\displaystyle \textbf{Question 1: }\text{Find the distance between the following pairs of points:}
\displaystyle \text{(i) }(-6,7)\text{ and }(-1,-5)\qquad\text{(ii) }(a+b,b+c)\text{ and }(a-b,c-b)
\displaystyle \text{(iii) }(a\sin\alpha,-b\cos\alpha)\text{ and }(-a\cos\alpha,b\sin\alpha)
\displaystyle \text{(iv) }(a,0)\text{ and }(0,b)
\displaystyle \text{Answer:}
\displaystyle \text{Using the distance formula, }d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
\displaystyle \text{(i) }d=\sqrt{(-1+6)^2+(-5-7)^2}
\displaystyle =\sqrt{25+144}=\sqrt{169}=13.

\displaystyle \text{(ii) }d=\sqrt{(a-b-a-b)^2+(c-b-b-c)^2}
\displaystyle =\sqrt{(-2b)^2+(c-2b-c)^2}
\displaystyle =\sqrt{4b^2+4b^2}=\sqrt{8b^2}=2\sqrt{2}|b|.

\displaystyle \text{(iii) }d=\sqrt{(-a\cos\alpha-a\sin\alpha)^2+(b\sin\alpha+b\cos\alpha)^2}
\displaystyle =\sqrt{a^2(\sin\alpha+\cos\alpha)^2+b^2(\sin\alpha+\cos\alpha)^2}
\displaystyle =|\sin\alpha+\cos\alpha|\sqrt{a^2+b^2}.

\displaystyle \text{(iv) }d=\sqrt{(0-a)^2+(b-0)^2}=\sqrt{a^2+b^2}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the value of }a\text{ when the distance between the points }(3,a)
\displaystyle \text{and }(4,1)\text{ is }\sqrt{10}.
\displaystyle \text{Answer:}
\displaystyle \text{Using the distance formula,}
\displaystyle \sqrt{(4-3)^2+(1-a)^2}=\sqrt{10}
\displaystyle 1+(1-a)^2=10
\displaystyle (1-a)^2=9
\displaystyle 1-a=\pm3
\displaystyle \therefore a=4\quad\text{or}\quad a=-2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The length of a line segment is }10\text{ units and the coordinates of one}
\displaystyle \text{end-point are }(2,-3).\text{ If the abscissa of the other end is }10,\text{ find the ordinate}
\displaystyle \text{of the other end.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the other end-point be }(10,y).
\displaystyle \text{Using the distance formula,}
\displaystyle \sqrt{(10-2)^2+(y+3)^2}=10
\displaystyle 64+(y+3)^2=100
\displaystyle (y+3)^2=36
\displaystyle y+3=\pm6
\displaystyle \therefore y=3\quad\text{or}\quad y=-9.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the points }(-4,-1),\,(-2,-4),\,(4,0)\text{ and }(2,3)\text{ are the}
\displaystyle \text{vertices of a rectangle.}\hfill\text{[CBSE 2006 C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-4,-1),\ B(-2,-4),\ C(4,0)\text{ and }D(2,3)\text{ be the given points.}
\displaystyle AB=\sqrt{(-2+4)^2+(-4+1)^2}=\sqrt{4+9}=\sqrt{13}
\displaystyle BC=\sqrt{(4+2)^2+(0+4)^2}=\sqrt{36+16}=2\sqrt{13}
\displaystyle CD=\sqrt{(2-4)^2+(3-0)^2}=\sqrt{4+9}=\sqrt{13}
\displaystyle DA=\sqrt{(-4-2)^2+(-1-3)^2}=\sqrt{36+16}=2\sqrt{13}
\displaystyle \therefore AB=CD\quad\text{and}\quad BC=DA.
\displaystyle \text{Thus, }ABCD\text{ is a parallelogram.}
\displaystyle AC=\sqrt{(4+4)^2+(0+1)^2}=\sqrt{64+1}=\sqrt{65}
\displaystyle BD=\sqrt{(2+2)^2+(3+4)^2}=\sqrt{16+49}=\sqrt{65}
\displaystyle \therefore AC=BD.
\displaystyle \text{Since }ABCD\text{ is a parallelogram with equal diagonals, it is a rectangle.}
\displaystyle \therefore \text{The given points are the vertices of a rectangle.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Show that the points }A(1,-2),\,B(3,6),\,C(5,10)\text{ and }D(3,2)\text{ are the}
\displaystyle \text{vertices of a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(3-1)^2+(6+2)^2}=\sqrt{4+64}=2\sqrt{17}
\displaystyle BC=\sqrt{(5-3)^2+(10-6)^2}=\sqrt{4+16}=2\sqrt{5}
\displaystyle CD=\sqrt{(3-5)^2+(2-10)^2}=\sqrt{4+64}=2\sqrt{17}
\displaystyle DA=\sqrt{(1-3)^2+(-2-2)^2}=\sqrt{4+16}=2\sqrt{5}
\displaystyle \therefore AB=CD\quad\text{and}\quad BC=DA.
\displaystyle \text{Since both pairs of opposite sides are equal, }ABCD\text{ is a parallelogram.}
\displaystyle \therefore \text{The given points are the vertices of a parallelogram.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{(i) Show that }\triangle ABC,\text{ where }A(-2,0),\,B(2,0),\,C(0,2)\text{ and}
\displaystyle \triangle PQR,\text{ where }P(-4,0),\,Q(4,0),\,R(0,4)\text{ are similar.}\hfill\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(2+2)^2+(0-0)^2}=4
\displaystyle BC=\sqrt{(0-2)^2+(2-0)^2}=2\sqrt{2}
\displaystyle CA=\sqrt{(-2-0)^2+(0-2)^2}=2\sqrt{2}
\displaystyle PQ=\sqrt{(4+4)^2+(0-0)^2}=8
\displaystyle QR=\sqrt{(0-4)^2+(4-0)^2}=4\sqrt{2}
\displaystyle RP=\sqrt{(-4-0)^2+(0-4)^2}=4\sqrt{2}
\displaystyle \therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}=\frac{1}{2}.
\displaystyle \therefore \triangle ABC\sim\triangle PQR\quad\text{(by SSS similarity criterion).}

\displaystyle \text{(ii) Show that the points }(-2,3),\,(8,3)\text{ and }(6,7)\text{ are the vertices of a}
\displaystyle \text{right-angled triangle.}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,3),\ B(8,3)\text{ and }C(6,7)\text{ be the given points.}
\displaystyle AB^2=(8+2)^2+(3-3)^2=100
\displaystyle BC^2=(6-8)^2+(7-3)^2=4+16=20
\displaystyle CA^2=(-2-6)^2+(3-7)^2=64+16=80
\displaystyle \therefore BC^2+CA^2=20+80=100=AB^2.
\displaystyle \therefore BC^2+CA^2=AB^2.
\displaystyle \text{Hence, by the converse of Pythagoras theorem, }\triangle ABC\text{ is right-angled at }C.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove that the points }(3,0),\,(6,4)\text{ and }(-1,3)\text{ are vertices of a}
\displaystyle \text{right-angled isosceles triangle.}\hfill\text{[CBSE 2006 C, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,0),\ B(6,4)\text{ and }C(-1,3)\text{ be the given points.}
\displaystyle AB^2=(6-3)^2+(4-0)^2=9+16=25
\displaystyle AC^2=(-1-3)^2+(3-0)^2=16+9=25
\displaystyle BC^2=(-1-6)^2+(3-4)^2=49+1=50
\displaystyle \therefore AB=AC.
\displaystyle \text{Hence, }\triangle ABC\text{ is isosceles.}
\displaystyle \text{Also, }AB^2+AC^2=25+25=50=BC^2.
\displaystyle \text{By the converse of Pythagoras theorem, }\triangle ABC\text{ is right-angled at }A.
\displaystyle \therefore \triangle ABC\text{ is a right-angled isosceles triangle.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Prove that }(2,-2),\,(-2,1)\text{ and }(5,2)\text{ are the vertices of a right-angled}
\displaystyle \text{triangle. Find the area of the triangle and the length of the hypotenuse.}\hfill\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,-2),\ B(-2,1)\text{ and }C(5,2)\text{ be the given points.}
\displaystyle AB^2=(-2-2)^2+(1+2)^2=16+9=25
\displaystyle AC^2=(5-2)^2+(2+2)^2=9+16=25
\displaystyle BC^2=(5+2)^2+(2-1)^2=49+1=50
\displaystyle \therefore AB^2+AC^2=25+25=50=BC^2.
\displaystyle \text{By the converse of Pythagoras theorem, }\triangle ABC\text{ is right-angled at }A.
\displaystyle AB=AC=5\text{ units and }BC=\sqrt{50}=5\sqrt{2}\text{ units.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times AB\times AC
\displaystyle =\frac{1}{2}\times5\times5=\frac{25}{2}\text{ sq. units.}
\displaystyle \therefore \text{Area of the triangle}=\frac{25}{2}\text{ sq. units.}
\displaystyle \therefore \text{Length of the hypotenuse}=5\sqrt{2}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove that the points }(2,3),\,(-4,-6)\text{ and }\left(1,\frac{3}{2}\right)\text{ do not form}
\displaystyle \text{a triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,3),\ B(-4,-6)\text{ and }C\left(1,\frac{3}{2}\right)\text{ be the given points.}
\displaystyle AB=\sqrt{(-4-2)^2+(-6-3)^2}=\sqrt{36+81}=\sqrt{117}=3\sqrt{13}
\displaystyle AC=\sqrt{(1-2)^2+\left(\frac{3}{2}-3\right)^2}=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}
\displaystyle BC=\sqrt{(1+4)^2+\left(\frac{3}{2}+6\right)^2}=\sqrt{25+\frac{225}{4}}=\frac{5\sqrt{13}}{2}
\displaystyle AC+BC=\frac{\sqrt{13}}{2}+\frac{5\sqrt{13}}{2}=3\sqrt{13}=AB.
\displaystyle \text{Hence, the three points are collinear and do not form a triangle.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The points }A(2,9),\,B(a,5)\text{ and }C(5,5)\text{ are the vertices of a triangle }ABC
\displaystyle \text{right-angled at }B.\text{ Find the values of }a\text{ and hence the area of }\triangle ABC.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\triangle ABC\text{ is right-angled at }B,
\displaystyle AC^2=AB^2+BC^2.
\displaystyle AC^2=(5-2)^2+(5-9)^2=9+16=25
\displaystyle AB^2=(a-2)^2+(5-9)^2=(a-2)^2+16
\displaystyle BC^2=(5-a)^2+(5-5)^2=(5-a)^2
\displaystyle \therefore 25=(a-2)^2+16+(5-a)^2
\displaystyle 25=a^2-4a+4+16+a^2-10a+25
\displaystyle 2a^2-14a+20=0
\displaystyle a^2-7a+10=0
\displaystyle (a-2)(a-5)=0
\displaystyle \therefore a=2\quad\text{or}\quad a=5.
\displaystyle \text{For }a=5,\ B=C,\text{ so no triangle is formed. Hence, }a=2.
\displaystyle AB=4\text{ units and }BC=3\text{ units.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times AB\times BC
\displaystyle =\frac{1}{2}\times4\times3=6\text{ sq. units.}
\displaystyle \therefore a=2\text{ and the area of }\triangle ABC=6\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If the point }P(2,2)\text{ is equidistant from the points }A(-2,k)\text{ and}
\displaystyle B(-2k,-3),\text{ find }k.\text{ Also, find the length of }AP.\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }P\text{ is equidistant from }A\text{ and }B,\quad PA=PB.
\displaystyle \therefore PA^2=PB^2
\displaystyle (2+2)^2+(2-k)^2=(2+2k)^2+(2+3)^2
\displaystyle 16+(2-k)^2=(2+2k)^2+25
\displaystyle 16+k^2-4k+4=4+8k+4k^2+25
\displaystyle 3k^2+12k+9=0
\displaystyle k^2+4k+3=0
\displaystyle (k+1)(k+3)=0
\displaystyle \therefore k=-1\quad\text{or}\quad k=-3.
\displaystyle \text{For }k=-1,\quad AP=\sqrt{(2+2)^2+(2+1)^2}=\sqrt{16+9}=5\text{ units.}
\displaystyle \text{For }k=-3,\quad AP=\sqrt{(2+2)^2+(2+3)^2}=\sqrt{16+25}=\sqrt{41}\text{ units.}
\displaystyle \therefore k=-1,\ AP=5\text{ units; or }k=-3,\ AP=\sqrt{41}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find a relation between }x\text{ and }y\text{ such that the point }(x,y)\text{ is equidistant}
\displaystyle \text{from the points }(3,6)\text{ and }(-3,4).\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y),\ A(3,6)\text{ and }B(-3,4).
\displaystyle \text{Since }P\text{ is equidistant from }A\text{ and }B,\quad PA=PB.
\displaystyle \therefore PA^2=PB^2
\displaystyle (x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2
\displaystyle x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16
\displaystyle -12x-4y+20=0
\displaystyle \therefore 3x+y=5.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Prove that the points }(-2,5),\,(0,1)\text{ and }(2,-3)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,5),\ B(0,1)\text{ and }C(2,-3)\text{ be the given points.}
\displaystyle AB=\sqrt{(0+2)^2+(1-5)^2}=\sqrt{4+16}=2\sqrt{5}
\displaystyle BC=\sqrt{(2-0)^2+(-3-1)^2}=\sqrt{4+16}=2\sqrt{5}
\displaystyle AC=\sqrt{(2+2)^2+(-3-5)^2}=\sqrt{16+64}=4\sqrt{5}
\displaystyle \therefore AB+BC=2\sqrt{5}+2\sqrt{5}=4\sqrt{5}=AC.
\displaystyle \therefore B\text{ lies on the line segment }AC.
\displaystyle \text{Hence, the given points are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If the point }A(2,-4)\text{ is equidistant from }P(3,8)\text{ and }Q(-10,y),
\displaystyle \text{find the values of }y.\text{ Also, find the distance }PQ. 
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ is equidistant from }P\text{ and }Q,\quad AP=AQ.
\displaystyle \therefore AP^2=AQ^2
\displaystyle (3-2)^2+(8+4)^2=(-10-2)^2+(y+4)^2
\displaystyle 1+144=144+(y+4)^2
\displaystyle (y+4)^2=1
\displaystyle y+4=\pm1
\displaystyle \therefore y=-3\quad\text{or}\quad y=-5.
\displaystyle \text{When }y=-3,\quad PQ=\sqrt{(-10-3)^2+(-3-8)^2}
\displaystyle =\sqrt{169+121}=\sqrt{290}\text{ units.}
\displaystyle \text{When }y=-5,\quad PQ=\sqrt{(-10-3)^2+(-5-8)^2}
\displaystyle =\sqrt{169+169}=\sqrt{338}=13\sqrt{2}\text{ units.}
\displaystyle \therefore y=-3,\ PQ=\sqrt{290}\text{ units; or }y=-5,\ PQ=13\sqrt{2}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The three vertices of a parallelogram are }(3,4),\,(3,8)\text{ and }(9,8).
\displaystyle \text{Find the fourth vertex.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,4),\ B(3,8)\text{ and }C(9,8)\text{ be three consecutive vertices.}
\displaystyle AB=8-4=4\text{ units and }BC=9-3=6\text{ units.}
\displaystyle \text{Since }ABCD\text{ is a parallelogram, }AD\parallel BC\text{ and }CD\parallel AB.
\displaystyle \therefore D=(9,4).
\displaystyle \therefore \text{The fourth vertex is }(9,4).
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Name the quadrilateral formed, if any, by the following points, and give}
\displaystyle \text{reasons for your answers:}
\displaystyle \text{(i) }A(2,-2),\,B(7,3),\,C(11,-1),\,D(6,-6)
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(7-2)^2+(3+2)^2}=5\sqrt{2}
\displaystyle BC=\sqrt{(11-7)^2+(-1-3)^2}=4\sqrt{2}
\displaystyle CD=\sqrt{(6-11)^2+(-6+1)^2}=5\sqrt{2}
\displaystyle DA=\sqrt{(2-6)^2+(-2+6)^2}=4\sqrt{2}
\displaystyle \therefore AB=CD\quad\text{and}\quad BC=DA.
\displaystyle \text{Hence, }ABCD\text{ is a parallelogram.}
\displaystyle AC=\sqrt{(11-2)^2+(-1+2)^2}=\sqrt{82}
\displaystyle BD=\sqrt{(6-7)^2+(-6-3)^2}=\sqrt{82}
\displaystyle \therefore AC=BD.
\displaystyle \text{A parallelogram with equal diagonals is a rectangle.}
\displaystyle \therefore ABCD\text{ is a rectangle.}

\displaystyle \text{(ii) }A(4,5),\,B(7,6),\,C(4,3),\,D(1,2)
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(7-4)^2+(6-5)^2}=\sqrt{10}
\displaystyle BC=\sqrt{(4-7)^2+(3-6)^2}=3\sqrt{2}
\displaystyle CD=\sqrt{(1-4)^2+(2-3)^2}=\sqrt{10}
\displaystyle DA=\sqrt{(4-1)^2+(5-2)^2}=3\sqrt{2}
\displaystyle \therefore AB=CD\quad\text{and}\quad BC=DA.
\displaystyle \text{Since both pairs of opposite sides are equal, }ABCD\text{ is a parallelogram.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Prove that the points }(3,0),\,(4,5),\,(-1,4)\text{ and }(-2,-1),\text{ taken in}
\displaystyle \text{order, form a rhombus. Also, find its area.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,0),\ B(4,5),\ C(-1,4)\text{ and }D(-2,-1)\text{ be the given points.}
\displaystyle AB=\sqrt{(4-3)^2+(5-0)^2}=\sqrt{26}
\displaystyle BC=\sqrt{(-1-4)^2+(4-5)^2}=\sqrt{26}
\displaystyle CD=\sqrt{(-2+1)^2+(-1-4)^2}=\sqrt{26}
\displaystyle DA=\sqrt{(3+2)^2+(0+1)^2}=\sqrt{26}
\displaystyle \therefore AB=BC=CD=DA.
\displaystyle \therefore ABCD\text{ is a rhombus.}
\displaystyle AC=\sqrt{(-1-3)^2+(4-0)^2}=\sqrt{32}=4\sqrt{2}
\displaystyle BD=\sqrt{(-2-4)^2+(-1-5)^2}=\sqrt{72}=6\sqrt{2}
\displaystyle \text{Area of rhombus }ABCD=\frac{1}{2}\times AC\times BD
\displaystyle =\frac{1}{2}\times4\sqrt{2}\times6\sqrt{2}=24\text{ sq. units.}
\displaystyle \therefore \text{Area of the rhombus}=24\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the seating arrangement of desks in a classroom, three students Rohini,}
\displaystyle \text{Sandhya and Bina are seated at }A(3,1),\,B(6,4)\text{ and }C(8,6).\text{ Do you think}
\displaystyle \text{they are seated in a line?}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(6-3)^2+(4-1)^2}=\sqrt{18}=3\sqrt{2}
\displaystyle BC=\sqrt{(8-6)^2+(6-4)^2}=\sqrt{8}=2\sqrt{2}
\displaystyle AC=\sqrt{(8-3)^2+(6-1)^2}=\sqrt{50}=5\sqrt{2}
\displaystyle \therefore AB+BC=3\sqrt{2}+2\sqrt{2}=5\sqrt{2}=AC.
\displaystyle \therefore A,\ B\text{ and }C\text{ are collinear.}
\displaystyle \text{Hence, Rohini, Sandhya and Bina are seated in a line.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find a point on }y\text{-axis which is equidistant from the points }(5,-2)
\displaystyle \text{and }(-3,2).\hfill\text{[CBSE 2009, 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(0,y)\text{ be the required point on }y\text{-axis.}
\displaystyle \text{Let }A(5,-2)\text{ and }B(-3,2).
\displaystyle \text{Since }P\text{ is equidistant from }A\text{ and }B,\quad PA=PB.
\displaystyle \therefore PA^2=PB^2
\displaystyle (0-5)^2+(y+2)^2=(0+3)^2+(y-2)^2
\displaystyle 25+y^2+4y+4=9+y^2-4y+4
\displaystyle 8y=-16
\displaystyle y=-2
\displaystyle \therefore \text{The required point is }(0,-2).
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Find a point on the }x\text{-axis which is equidistant from the points }(7,6)
\displaystyle \text{and }(-3,4).\hfill\text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,0)\text{ be the required point on the }x\text{-axis.}
\displaystyle \text{Let }A(7,6)\text{ and }B(-3,4).
\displaystyle \text{Since }P\text{ is equidistant from }A\text{ and }B,\quad PA=PB.
\displaystyle \therefore PA^2=PB^2
\displaystyle (x-7)^2+(0-6)^2=(x+3)^2+(0-4)^2
\displaystyle x^2-14x+49+36=x^2+6x+9+16
\displaystyle -20x=-60
\displaystyle x=3
\displaystyle \therefore \text{The required point is }(3,0).
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{(i) Prove that the points }A(2,3),\,B(-2,2),\,C(-1,-2)\text{ and }D(3,-1)
\displaystyle \text{are the vertices of a square }ABCD.\hfill\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(-2-2)^2+(2-3)^2}=\sqrt{17}
\displaystyle BC=\sqrt{(-1+2)^2+(-2-2)^2}=\sqrt{17}
\displaystyle CD=\sqrt{(3+1)^2+(-1+2)^2}=\sqrt{17}
\displaystyle DA=\sqrt{(2-3)^2+(3+1)^2}=\sqrt{17}
\displaystyle \therefore AB=BC=CD=DA.
\displaystyle \text{Hence, }ABCD\text{ is a rhombus.}
\displaystyle AC=\sqrt{(-1-2)^2+(-2-3)^2}=\sqrt{34}
\displaystyle BD=\sqrt{(3+2)^2+(-1-2)^2}=\sqrt{34}
\displaystyle \therefore AC=BD.
\displaystyle \text{A rhombus with equal diagonals is a square.}
\displaystyle \therefore ABCD\text{ is a square.}

\displaystyle \text{(ii) Name the type of triangle }PQR\text{ formed by the points }P(\sqrt{2},\sqrt{2}),
\displaystyle Q(-\sqrt{2},-\sqrt{2})\text{ and }R(-\sqrt{6},\sqrt{6}). 
\displaystyle \text{Answer:}
\displaystyle PQ=\sqrt{(-\sqrt{2}-\sqrt{2})^2+(-\sqrt{2}-\sqrt{2})^2}=4
\displaystyle QR=\sqrt{(-\sqrt{6}+\sqrt{2})^2+(\sqrt{6}+\sqrt{2})^2}=4
\displaystyle RP=\sqrt{(\sqrt{2}+\sqrt{6})^2+(\sqrt{2}-\sqrt{6})^2}=4
\displaystyle \therefore PQ=QR=RP.
\displaystyle \therefore \triangle PQR\text{ is an equilateral triangle.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Find the value of }x\text{ such that }PQ=QR,\text{ where the coordinates of }P,Q
\displaystyle \text{and }R\text{ are }(6,-1),\,(1,3)\text{ and }(x,8)\text{ respectively.}\hfill\text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }PQ=QR,\quad PQ^2=QR^2.
\displaystyle (1-6)^2+(3+1)^2=(x-1)^2+(8-3)^2
\displaystyle 25+16=(x-1)^2+25
\displaystyle (x-1)^2=16
\displaystyle x-1=\pm4
\displaystyle \therefore x=5\quad\text{or}\quad x=-3.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }Q(0,1)\text{ is equidistant from }P(5,-3)\text{ and }R(x,6),\text{ find the}
\displaystyle \text{values of }x.\text{ Also, find the distances }QR\text{ and }PR.
\displaystyle \text{Answer:}
\displaystyle \text{Since }Q\text{ is equidistant from }P\text{ and }R,\quad QP=QR.
\displaystyle \therefore QP^2=QR^2
\displaystyle (5-0)^2+(-3-1)^2=(x-0)^2+(6-1)^2
\displaystyle 25+16=x^2+25
\displaystyle x^2=16
\displaystyle \therefore x=4\quad\text{or}\quad x=-4.
\displaystyle QR=QP=\sqrt{25+16}=\sqrt{41}\text{ units.}
\displaystyle \text{When }x=4,\quad PR=\sqrt{(4-5)^2+(6+3)^2}
\displaystyle =\sqrt{1+81}=\sqrt{82}\text{ units.}
\displaystyle \text{When }x=-4,\quad PR=\sqrt{(-4-5)^2+(6+3)^2}
\displaystyle =\sqrt{81+81}=9\sqrt{2}\text{ units.}
\displaystyle \therefore x=4,\ QR=\sqrt{41},\ PR=\sqrt{82}\text{ units; or}
\displaystyle x=-4,\ QR=\sqrt{41},\ PR=9\sqrt{2}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Find the values of }y\text{ for which the distance between the points }P(2,-3)
\displaystyle \text{and }Q(10,y)\text{ is }10\text{ units.}
\displaystyle \text{Answer:}
\displaystyle \text{Using the distance formula,}
\displaystyle \sqrt{(10-2)^2+(y+3)^2}=10
\displaystyle 64+(y+3)^2=100
\displaystyle (y+3)^2=36
\displaystyle y+3=\pm6
\displaystyle \therefore y=3\quad\text{or}\quad y=-9.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }A(3,y)\text{ is equidistant from points }P(8,-3)\text{ and }Q(7,6),\text{ find}
\displaystyle \text{the value of }y\text{ and find the distance }AQ.\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ is equidistant from }P\text{ and }Q,\quad AP=AQ.
\displaystyle \therefore AP^2=AQ^2
\displaystyle (8-3)^2+(-3-y)^2=(7-3)^2+(6-y)^2
\displaystyle 25+(y+3)^2=16+(6-y)^2
\displaystyle 25+y^2+6y+9=16+y^2-12y+36
\displaystyle 18y=18
\displaystyle \therefore y=1.
\displaystyle AQ=\sqrt{(7-3)^2+(6-1)^2}
\displaystyle =\sqrt{16+25}=\sqrt{41}\text{ units.}
\displaystyle \therefore y=1\text{ and }AQ=\sqrt{41}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Prove that abscissa of a point }P\text{ which is equidistant from points with}
\displaystyle \text{coordinates }A(7,1)\text{ and }B(3,5)\text{ is }2\text{ more than its ordinate.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be equidistant from }A(7,1)\text{ and }B(3,5).
\displaystyle \therefore PA=PB
\displaystyle \therefore PA^2=PB^2
\displaystyle (x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2
\displaystyle x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25
\displaystyle -8x+8y+16=0
\displaystyle x-y=2
\displaystyle \therefore x=y+2.
\displaystyle \text{Hence, the abscissa of }P\text{ is }2\text{ more than its ordinate.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Find the equation of the perpendicular bisector of the line segment joining}
\displaystyle \text{the points }(7,1)\text{ and }(3,5). 
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be any point on the perpendicular bisector of the line segment.}
\displaystyle \therefore P\text{ is equidistant from }A(7,1)\text{ and }B(3,5).
\displaystyle PA=PB
\displaystyle \therefore PA^2=PB^2
\displaystyle (x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2
\displaystyle x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25
\displaystyle -8x+8y+16=0
\displaystyle \therefore x-y-2=0.
\displaystyle \text{Hence, the equation of the perpendicular bisector is }x-y-2=0.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{The centre of a circle is }(2a,a-7).\text{ Find the values of }a\text{ if the circle}
\displaystyle \text{passes through the point }(11,-9)\text{ and has diameter }10\sqrt{2}\text{ units.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter}=10\sqrt{2}\text{ units.}
\displaystyle \therefore \text{Radius}=5\sqrt{2}\text{ units.}
\displaystyle \text{Since }(11,-9)\text{ lies on the circle,}
\displaystyle \sqrt{(2a-11)^2+(a-7+9)^2}=5\sqrt{2}
\displaystyle (2a-11)^2+(a+2)^2=50
\displaystyle 4a^2-44a+121+a^2+4a+4=50
\displaystyle 5a^2-40a+75=0
\displaystyle a^2-8a+15=0
\displaystyle (a-3)(a-5)=0
\displaystyle \therefore a=3\quad\text{or}\quad a=5.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Ayush starts walking from his house to office. Instead of going to the office}
\displaystyle \text{directly, he goes to a bank first, from there to his daughter's school and then reaches}
\displaystyle \text{the office. What is the extra distance travelled by Ayush in reaching the office? Assume}
\displaystyle \text{that all distances covered are in straight lines. The house is situated at }(2,4),\text{ bank}
\displaystyle \text{at }(5,8),\text{ school at }(13,14)\text{ and office at }(13,26).\text{ Coordinates are in kilometres.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the house, bank, school and office be }H,\ B,\ S\text{ and }O\text{ respectively.}
\displaystyle HB=\sqrt{(5-2)^2+(8-4)^2}=\sqrt{9+16}=5\text{ km}
\displaystyle BS=\sqrt{(13-5)^2+(14-8)^2}=\sqrt{64+36}=10\text{ km}
\displaystyle SO=\sqrt{(13-13)^2+(26-14)^2}=12\text{ km}
\displaystyle \therefore \text{Total distance travelled}=5+10+12=27\text{ km}
\displaystyle HO=\sqrt{(13-2)^2+(26-4)^2}
\displaystyle =\sqrt{121+484}=\sqrt{605}=11\sqrt{5}\text{ km}
\displaystyle \therefore \text{Extra distance travelled}=27-11\sqrt{5}\text{ km}
\displaystyle \approx27-24.60=2.40\text{ km}
\displaystyle \therefore \text{Ayush travels approximately }2.40\text{ km extra.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{(i) If }(0,-3)\text{ and }(0,3)\text{ are the two vertices of an equilateral}
\displaystyle \text{triangle, find the coordinates of its third vertex.}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(0,-3),\ B(0,3)\text{ and }C(x,y)\text{ be the vertices of the equilateral triangle.}
\displaystyle \text{Since }AC=BC,
\displaystyle (x-0)^2+(y+3)^2=(x-0)^2+(y-3)^2
\displaystyle (y+3)^2=(y-3)^2
\displaystyle y=0
\displaystyle \text{Also, }AB=6\text{ units and }AC=AB.
\displaystyle x^2+(0+3)^2=6^2
\displaystyle x^2+9=36
\displaystyle x^2=27
\displaystyle x=\pm3\sqrt{3}
\displaystyle \therefore \text{The third vertex is }(3\sqrt{3},0)\text{ or }(-3\sqrt{3},0).

\displaystyle \text{(ii) If }(-5,3)\text{ and }(5,3)\text{ are two vertices of an equilateral triangle, find the}
\displaystyle \text{coordinates of the third vertex, given that the origin lies inside the triangle.}
\displaystyle \text{Take }\sqrt{3}=1.7.\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-5,3),\ B(5,3)\text{ and }C(x,y)\text{ be the vertices of the equilateral triangle.}
\displaystyle \text{Since }AC=BC,
\displaystyle (x+5)^2+(y-3)^2=(x-5)^2+(y-3)^2
\displaystyle (x+5)^2=(x-5)^2
\displaystyle x=0
\displaystyle \text{Also, }AB=10\text{ units and }AC=AB.
\displaystyle (0+5)^2+(y-3)^2=10^2
\displaystyle 25+(y-3)^2=100
\displaystyle (y-3)^2=75
\displaystyle y-3=\pm5\sqrt{3}
\displaystyle y=3\pm5\sqrt{3}
\displaystyle \text{Since the origin lies inside the triangle, the third vertex must lie below the }x\text{-axis.}
\displaystyle \therefore y=3-5\sqrt{3}=3-5(1.7)=-5.5
\displaystyle \therefore \text{The third vertex is }(0,-5.5).
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{An equilateral triangle has two vertices at the points }(3,4)\text{ and }(-2,3).
\displaystyle \text{Find the coordinates of the third vertex.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,4),\ B(-2,3)\text{ and }C(x,y)\text{ be the vertices of the equilateral triangle.}
\displaystyle \text{Since }AC=BC,
\displaystyle (x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2
\displaystyle 5x+y-6=0
\displaystyle \therefore y=6-5x\qquad\ldots\text{(i)}
\displaystyle \text{Also, }AC=AB\text{ and }AB^2=(-2-3)^2+(3-4)^2=26.
\displaystyle \therefore (x-3)^2+(y-4)^2=26
\displaystyle \text{Using }y=6-5x,
\displaystyle (x-3)^2+(2-5x)^2=26
\displaystyle 2x^2-2x-1=0
\displaystyle x=\frac{1\pm\sqrt{3}}{2}
\displaystyle \text{From (i), }y=\frac{7\mp5\sqrt{3}}{2}.
\displaystyle \therefore \text{The third vertex is }\left(\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2}\right)\text{ or}
\displaystyle \left(\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Find the circumcentre of the triangle whose vertices are }(-2,-3),\,(-1,0)
\displaystyle \text{and }(7,-6).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be the circumcentre and }A(-2,-3),\ B(-1,0),\ C(7,-6).
\displaystyle \text{Since the circumcentre is equidistant from the vertices, }PA=PB=PC.
\displaystyle PA^2=PB^2
\displaystyle (x+2)^2+(y+3)^2=(x+1)^2+y^2
\displaystyle x+3y+6=0\qquad\ldots\text{(i)}
\displaystyle PB^2=PC^2
\displaystyle (x+1)^2+y^2=(x-7)^2+(y+6)^2
\displaystyle 4x-3y-21=0\qquad\ldots\text{(ii)}
\displaystyle \text{Solving (i) and (ii), we get }x=3,\quad y=-3.
\displaystyle \therefore \text{The circumcentre of the triangle is }(3,-3).
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Find the angle subtended at the origin by the line segment whose end points}
\displaystyle \text{are }(0,100)\text{ and }(10,0).
\displaystyle \text{Answer:}
\displaystyle \text{Let }O(0,0),\ P(0,100)\text{ and }Q(10,0).
\displaystyle OP=100,\quad OQ=10
\displaystyle PQ=\sqrt{(10-0)^2+(0-100)^2}
\displaystyle =\sqrt{100+10000}=10\sqrt{101}
\displaystyle OP^2+OQ^2=100^2+10^2=10100=PQ^2.
\displaystyle \text{By the converse of Pythagoras theorem, }\angle POQ=90^\circ.
\displaystyle \therefore \text{The angle subtended at the origin is }90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Two opposite vertices of a square are }(-1,2)\text{ and }(3,2).\text{ Find the}
\displaystyle \text{coordinates of the other two vertices.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-1,2)\text{ and }C(3,2)\text{ be the opposite vertices of the square.}
\displaystyle AC=\sqrt{(3+1)^2+(2-2)^2}=4\text{ units.}
\displaystyle \text{Mid-point of }AC=\left(\frac{-1+3}{2},\frac{2+2}{2}\right)=(1,2).
\displaystyle \text{The diagonals of a square are equal, bisect each other and are perpendicular.}
\displaystyle \therefore BD=AC=4\text{ units and }\frac{BD}{2}=2\text{ units.}
\displaystyle \text{Since }AC\text{ is horizontal, }BD\text{ is vertical through }(1,2).
\displaystyle \therefore B=(1,4)\quad\text{and}\quad D=(1,0).
\displaystyle \therefore \text{The other two vertices are }(1,4)\text{ and }(1,0).
\displaystyle \\

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