\displaystyle \textbf{Question 1: }\text{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) }\left(\frac{3}{5},2\right)\text{ and }\left(-\frac{1}{5},1\frac{2}{5}\right)
\displaystyle \text{(iv) }(\sqrt3+1,1)\text{ and }(0,\sqrt3)
\displaystyle \text{Answer:}
\displaystyle \text{Distance between two points }(x_1,y_1)\text{ and }(x_2,y_2)
\displaystyle =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
\displaystyle \text{(i) Distance}=\sqrt{(2+3)^2+(-6-6)^2}
\displaystyle =\sqrt{25+144}=\sqrt{169}=13.

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

\displaystyle \text{(iii) Distance}=\sqrt{\left(-\frac15-\frac35\right)^2+\left(\frac75-2\right)^2}
\displaystyle =\sqrt{\left(-\frac45\right)^2+\left(-\frac35\right)^2}
\displaystyle =\sqrt{\frac{16}{25}+\frac{9}{25}}=\sqrt{1}=1.

\displaystyle \text{(iv) Distance}=\sqrt{\{0-(\sqrt3+1)\}^2+(\sqrt3-1)^2}
\displaystyle =\sqrt{(\sqrt3+1)^2+(\sqrt3-1)^2}
\displaystyle =\sqrt{(4+2\sqrt3)+(4-2\sqrt3)}
\displaystyle =\sqrt8=2\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the distance between the origin and the point:}
\displaystyle \text{(i) }(-8,6)\qquad\text{(ii) }(-5,-12)\qquad\text{(iii) }(8,-15)
\displaystyle \text{Answer:}
\displaystyle \text{Distance of a point }(x,y)\text{ from the origin}=\sqrt{x^2+y^2}.
\displaystyle \text{(i) Distance}=\sqrt{(-8)^2+6^2}
\displaystyle =\sqrt{64+36}=\sqrt{100}=10\text{ units.}

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

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

\displaystyle \textbf{Question 3: }\text{The distance between the points }(3,1)\text{ and }(0,x)\text{ is }5.
\displaystyle \text{Find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Using the distance formula,}
\displaystyle \sqrt{(0-3)^2+(x-1)^2}=5
\displaystyle \Rightarrow 9+(x-1)^2=25
\displaystyle \Rightarrow (x-1)^2=16
\displaystyle \Rightarrow x-1=\pm4
\displaystyle \therefore x=5\text{ or }x=-3.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the co-ordinates of points on the }x\text{-axis which are at a distance of}
\displaystyle 17\text{ units from the point }(11,-8).
\displaystyle \text{Answer:}
\displaystyle \text{Let the co-ordinates of the point on the }x\text{-axis be }(x,0).
\displaystyle \text{Using the distance formula,}
\displaystyle \sqrt{(x-11)^2+\{0-(-8)\}^2}=17
\displaystyle \Rightarrow (x-11)^2+64=289
\displaystyle \Rightarrow (x-11)^2=225
\displaystyle \Rightarrow x-11=\pm15
\displaystyle \Rightarrow x=26\text{ or }x=-4.
\displaystyle \therefore \text{The required points are }(26,0)\text{ and }(-4,0).
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the co-ordinates of the points on the }y\text{-axis, which are at a distance}
\displaystyle \text{of }10\text{ units from the point }(-8,4).
\displaystyle \text{Answer:}
\displaystyle \text{Let the co-ordinates of the point on the }y\text{-axis be }(0,y).
\displaystyle \text{Using the distance formula,}
\displaystyle \sqrt{\{0-(-8)\}^2+(y-4)^2}=10
\displaystyle \Rightarrow 64+(y-4)^2=100
\displaystyle \Rightarrow (y-4)^2=36
\displaystyle \Rightarrow y-4=\pm6
\displaystyle \Rightarrow y=10\text{ or }y=-2.
\displaystyle \therefore \text{The required points are }(0,10)\text{ and }(0,-2).
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A point }A\text{ is at a distance of }\sqrt{10}\text{ unit from the point }(4,3).\text{ Find the}
\displaystyle \text{co-ordinates of point }A,\text{ if its ordinate is twice its abscissa.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the abscissa of }A=x.
\displaystyle \therefore \text{Its ordinate}=2x.
\displaystyle \therefore A=(x,2x).
\displaystyle \text{Using the distance formula,}
\displaystyle \sqrt{(x-4)^2+(2x-3)^2}=\sqrt{10}
\displaystyle \Rightarrow (x-4)^2+(2x-3)^2=10
\displaystyle \Rightarrow x^2-8x+16+4x^2-12x+9=10
\displaystyle \Rightarrow 5x^2-20x+15=0
\displaystyle \Rightarrow x^2-4x+3=0
\displaystyle \Rightarrow (x-1)(x-3)=0
\displaystyle \Rightarrow x=1\text{ or }x=3.
\displaystyle \text{When }x=1,\quad A=(1,2).
\displaystyle \text{When }x=3,\quad A=(3,6).
\displaystyle \therefore \text{The co-ordinates of }A\text{ are }(1,2)\text{ and }(3,6).
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A point }P(2,-1)\text{ is equidistant from the points }(a,7)\text{ and }(-3,a).
\displaystyle \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 \Rightarrow (a-2)^2+64=25+(a+1)^2
\displaystyle \Rightarrow a^2-4a+4+64=25+a^2+2a+1
\displaystyle \Rightarrow -4a+68=2a+26
\displaystyle \Rightarrow 6a=42
\displaystyle \therefore a=7.
\displaystyle \\

\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 on the }x\text{-axis be }P(x,0).
\displaystyle \text{Since }P\text{ is equidistant from }(7,6)\text{ and }(-3,4),
\displaystyle \sqrt{(x-7)^2+(0-6)^2}=\sqrt{\{x-(-3)\}^2+(0-4)^2}
\displaystyle \Rightarrow (x-7)^2+36=(x+3)^2+16
\displaystyle \Rightarrow x^2-14x+49+36=x^2+6x+9+16
\displaystyle \Rightarrow -14x+85=6x+25
\displaystyle \Rightarrow 20x=60
\displaystyle \Rightarrow x=3.
\displaystyle \therefore \text{The required point is }(3,0).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find a point on the }y\text{-axis which is equidistant from the points }(5,2)
\displaystyle \text{and }(-4,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point on the }y\text{-axis be }P(0,y).
\displaystyle \text{Since }P\text{ is equidistant from }(5,2)\text{ and }(-4,3),
\displaystyle \sqrt{(0-5)^2+(y-2)^2}=\sqrt{\{0-(-4)\}^2+(y-3)^2}
\displaystyle \Rightarrow 25+(y-2)^2=16+(y-3)^2
\displaystyle \Rightarrow 25+y^2-4y+4=16+y^2-6y+9
\displaystyle \Rightarrow -4y+29=-6y+25
\displaystyle \Rightarrow 2y=-4
\displaystyle \therefore y=-2.
\displaystyle \therefore \text{The required point is }(0,-2).
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A point }P\text{ lies on the }x\text{-axis and another point }Q\text{ lies on the }y\text{-axis.}
\displaystyle \text{(i) Write the ordinate of point }P.
\displaystyle \text{(ii) Write the abscissa of point }Q.
\displaystyle \text{(iii) If the abscissa of point }P\text{ is }-12\text{ and the ordinate of point }Q\text{ is }-16,
\displaystyle \text{calculate the length of line segment }PQ.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }P\text{ lies on the }x\text{-axis, its ordinate is }0.

\displaystyle \text{(ii) Since }Q\text{ lies on the }y\text{-axis, its abscissa is }0.

\displaystyle \text{(iii) }P=(-12,0)\text{ and }Q=(0,-16).
\displaystyle PQ=\sqrt{\{0-(-12)\}^2+(-16-0)^2}
\displaystyle =\sqrt{12^2+(-16)^2}
\displaystyle =\sqrt{144+256}=\sqrt{400}=20\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Show that the points }P(0,5),\ Q(5,10)\text{ and }R(6,3)\text{ are the vertices}
\displaystyle \text{of an isosceles triangle.}
\displaystyle \text{Answer:}
\displaystyle PQ=\sqrt{(5-0)^2+(10-5)^2}
\displaystyle =\sqrt{25+25}=\sqrt{50}=5\sqrt2.
\displaystyle QR=\sqrt{(6-5)^2+(3-10)^2}
\displaystyle =\sqrt{1+49}=\sqrt{50}=5\sqrt2.
\displaystyle PR=\sqrt{(6-0)^2+(3-5)^2}
\displaystyle =\sqrt{36+4}=\sqrt{40}=2\sqrt{10}.
\displaystyle \therefore PQ=QR.
\displaystyle \therefore \triangle PQR\text{ is an isosceles triangle.}
\displaystyle \\

\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 the}
\displaystyle \text{vertices of a rectangle }PQRS.
\displaystyle \text{Answer:}
\displaystyle PQ=\sqrt{(6-0)^2+\{2-(-4)\}^2}
\displaystyle =\sqrt{36+36}=\sqrt{72}=6\sqrt2.
\displaystyle QR=\sqrt{(3-6)^2+(5-2)^2}
\displaystyle =\sqrt{9+9}=\sqrt{18}=3\sqrt2.
\displaystyle RS=\sqrt{(-3-3)^2+(-1-5)^2}
\displaystyle =\sqrt{36+36}=\sqrt{72}=6\sqrt2.
\displaystyle PS=\sqrt{(-3-0)^2+\{-1-(-4)\}^2}
\displaystyle =\sqrt{9+9}=\sqrt{18}=3\sqrt2.
\displaystyle \therefore PQ=RS\text{ and }QR=PS.
\displaystyle PR=\sqrt{(3-0)^2+\{5-(-4)\}^2}
\displaystyle =\sqrt{9+81}=\sqrt{90}=3\sqrt{10}.
\displaystyle QS=\sqrt{(-3-6)^2+(-1-2)^2}
\displaystyle =\sqrt{81+9}=\sqrt{90}=3\sqrt{10}.
\displaystyle \therefore PR=QS.
\displaystyle \text{Since both pairs of opposite sides are equal and the diagonals are equal,}
\displaystyle \therefore PQRS\text{ is a rectangle.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Prove that the points }A(1,-3),\ B(-3,0)\text{ and }C(4,1)\text{ are the vertices}
\displaystyle \text{of an isosceles right-angled triangle. Find the area of the triangle.}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(-3-1)^2+\{0-(-3)\}^2}
\displaystyle =\sqrt{16+9}=\sqrt{25}=5.
\displaystyle AC=\sqrt{(4-1)^2+\{1-(-3)\}^2}
\displaystyle =\sqrt{9+16}=\sqrt{25}=5.
\displaystyle BC=\sqrt{\{4-(-3)\}^2+(1-0)^2}
\displaystyle =\sqrt{49+1}=\sqrt{50}=5\sqrt2.
\displaystyle \therefore AB=AC.
\displaystyle \therefore \triangle ABC\text{ is isosceles.}
\displaystyle AB^2+AC^2=5^2+5^2=25+25=50
\displaystyle BC^2=(5\sqrt2)^2=50.
\displaystyle \therefore AB^2+AC^2=BC^2.
\displaystyle \therefore \triangle ABC\text{ is right-angled at }A.
\displaystyle \therefore \triangle ABC\text{ is an isosceles right-angled triangle.}
\displaystyle \text{Area of }\triangle ABC=\frac12\times AB\times AC
\displaystyle =\frac12\times5\times5=\frac{25}{2}\text{ sq. units.}
\displaystyle \\

\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}
\displaystyle \text{vertices of a square }ABCD.
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(1-5)^2+(5-6)^2}
\displaystyle =\sqrt{16+1}=\sqrt{17}.
\displaystyle BC=\sqrt{(2-1)^2+(1-5)^2}
\displaystyle =\sqrt{1+16}=\sqrt{17}.
\displaystyle CD=\sqrt{(6-2)^2+(2-1)^2}
\displaystyle =\sqrt{16+1}=\sqrt{17}.
\displaystyle DA=\sqrt{(5-6)^2+(6-2)^2}
\displaystyle =\sqrt{1+16}=\sqrt{17}.
\displaystyle \therefore AB=BC=CD=DA.
\displaystyle AC=\sqrt{(2-5)^2+(1-6)^2}
\displaystyle =\sqrt{9+25}=\sqrt{34}.
\displaystyle BD=\sqrt{(6-1)^2+(2-5)^2}
\displaystyle =\sqrt{25+9}=\sqrt{34}.
\displaystyle \therefore AC=BD.
\displaystyle \text{Since all the sides are equal and the diagonals are also equal,}
\displaystyle \therefore ABCD\text{ is a square.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Show that }(-3,2),\ (-5,-5),\ (2,-3)\text{ and }(4,4)\text{ are the vertices}
\displaystyle \text{of a rhombus.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-3,2),\ B(-5,-5),\ C(2,-3)\text{ and }D(4,4).
\displaystyle AB=\sqrt{\{-5-(-3)\}^2+(-5-2)^2}
\displaystyle =\sqrt{4+49}=\sqrt{53}.
\displaystyle BC=\sqrt{\{2-(-5)\}^2+\{-3-(-5)\}^2}
\displaystyle =\sqrt{49+4}=\sqrt{53}.
\displaystyle CD=\sqrt{(4-2)^2+\{4-(-3)\}^2}
\displaystyle =\sqrt{4+49}=\sqrt{53}.
\displaystyle DA=\sqrt{\{-3-4\}^2+(2-4)^2}
\displaystyle =\sqrt{49+4}=\sqrt{53}.
\displaystyle \therefore AB=BC=CD=DA.
\displaystyle \therefore ABCD\text{ is a rhombus.}
\displaystyle \\

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

\displaystyle \textbf{Question 17: }\text{The vertices of a triangle are }(5,1),\ (11,1)\text{ and }(11,9).\text{ Find the}
\displaystyle \text{co-ordinates of the circumcentre of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the circumcentre be }P(x,y).
\displaystyle \text{Let }A=(5,1),\ B=(11,1)\text{ and }C=(11,9).
\displaystyle \text{Since }P\text{ is the circumcentre, }PA=PB=PC.
\displaystyle PA=PB
\displaystyle \Rightarrow \sqrt{(x-5)^2+(y-1)^2}=\sqrt{(x-11)^2+(y-1)^2}
\displaystyle \Rightarrow (x-5)^2=(x-11)^2
\displaystyle \Rightarrow x^2-10x+25=x^2-22x+121
\displaystyle \Rightarrow 12x=96
\displaystyle \Rightarrow x=8.
\displaystyle PB=PC
\displaystyle \Rightarrow \sqrt{(x-11)^2+(y-1)^2}=\sqrt{(x-11)^2+(y-9)^2}
\displaystyle \Rightarrow (y-1)^2=(y-9)^2
\displaystyle \Rightarrow y^2-2y+1=y^2-18y+81
\displaystyle \Rightarrow 16y=80
\displaystyle \Rightarrow y=5.
\displaystyle \therefore \text{The circumcentre of the triangle is }(8,5).
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Given }A=(3,1)\text{ and }B=(0,y-1).\text{ Find }y\text{ if }AB=5.
\displaystyle \text{Answer:}
\displaystyle \text{Using the distance formula,}
\displaystyle \sqrt{(0-3)^2+\{(y-1)-1\}^2}=5
\displaystyle \Rightarrow 9+(y-2)^2=25
\displaystyle \Rightarrow (y-2)^2=16
\displaystyle \Rightarrow y-2=\pm4
\displaystyle \therefore y=6\text{ or }y=-2.
\displaystyle \\

\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 \text{Using the distance formula,}
\displaystyle \sqrt{\{11-(x+2)\}^2+\{6-(-2)\}^2}=17
\displaystyle \Rightarrow (9-x)^2+64=289
\displaystyle \Rightarrow (9-x)^2=225
\displaystyle \Rightarrow 9-x=\pm15
\displaystyle \therefore x=-6\text{ or }x=24.
\displaystyle \\

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

\displaystyle \textbf{Question 21: }\text{The length of line }PQ\text{ is }10\text{ units and the co-ordinates of }P\text{ are }(2,-3);
\displaystyle \text{calculate the co-ordinates of point }Q,\text{ if its abscissa is }10.
\displaystyle \text{Answer:}
\displaystyle \text{Let the co-ordinates of }Q\text{ be }(10,y).
\displaystyle \text{Given, }PQ=10.
\displaystyle \therefore \sqrt{(10-2)^2+\{y-(-3)\}^2}=10
\displaystyle \Rightarrow 64+(y+3)^2=100
\displaystyle \Rightarrow (y+3)^2=36
\displaystyle \Rightarrow y+3=\pm6
\displaystyle \Rightarrow y=3\text{ or }y=-9.
\displaystyle \therefore \text{The co-ordinates of }Q\text{ are }(10,3)\text{ and }(10,-9).
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Point }P(2,-7)\text{ is the centre of a circle with radius }13\text{ unit. }PT\text{ is}
\displaystyle \text{perpendicular to chord }AB\text{ and }T=(-2,-4).\text{ Calculate the length of:}
\displaystyle \text{(i) }AT\qquad\text{(ii) }AB \displaystyle \text{Answer:}
\displaystyle PT=\sqrt{\{2-(-2)\}^2+\{-7-(-4)\}^2}
\displaystyle =\sqrt{4^2+(-3)^2}
\displaystyle =\sqrt{16+9}=5\text{ units.}
\displaystyle \text{Since }PT\perp AB,\text{ the perpendicular from the centre to a chord bisects the chord.}
\displaystyle \therefore AT=TB.
\displaystyle \text{Also, }PA=13\text{ units.}
\displaystyle \text{In right-angled }\triangle PTA,
\displaystyle PA^2=PT^2+AT^2
\displaystyle \Rightarrow 13^2=5^2+AT^2
\displaystyle \Rightarrow 169=25+AT^2
\displaystyle \Rightarrow AT^2=144
\displaystyle \therefore AT=12\text{ units.}
\displaystyle \text{(i) }AT=12\text{ units.}

\displaystyle \text{(ii) }AB=AT+TB
\displaystyle =12+12=24\text{ units.}
\displaystyle \\

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

\displaystyle \text{Answer:}
\displaystyle PQ=\sqrt{(5-2)^2+(4-2)^2}
\displaystyle =\sqrt{9+4}=\sqrt{13}
\displaystyle =3.6055\ldots
\displaystyle \therefore PQ=3.61\text{ units (correct to three significant figures).}
\displaystyle \\

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

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

\displaystyle \textbf{Question 26: }\text{Find the point on }y\text{-axis whose distances from the points }A(6,7)\text{ and}
\displaystyle B(4,-3)\text{ are in the ratio }1:2.
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point on the }y\text{-axis be }P(0,y).
\displaystyle \text{Given, }\frac{PA}{PB}=\frac{1}{2}
\displaystyle \Rightarrow \frac{\sqrt{(0-6)^2+(y-7)^2}}{\sqrt{(0-4)^2+\{y-(-3)\}^2}}=\frac{1}{2}
\displaystyle \Rightarrow \frac{\sqrt{36+(y-7)^2}}{\sqrt{16+(y+3)^2}}=\frac{1}{2}
\displaystyle \Rightarrow 4\{36+(y-7)^2\}=16+(y+3)^2
\displaystyle \Rightarrow 144+4(y^2-14y+49)=16+y^2+6y+9
\displaystyle \Rightarrow 3y^2-62y+315=0
\displaystyle \Rightarrow 3y^2-27y-35y+315=0
\displaystyle \Rightarrow 3y(y-9)-35(y-9)=0
\displaystyle \Rightarrow (y-9)(3y-35)=0
\displaystyle \Rightarrow y=9\text{ or }y=\frac{35}{3}.
\displaystyle \therefore \text{The required points are }(0,9)\text{ and }\left(0,\frac{35}{3}\right).
\displaystyle \\

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

\displaystyle \textbf{Question 28: }\text{The points }A(3,0),\ B(a,-2)\text{ and }C(4,-1)\text{ are the vertices of triangle}
\displaystyle ABC\text{ right angled at vertex }A.\text{ Find the value of }a.
\displaystyle \text{Answer:}
\displaystyle AB^2=(a-3)^2+(-2-0)^2
\displaystyle =(a-3)^2+4.
\displaystyle AC^2=(4-3)^2+(-1-0)^2
\displaystyle =1+1=2.
\displaystyle BC^2=(4-a)^2+\{-1-(-2)\}^2
\displaystyle =(4-a)^2+1.
\displaystyle \text{Since }\triangle ABC\text{ is right angled at }A,
\displaystyle AB^2+AC^2=BC^2
\displaystyle \Rightarrow (a-3)^2+4+2=(4-a)^2+1
\displaystyle \Rightarrow a^2-6a+9+6=a^2-8a+16+1
\displaystyle \Rightarrow -6a+15=-8a+17
\displaystyle \Rightarrow 2a=2
\displaystyle \therefore a=1.
\displaystyle \\

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