axis whose ordinate is }9.
\displaystyle \text{Answer:}
\displaystyle \text{Since }B\text{ lies on the }y\text{-axis and its ordinate is }9,\ B=(0,9).
\displaystyle \text{Distance }=\sqrt{(0-5)^2+(9+3)^2}=\sqrt{25+144}=\sqrt{169}=13\text{ units}
\\

\displaystyle \textbf{Question 26: } \text{Find the point on the }y\text{-axis whose distances from }A(6,7)\text{ and } \\ B(4,-3)\text{ are in the ratio }1:2.
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point be }P(0,y).
\displaystyle \frac{AP}{PB}=\frac{1}{2}
\displaystyle \frac{\sqrt{(0-6)^2+(y-7)^2}}{\sqrt{(0-4)^2+(y+3)^2}}=\frac{1}{2}
\displaystyle 2\sqrt{36+(y-7)^2}=\sqrt{16+(y+3)^2}
\displaystyle 4\{36+(y-7)^2\}=16+(y+3)^2
\displaystyle 4(y^2-14y+85)=y^2+6y+25
\displaystyle 4y^2-56y+340=y^2+6y+25
\displaystyle 3y^2-62y+315=0
\displaystyle (3y-35)(y-9)=0
\displaystyle \therefore y=\frac{35}{3}\text{ or }y=9.
\displaystyle \therefore \text{The required points are }(0,9)\text{ and }\left(0,\frac{35}{3}\right).
\\

\displaystyle \textbf{Question 27: } \text{The distances of point }P(x,y)\text{ from the points }A(1,-3)\text{ and } \\ B(-2,2)\text{ are in the ratio }2:3.\text{ Show that }5x^2+5y^2-34x+70y+58=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }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+y^2-2x+6y+10)=4(x^2+y^2+4x-4y+8)
\displaystyle 9x^2+9y^2-18x+54y+90=4x^2+4y^2+16x-16y+32
\displaystyle 5x^2+5y^2-34x+70y+58=0
\displaystyle \text{Hence proved.}
\\

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