\displaystyle \textbf{Question 1: } \text{A line segment joining }A(-1,\frac{5}{3})\text{ and }B(a,5)\text{ is divided in the ratio } \\ 1:3\text{ at }P,
\displaystyle \text{the point where the line segment }AB\text{ intersects the }y\text{-axis.}
\displaystyle \text{(i) Calculate the value of }a
\displaystyle \text{(ii) Calculate the coordinates of }P.\hfill \text{[ICSE 1994]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }P\text{ divides }AB\text{ internally in the ratio }1:3,\text{ by the section formula,}
\displaystyle P\left(\frac{1(a)+3(-1)}{1+3},\frac{1(5)+3\left(\frac{5}{3}\right)}{1+3}\right)
\displaystyle \text{As }P\text{ lies on the }y\text{-axis, its }x\text{-coordinate is }0.
\displaystyle \frac{a-3}{4}=0
\displaystyle a=3
\displaystyle y=\frac{5+5}{4}=\frac{10}{4}=\frac{5}{2}
\displaystyle \therefore \text{(i) }a=3
\displaystyle \therefore \text{(ii) }P\left(0,\frac{5}{2}\right)
\\

\displaystyle \textbf{Question 2:} \text{In what ratio is the line joining }A(0,3)\text{ and }B(4,-1)\text{ divided by the }x\text{-axis?}
\displaystyle \text{Write the coordinates of the point where }AB\text{ intersects the }x\text{-axis.}\hfill \text{[ICSE 1993]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the }x\text{-axis divide }AB\text{ in the ratio }k:1\text{ at }P(x,0).
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle 0=\frac{k(-1)+1(3)}{k+1}
\displaystyle -k+3=0
\displaystyle k=3
\displaystyle \therefore \text{The required ratio is }3:1.
\displaystyle x=\frac{3(4)+1(0)}{3+1}=\frac{12}{4}=3
\displaystyle \therefore P(3,0)
\\

\displaystyle \textbf{Question 3: } \text{The mid-point of the segment }AB,\text{ as shown in the diagram, is }C(4,-3).
\displaystyle \text{Write down the coordinates of }A\text{ and }B.\hfill \text{[ICSE 1996]}
\displaystyle \text{Answer:}
\displaystyle \text{From the diagram, let }A(x,0)\text{ and }B(0,y).
\displaystyle \text{Since the midpoint of }AB\text{ is }(4,-3),
\displaystyle 4=\frac{x+0}{2}
\displaystyle x=8
\displaystyle -3=\frac{0+y}{2}
\displaystyle y=-6
\displaystyle \therefore A(8,0)\text{ and }B(0,-6)
\\

\displaystyle \textbf{Question 4: }AB\text{ is a diameter of a circle with center }C(-2,5).
\displaystyle \text{If }A(3,-7),\text{ find}
\displaystyle \text{(i) the length of radius }AC
\displaystyle \text{(ii) the coordinates of }B.\hfill \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }C\text{ is the midpoint of diameter }AB,
\displaystyle -2=\frac{x+3}{2}
\displaystyle x=-7
\displaystyle 5=\frac{y+(-7)}{2}
\displaystyle y=17
\displaystyle \therefore \text{(ii) }B(-7,17)
\displaystyle \text{Now, }AC=\sqrt{(-2-3)^2+\left(5-(-7)\right)^2}
\displaystyle =\sqrt{(-5)^2+12^2}
\displaystyle =\sqrt{25+144}
\displaystyle =\sqrt{169}=13
\displaystyle \therefore \text{(i) Radius }AC=13
\\

\displaystyle \textbf{Question 5: } \text{Find the coordinates of the centroid of a triangle }ABC
\displaystyle \text{whose vertices are }A(-1,3),\ B(1,-1)\text{ and }C(5,1).\hfill \text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }O(x,y)\text{ be the centroid of }\triangle ABC.
\displaystyle x=\frac{-1+1+5}{3}=\frac{5}{3}
\displaystyle y=\frac{3+(-1)+1}{3}=1
\displaystyle \therefore \text{The coordinates of the centroid are }\left(\frac{5}{3},1\right)
\\

\displaystyle \textbf{Question 6: } \text{The mid-point of the line segment joining }(2a,4)\text{ and }
\displaystyle (-2,2b)\text{ is }(1,2a+1). \ \text{Find the values of }a\text{ and }b.\hfill \text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{The given midpoint is }(1,2a+1).
\displaystyle \text{Using the midpoint formula,}
\displaystyle 1=\frac{2a+(-2)}{2}
\displaystyle 2=2a-2
\displaystyle 2a=4
\displaystyle a=2
\displaystyle \text{Also, }2a+1=\frac{4+2b}{2}
\displaystyle 2(2)+1=\frac{4+2b}{2}
\displaystyle 5=\frac{4+2b}{2}
\displaystyle 10=4+2b
\displaystyle 2b=6
\displaystyle b=3
\displaystyle \therefore a=2\text{ and }b=3
\\

\displaystyle \textbf{Question 7: (i) } \text{Write down the coordinates of the point }P\text{ that divides the line joining }
\displaystyle A(-4,1)\text{ and }B(17,10) \ \text{in the ratio }1:2.
\displaystyle \text{(ii) Calculate the distance }OP,\text{ where }O\text{ is the origin.}
\displaystyle \text{(iii) In what ratio does the }y\text{-axis divide the line }AB?\hfill \text{[ICSE 1995]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }P\text{ divides }A(-4,1)\text{ and }B(17,10)\text{ in the ratio }1:2.
\displaystyle P\left(\frac{1(17)+2(-4)}{1+2},\frac{1(10)+2(1)}{1+2}\right)
\displaystyle P\left(\frac{17-8}{3},\frac{10+2}{3}\right)
\displaystyle P(3,4)
\displaystyle \therefore \text{The coordinates of }P\text{ are }(3,4).
\displaystyle \text{(ii) }OP=\sqrt{(3-0)^2+(4-0)^2}
\displaystyle =\sqrt{9+16}
\displaystyle =\sqrt{25}=5
\displaystyle \therefore OP=5
\displaystyle \text{(iii) Let the }y\text{-axis divide }AB\text{ in the ratio }k:1\text{ at }Q(0,y).
\displaystyle \text{Since }Q\text{ lies on the }y\text{-axis, its }x\text{-coordinate is }0.
\displaystyle 0=\frac{k(17)+1(-4)}{k+1}
\displaystyle 17k-4=0
\displaystyle k=\frac{4}{17}
\displaystyle \therefore \text{The required ratio is }\frac{4}{17}:1=4:17
\\

\displaystyle \textbf{Question 8: } \text{Prove that the points }A(-5,4),\ B(-1,-2)\text{ and }C(5,2)
\displaystyle \text{are the vertices of an isosceles right-angled triangle. Find the coordinates of }D
\displaystyle \text{so that }ABCD\text{ is a square.}\hfill \text{[ICSE 1992]}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{\left(-1-(-5)\right)^2+(-2-4)^2}
\displaystyle =\sqrt{4^2+(-6)^2}=\sqrt{16+36}=\sqrt{52}
\displaystyle BC=\sqrt{\left(5-(-1)\right)^2+\left(2-(-2)\right)^2}
\displaystyle =\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}
\displaystyle AC=\sqrt{\left(5-(-5)\right)^2+(2-4)^2}
\displaystyle =\sqrt{10^2+(-2)^2}=\sqrt{100+4}=\sqrt{104}
\displaystyle \text{Now, }AB^2+BC^2=52+52=104
\displaystyle \text{and }AC^2=104
\displaystyle \therefore AB^2+BC^2=AC^2
\displaystyle \text{Hence, }\triangle ABC\text{ is right-angled at }B.
\displaystyle \text{Also, }AB=BC.
\displaystyle \therefore \triangle ABC\text{ is an isosceles right-angled triangle.}
\displaystyle \text{For square }ABCD,\text{ the fourth vertex is given by }D=A+C-B.
\displaystyle D=(-5,4)+(5,2)-(-1,-2)
\displaystyle D=(0,6)+(1,2)
\displaystyle D=(1,8)
\displaystyle \therefore \text{The coordinates of }D\text{ are }(1,8).
\\

\displaystyle \textbf{Question 9: } \text{Calculate the ratio in which the line joining }A(-4,2)\text{ and }B(3,6)
\displaystyle \text{is divided by point }P(x,3).\text{ Also, find (i) }x\text{ (ii) length of }AP.\hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,3)\text{ divide }AB\text{ in the ratio }k:1.
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle 3=\frac{k(6)+1(2)}{k+1}
\displaystyle 3k+3=6k+2
\displaystyle 3k=1
\displaystyle k=\frac{1}{3}
\displaystyle \therefore \text{The required ratio is }\frac{1}{3}:1=1:3.
\displaystyle x=\frac{1(3)+3(-4)}{1+3}
\displaystyle x=\frac{3-12}{4}=-\frac{9}{4}
\displaystyle \therefore \text{(i) }x=-\frac{9}{4}
\displaystyle AP=\sqrt{\left(-\frac{9}{4}-(-4)\right)^2+(3-2)^2}
\displaystyle =\sqrt{\left(\frac{7}{4}\right)^2+1^2}
\displaystyle =\sqrt{\frac{49}{16}+\frac{16}{16}}
\displaystyle =\sqrt{\frac{65}{16}}=\frac{\sqrt{65}}{4}
\displaystyle \therefore \text{(ii) }AP=\frac{\sqrt{65}}{4}
\\

\displaystyle \textbf{Question 10: } \text{Calculate the ratio in which the line joining }A(6,5)\text{ and } \\ B(4,-3)\text{ is divided by the line }y=2.\hfill \text{[2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line }y=2\text{ divide }AB\text{ in the ratio }k:1\text{ at }P(x,2).
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle 2=\frac{k(-3)+1(5)}{k+1}
\displaystyle 2k+2=-3k+5
\displaystyle 5k=3
\displaystyle k=\frac{3}{5}
\displaystyle \therefore \text{The required ratio is }\frac{3}{5}:1=3:5.
\displaystyle x=\frac{3(4)+5(6)}{3+5}
\displaystyle =\frac{12+30}{8}=\frac{21}{4}
\displaystyle \therefore \text{The point of intersection is }\left(\frac{21}{4},2\right)
\\

\displaystyle \textbf{Question 11: } \text{If }A=(-4,3)\text{ and }B=(8,-6),
\displaystyle \text{(i) find the length of }AB
\displaystyle \text{(ii) In what ratio is the line joining }A\text{ and }B\text{ divided by the }x\text{-axis?}\hfill \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }AB=\sqrt{(8-(-4))^2+(-6-3)^2}
\displaystyle =\sqrt{12^2+(-9)^2}
\displaystyle =\sqrt{144+81}
\displaystyle =\sqrt{225}=15
\displaystyle \therefore AB=15
\displaystyle \text{(ii) Let the }x\text{-axis divide }AB\text{ in the ratio }k:1\text{ at }P(x,0).
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle 0=\frac{k(-6)+1(3)}{k+1}
\displaystyle -6k+3=0
\displaystyle 6k=3
\displaystyle k=\frac{1}{2}
\displaystyle \therefore \text{The required ratio is }\frac{1}{2}:1=1:2
\\

\displaystyle \textbf{Question 12: } \text{The line segment joining }A(2,3)\text{ and }B(6,-5)\text{ is intercepted by the }x\text{-axis at the point }K.
\displaystyle \text{Write down the ordinate of the point }K.\text{ Hence, find the ratio in which }K\text{ divides }AB.
\displaystyle \text{Also, find the coordinates of the point }K.\hfill \text{[ICSE 1990, 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }K\text{ lies on the }x\text{-axis, its ordinate is }0.
\displaystyle \text{Let }K(x,0)\text{ divide }AB\text{ in the ratio }k:1.
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle 0=\frac{k(-5)+1(3)}{k+1}
\displaystyle -5k+3=0
\displaystyle 5k=3
\displaystyle k=\frac{3}{5}
\displaystyle \therefore \text{The required ratio is }\frac{3}{5}:1=3:5.
\displaystyle x=\frac{3(6)+5(2)}{3+5}
\displaystyle =\frac{18+10}{8}=\frac{28}{8}=\frac{7}{2}
\displaystyle \therefore K\left(\frac{7}{2},0\right)
\\

\displaystyle \textbf{Question 13: } \text{In the given figure, line }APB\text{ meets the }x\text{-axis at point }A\text{ and the }y\text{-axis at point }B.
\displaystyle P\text{ is the point }(-4,2)\text{ and }AP:PB=1:2.
\displaystyle \text{Find the coordinates of }A\text{ and }B.\hfill \text{[ICSE 1999, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(x,0)\text{ and }B(0,y).
\displaystyle \text{Since }P(-4,2)\text{ divides }AB\text{ internally in the ratio }1:2,
\displaystyle -4=\frac{1(0)+2x}{1+2}
\displaystyle -12=2x
\displaystyle x=-6
\displaystyle 2=\frac{1(y)+2(0)}{1+2}
\displaystyle 6=y
\displaystyle \therefore A(-6,0)\text{ and }B(0,6)
\\

\displaystyle \textbf{Question 14: } \text{Given a line segment }AB\text{ joining the points }A(-4,6)\text{ and }B(8,-3).\text{ Find:}
\displaystyle \text{(i) The ratio in which }AB\text{ is divided by the }y\text{-axis}
\displaystyle \text{(ii) The coordinates of the point of intersection}
\displaystyle \text{(iii) The length of }AB.\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the }y\text{-axis divide }AB\text{ in the ratio }k:1\text{ at }P(0,y).
\displaystyle \text{Using the section formula for the }x\text{-coordinate,}
\displaystyle 0=\frac{k(8)+1(-4)}{k+1}
\displaystyle 8k-4=0
\displaystyle 8k=4
\displaystyle k=\frac{1}{2}
\displaystyle \therefore \text{(i) The required ratio is }\frac{1}{2}:1=1:2.
\displaystyle y=\frac{1(-3)+2(6)}{1+2}
\displaystyle =\frac{-3+12}{3}=3
\displaystyle \therefore \text{(ii) The point of intersection is }(0,3).
\displaystyle \text{(iii) }AB=\sqrt{(8-(-4))^2+(-3-6)^2}
\displaystyle =\sqrt{12^2+(-9)^2}
\displaystyle =\sqrt{144+81}
\displaystyle =\sqrt{225}=15\text{ units}
\\

\displaystyle \textbf{Question 15: }KM\text{ is a straight line of }13\text{ units. If }K\text{ has the coordinates }(2,5)
\displaystyle \text{and }M\text{ has the coordinates }(x,-7),\text{ find the value of }x.\hfill \text{[ICSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }K(2,5),\ M(x,-7)\text{ and }KM=13\text{ units.}
\displaystyle \sqrt{(x-2)^2+(-7-5)^2}=13
\displaystyle \sqrt{(x-2)^2+(-12)^2}=13
\displaystyle \text{Squaring both sides,}
\displaystyle (x-2)^2+144=169
\displaystyle x^2-4x+4+144=169
\displaystyle x^2-4x-21=0
\displaystyle (x-7)(x+3)=0
\displaystyle x=7\text{ or }x=-3
\\

\displaystyle \textbf{Question 16: } \text{The mid-point of the line segment joining }(3m,6)\text{ and }
\displaystyle (-4,3n)\text{ is }(1,2m-1). \ \text{Find the values of }m\text{ and }n.\hfill \text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Using the midpoint formula,}
\displaystyle 1=\frac{3m+(-4)}{2}
\displaystyle 2=3m-4
\displaystyle 3m=6
\displaystyle m=2
\displaystyle 2m-1=\frac{6+3n}{2}
\displaystyle 2(2)-1=\frac{6+3n}{2}
\displaystyle 3=\frac{6+3n}{2}
\displaystyle 6=6+3n
\displaystyle 3n=0
\displaystyle n=0
\displaystyle \therefore m=2\text{ and }n=0
\\

\displaystyle \textbf{Question 17: }ABC\text{ is a triangle and }G(4,3)\text{ is the centroid of the triangle.}
\displaystyle \text{If }A(1,3),\ B(4,b)\text{ and }C(a,1),\text{ find }a\text{ and }b.
\displaystyle \text{Also find the length of the side }BC.\hfill \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }G(4,3)\text{ is the centroid of }\triangle ABC,
\displaystyle 4=\frac{1+4+a}{3}
\displaystyle 12=5+a
\displaystyle a=7
\displaystyle 3=\frac{3+b+1}{3}
\displaystyle 9=4+b
\displaystyle b=5
\displaystyle \therefore B(4,5)\text{ and }C(7,1)
\displaystyle BC=\sqrt{(7-4)^2+(1-5)^2}
\displaystyle =\sqrt{3^2+(-4)^2}
\displaystyle =\sqrt{9+16}
\displaystyle =\sqrt{25}=5\text{ units}
\\

\displaystyle \textbf{Question 18.}~\text{Calculate the ratio in which the line joining }A(6,5)\text{ and }B(4,-3) \\ \text{ is divided by the line }y=2.~\text{[ICSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line joining }A\text{ and }B\text{ be divided by the line }y=2\text{ in the ratio }m_1:m_2.
\displaystyle \text{Let the coordinate of intersection point be }M(x,y).

\displaystyle \text{By using section formula, }M(x,y)=\left(\frac{m_1\cdot4+m_2\cdot6}{m_1+m_2},\frac{m_1\cdot(-3)+m_2\cdot5}{m_1+m_2}\right).
\displaystyle \text{Using section formula: }M(x,y)=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right).
\displaystyle =\left(\frac{4m_1+6m_2}{m_1+m_2},\frac{-3m_1+5m_2}{m_1+m_2}\right).
\displaystyle \text{Since, line }y=2\text{ divides }AB,\text{ so y-coordinate of }M\text{ is }2.
\displaystyle \therefore 2=\frac{-3m_1+5m_2}{m_1+m_2}.
\displaystyle \Rightarrow 2(m_1+m_2)=-3m_1+5m_2.
\displaystyle \Rightarrow 2m_1+2m_2=-3m_1+5m_2.
\displaystyle \Rightarrow 2m_1+3m_1=5m_2-2m_2.
\displaystyle \Rightarrow 5m_1=3m_2\Rightarrow \frac{m_1}{m_2}=\frac{3}{5}.
\displaystyle \text{So, the line }y=2\text{ divides the line joining }A(6,5)\text{ and }B(4,-3)\text{ in the ratio }3:5.

\displaystyle \textbf{Question 19.}~\text{The line segment joining }A(2,3)\text{ and }B(6,-5)\text{ is intersected} \\ \text{by the X-axis at the point }K.\text{ Write the coordinates of the point }K.\text{ Hence, find} \\ \text{the ratio in which }K\text{ divides }AB.~\text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Since, the line segment joining }A\text{ and }B\text{ is intersected by the X-axis at point }K, \\ \text{so its coordinates will be }(x,0).
\displaystyle \text{Let }K\text{ divides }AB\text{ in the ratio }m_1:m_2.
\displaystyle \text{Then, coordinates of }K\text{ are }\left(\frac{6m_1+2m_2}{m_1+m_2},\frac{-5m_1+3m_2}{m_1+m_2}\right)
\displaystyle \text{Since, y-coordinate of }K\text{ is zero, }
\displaystyle \frac{-5m_1+3m_2}{m_1+m_2}=0
\displaystyle \Rightarrow -5m_1+3m_2=0\Rightarrow 5m_1=3m_2
\displaystyle \Rightarrow \frac{m_1}{m_2}=\frac{3}{5}
\displaystyle \text{Hence, the required ratio is }3:5.
\displaystyle \text{Now, }x=\frac{6m_1+2m_2}{m_1+m_2}=\frac{6\cdot3+2\cdot5}{3+5}
\displaystyle =\frac{18+10}{8}=\frac{28}{8}=\frac{7}{2}
\displaystyle \text{So, coordinates of point }K\text{ are }\left(\frac{7}{2},0\right).

\displaystyle \textbf{Question 20.}~\text{Find the coordinates of the centroid of a triangle, whose vertices} \\ \text{are }A(-1,3),~B(1,-1)\text{ and }C(5,1).~\text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, vertices of the triangle are }A(-1,3),~B(1,-1)\text{ and }C(5,1).

\displaystyle \text{Let the coordinates of centroid be }G(x,y).
\displaystyle x=\frac{-1+1+5}{3}=\frac{5}{3}\text{ and }y=\frac{3-1+1}{3}=\frac{3}{3}=1
\displaystyle \text{Hence, the required coordinates of centroid are }\left(\frac{5}{3},1\right).

\displaystyle \textbf{Question 21.}~\text{The mid-point of the line segment joining }(2a,4)\text{ and }(-2,2b) \\ \text{ is }(1,2a+1).\text{ Find the values of }a\text{ and }b.~\text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{The mid-point of the line segment joining }(2a,4)\text{ and }(-2,2b)\text{ is }\left(\frac{2a-2}{2},\frac{4+2b}{2}\right).
\displaystyle \text{Coordinates of mid-point }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)
\displaystyle =\left(a-1,2+b\right)
\displaystyle \text{But coordinates of mid-point is given }(1,2a+1).
\displaystyle \therefore (1,2a+1)=(a-1,2+b)
\displaystyle \text{On comparing coordinates from both sides, we get }
\displaystyle 1=a-1\text{ and }2a+1=2+b
\displaystyle \Rightarrow a=2\text{ and }2a=1+b
\displaystyle \Rightarrow 4=1+b\Rightarrow b=3
\displaystyle \text{Hence, the values of }a\text{ and }b\text{ are }2\text{ and }3\text{ respectively.}

\displaystyle \textbf{Question 22.}~\text{If the line joining the points }A(4,-5)\text{ and }B(4,5)\text{ is divided} \\ \text{by the point }P,\text{ such that }\dfrac{AP}{AB}=\dfrac{2}{5},\text{ then find the coordinates of }P.~\text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{AP}{AB}=\frac{2}{5}\Rightarrow \frac{AB}{AP}=\frac{5}{2}
\displaystyle \text{Subtracting }1\text{ from both sides, }
\displaystyle \frac{AB}{AP}-1=\frac{5}{2}-1
\displaystyle \Rightarrow \frac{AB-AP}{AP}=\frac{3}{2}
\displaystyle \Rightarrow \frac{PB}{AP}=\frac{3}{2}
\displaystyle \text{Since, }P\text{ divides }AB\text{ internally in the ratio }2:3.
\displaystyle \text{Coordinates of }P=\left(\frac{2\cdot4+3\cdot4}{2+3},\frac{2\cdot5+3\cdot(-5)}{2+3}\right)
\displaystyle =\left(\frac{8+12}{5},\frac{10-15}{5}\right)=\left(\frac{20}{5},\frac{-5}{5}\right)=(4,-1)

\displaystyle \textbf{Question 23.}~\text{Calculate the ratio in which the line joining }A(-4,2)\text{ and } \\ B(3,6)\text{ is divided by }P(x,3).\text{ Also, find (i) }x,\text{ (ii) length of }AP.~\text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,3)\text{ divides the line segment joining the points }A(-4,2)\text{ and }B(3,6)\text{ in the ratio }k:1.

\displaystyle \text{Using the section formula, coordinates of }P\text{ are }
\displaystyle \left(\frac{k\cdot3+1\cdot(-4)}{k+1},\frac{k\cdot6+1\cdot2}{k+1}\right)
\displaystyle =\left(\frac{3k-4}{k+1},\frac{6k+2}{k+1}\right)
\displaystyle \text{But coordinates of }P\text{ are given as }(x,3).
\displaystyle \therefore (x,3)=\left(\frac{3k-4}{k+1},\frac{6k+2}{k+1}\right)
\displaystyle \text{On equating y-coordinates, }
\displaystyle 3=\frac{6k+2}{k+1}
\displaystyle \Rightarrow 3(k+1)=6k+2
\displaystyle \Rightarrow 3k+3=6k+2
\displaystyle \Rightarrow 6k-3k=3-2
\displaystyle \Rightarrow 3k=1\Rightarrow k=\frac{1}{3}
\displaystyle \text{Hence, the required ratio is }1:3\text{ (internally).}
\displaystyle \text{From }(x,3)=\left(\frac{3k-4}{k+1},\frac{6k+2}{k+1}\right),
\displaystyle x=\frac{3k-4}{k+1}=\frac{3\cdot\frac{1}{3}-4}{\frac{1}{3}+1}
\displaystyle =\frac{1-4}{\frac{4}{3}}=\frac{-3}{\frac{4}{3}}=\frac{-9}{4}
\displaystyle \text{(ii) Length of }AP=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\displaystyle =\sqrt{\left(-\frac{9}{4}+4\right)^2+(3-2)^2}
\displaystyle =\sqrt{\left(\frac{-9+16}{4}\right)^2+1^2}
\displaystyle =\sqrt{\left(\frac{7}{4}\right)^2+1}
\displaystyle =\sqrt{\frac{49}{16}+1}=\sqrt{\frac{65}{16}}=\frac{\sqrt{65}}{4}\text{ units}

\displaystyle \textbf{Question 24.}~AB\text{ is a diameter of a circle with centre }C(-2,5)\text{ and }A(3,-7). \\ \text{ Find (i) the length of radius }AC,\text{ (ii) the coordinates of }B.~\text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }A(3,-7)=(x_1,y_1)\text{ and }C(-2,5)=(x_2,y_2).

\displaystyle \text{Length of radius }AC=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\displaystyle =\sqrt{(-2-3)^2+(5+7)^2}
\displaystyle =\sqrt{(-5)^2+12^2}=\sqrt{25+144}=\sqrt{169}=13\text{ units}
\displaystyle \text{(ii) Let coordinates of }B\text{ be }(x_3,y_3).
\displaystyle \text{Since, }C\text{ is the centre of circle and mid-point of diameter }AB,
\displaystyle \text{Coordinates of }C=\left(\frac{3+x_3}{2},\frac{-7+y_3}{2}\right)
\displaystyle \text{But coordinates of }C\text{ are }(-2,5).
\displaystyle \therefore (-2,5)=\left(\frac{3+x_3}{2},\frac{-7+y_3}{2}\right)
\displaystyle \text{On comparing coordinates, }
\displaystyle -2=\frac{3+x_3}{2}\text{ and }5=\frac{-7+y_3}{2}
\displaystyle \Rightarrow -4=3+x_3\Rightarrow x_3=-7
\displaystyle \Rightarrow 10=-7+y_3\Rightarrow y_3=17
\displaystyle \text{Thus, coordinates of }B\text{ are }(-7,17).
\displaystyle \Rightarrow -2\times2=3+x_3\text{ and }5\times2=-7+y_3
\displaystyle \Rightarrow -4=3+x_3\Rightarrow x_3=-7\text{ and }10=-7+y_3\Rightarrow y_3=17
\displaystyle \text{Hence, the coordinates of }B\text{ are }(-7,17).

\displaystyle \textbf{Question 25.}~\text{Given a line segment }AB\text{ joining the points }A(-4,6)\text{ and }B(8,-3). \\ \text{ Find (i) the ratio in which }AB\text{ is divided by the Y-axis, (ii) the coordinates of the} \\ \text{point of intersection, (iii) the length of }AB.~\text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, coordinates of }A\text{ and }B\text{ are }(-4,6)\text{ and }(8,-3).
\displaystyle \text{(i) Let }AB\text{ be divided by the Y-axis at point }M(0,y)\text{ in the ratio }m_1:m_2.
\displaystyle \text{Then, coordinates of }M\text{ are }\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)
\displaystyle =\left(\frac{8m_1-4m_2}{m_1+m_2},\frac{-3m_1+6m_2}{m_1+m_2}\right)
\displaystyle \text{But x-coordinate of }M\text{ is }0.
\displaystyle \therefore \frac{8m_1-4m_2}{m_1+m_2}=0\Rightarrow 8m_1-4m_2=0
\displaystyle \Rightarrow 4m_2=8m_1\Rightarrow \frac{m_1}{m_2}=\frac{1}{2}
\displaystyle \text{Hence, required ratio is }1:2.
\displaystyle \text{(ii) At the point of intersection }x=0,
\displaystyle y=\frac{-3m_1+6m_2}{m_1+m_2}
\displaystyle =\frac{-3\cdot1+6\cdot2}{1+2}
\displaystyle =\frac{-3+12}{3}=\frac{9}{3}=3
\displaystyle \text{Hence, coordinates of point of intersection are }(0,3).
\displaystyle \text{(iii) Length of }AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\displaystyle =\sqrt{(8+4)^2+(-3-6)^2}
\displaystyle =\sqrt{12^2+(-9)^2}
\displaystyle =\sqrt{144+81}=\sqrt{225}=15\text{ units}

\displaystyle \textbf{Question 26.}~ABC\text{ is a triangle and }G(4,3)\text{ is the centroid of the triangle.} \\ \text{If }A(1,3),~B(4,b)\text{ and }C(a,1),\text{ then find }a\text{ and }b.\text{ Find the length of side }BC.~\text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, vertices and centroid of triangle are }A(1,3),~B(4,b),~C(a,1)\text{ and }G(4,3)\text{ respectively.}

\displaystyle \text{We know that coordinates of centroid are }\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)
\displaystyle \text{Here }(x_1,y_1)=(1,3),~(x_2,y_2)=(4,b)\text{ and }(x_3,y_3)=(a,1).
\displaystyle \therefore G=\left(\frac{1+4+a}{3},\frac{3+b+1}{3}\right)=\left(\frac{5+a}{3},\frac{4+b}{3}\right)
\displaystyle \text{But centroid is given as }G(4,3).
\displaystyle \therefore (4,3)=\left(\frac{5+a}{3},\frac{4+b}{3}\right)
\displaystyle \text{On comparing x and y-coordinates, we get }
\displaystyle 4=\frac{5+a}{3}\text{ and }3=\frac{4+b}{3}
\displaystyle \Rightarrow 12=5+a\Rightarrow a=7\text{ and }9=4+b\Rightarrow b=5
\displaystyle \text{Coordinates of }B\text{ and }C\text{ are }(4,5)\text{ and }(7,1).
\displaystyle \text{Now, length of side }BC=\sqrt{(7-4)^2+(1-5)^2}
\displaystyle =\sqrt{3^2+(-4)^2}
\displaystyle =\sqrt{9+16}=\sqrt{25}=5\text{ units}

\displaystyle \textbf{Question 27.}~\text{If }A(-4,3)\text{ and }B(8,-6).\text{ (i) Find the length of }AB. \\ \text{ (ii) In what ratio is the line joining }AB,\text{ divided by the X-axis?}~\text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A(-4,3),~B(8,-6).
\displaystyle \text{(i) Length of }AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\displaystyle =\sqrt{(8+4)^2+(-6-3)^2}
\displaystyle =\sqrt{12^2+(-9)^2}
\displaystyle =\sqrt{144+81}=\sqrt{225}=15\text{ units}
\displaystyle \text{(ii) Let the point at which }AB\text{ is divided by X-axis be }M(x,0).
\displaystyle \text{Let }AM:MB=m_1:m_2

\displaystyle \text{Clearly, the coordinates of }M=\left(\frac{8m_1-4m_2}{m_1+m_2},\frac{-6m_1+3m_2}{m_1+m_2}\right)
\displaystyle \text{But coordinate of }M\text{ is }(x,0).
\displaystyle \therefore \left(\frac{8m_1-4m_2}{m_1+m_2},\frac{-6m_1+3m_2}{m_1+m_2}\right)=(x,0)
\displaystyle \text{On equating the y-coordinate from both sides, we get}
\displaystyle \frac{-6m_1+3m_2}{m_1+m_2}=0
\displaystyle \Rightarrow -6m_1+3m_2=0
\displaystyle \Rightarrow 3m_2=6m_1
\displaystyle \Rightarrow \frac{m_1}{m_2}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{Hence, the line joining }AB\text{ is divided by the X-axis in the ratio }1:2.

\displaystyle \textbf{Question 28.}~\text{The line joining }P(-4,5)\text{ and }Q(3,2)\text{ intersect the Y-axis at }R.\text{ }PM \\ \text{and }QN\text{ are perpendicular from }P\text{ and }Q\text{ on the X-axis. Find (i) the ratio }PR:PQ,\text{ (ii) the coordinates of }R,\text{ (iii) the area of the quadrilateral }PMNQ.~\text{[ICSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }PR:RQ=m_1:m_2
\displaystyle \text{Since, the line }PQ\text{ intersects the Y-axis at }R.

\displaystyle \text{The coordinates of }R\text{ will be }(0,y).
\displaystyle \text{By using section formula, x-coordinate of }R=\frac{m_1x_2+m_2x_1}{m_1+m_2}
\displaystyle \therefore 0=\frac{m_1\cdot3+m_2\cdot(-4)}{m_1+m_2}
\displaystyle \Rightarrow 0=3m_1-4m_2
\displaystyle \Rightarrow 3m_1=4m_2\Rightarrow \frac{m_1}{m_2}=\frac{4}{3}
\displaystyle \Rightarrow \frac{PR}{QR}=\frac{4}{3}\Rightarrow \frac{QR}{PR}=\frac{3}{4}
\displaystyle \Rightarrow \frac{QR}{PR}+1=\frac{3}{4}+1
\displaystyle \Rightarrow \frac{QR+PR}{PR}=\frac{3+4}{4}
\displaystyle \Rightarrow \frac{PQ}{PR}=\frac{7}{4}\Rightarrow \frac{PR}{PQ}=\frac{4}{7}
\displaystyle \therefore PR:PQ=4:7
\displaystyle \text{Hence, the required ratio is }4:7.
\displaystyle \text{(ii) y-coordinate of }R=\frac{m_1y_2+m_2y_1}{m_1+m_2}
\displaystyle =\frac{4\cdot2+3\cdot5}{4+3}=\frac{8+15}{7}=\frac{23}{7}
\displaystyle \text{Hence, the coordinates of }R\text{ are }\left(0,\frac{23}{7}\right).
\displaystyle \text{(iii) Since, quadrilateral }PMQN\text{ is a trapezium.}
\displaystyle \text{Area}=\frac{1}{2}\left(\text{Sum of parallel sides}\times\text{Height}\right)
\displaystyle =\frac{1}{2}(PM+NQ)\times MN
\displaystyle =\frac{1}{2}(5+2)\times7=\frac{49}{2}=24.5\text{ sq units}

\displaystyle \textbf{Question 29.}~\text{In the figure given below, the line segment }AB\text{ meets the} \\ \text{X-axis at }A\text{ and Y-axis at }B.\text{ The point }P(-3,4)\text{ on }AB\text{ divides it in the ratio } \\ 2:3.\text{ Find the coordinates of }A\text{ and }B.~\text{[ICSE 2013]}

\displaystyle \text{Answer:}
\displaystyle \text{Given points }A\text{ and }B\text{ lie on X and Y-axes, respectively.}
\displaystyle \text{So, let the coordinates of }A\text{ and }B\text{ be }(x,0)\text{ and }(0,y)\text{ respectively.}
\displaystyle \text{Since, the point }P(-3,4)\text{ on }AB\text{ divides it in the ratio }2:3.
\displaystyle \therefore AP:BP=2:3.
\displaystyle \text{By using section formula, coordinates of }P
\displaystyle =\left(\frac{2\cdot0+3x}{2+3},\frac{2y+3\cdot0}{2+3}\right).
\displaystyle \text{But coordinates of }P\text{ is given }(-3,4).
\displaystyle \therefore (-3,4)=\left(\frac{3x}{5},\frac{2y}{5}\right).
\displaystyle \text{On comparing the coordinates from both sides, we get}
\displaystyle -3=\frac{3x}{5}\text{ and }4=\frac{2y}{5}.
\displaystyle \Rightarrow -3\times5=3x\text{ and }4\times5=2y.
\displaystyle \Rightarrow 3x=-15\text{ and }20=2y.
\displaystyle \Rightarrow x=-5\text{ and }y=10.
\displaystyle \text{Hence, the coordinates of }A\text{ and }B\text{ are }(-5,0)\text{ and }(0,10).

\displaystyle \textbf{Question 30.}~P(1,-2)\text{ is a point on the line segment joining }A(3,-6)\text{ and } \\ B(x,y)\text{ such that }AP:PB=2:3.\text{ Find the coordinates of }B.~\text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, a point }P\text{ divides a line segment joining }A(3,-6)\text{ and }B(x,y)\text{ such that }AP:PB=2:3.
\displaystyle \text{i.e. }P\text{ divides }AB\text{ in the ratio }2:3\text{ internally.}
\displaystyle \therefore \text{Coordinates of }P=\left(\frac{2x+3\cdot3}{2+3},\frac{2y+3\cdot(-6)}{2+3}\right)
\displaystyle =\left(\frac{2x+9}{5},\frac{2y-18}{5}\right)
\displaystyle \text{But given coordinates of }P\text{ are }(1,-2).
\displaystyle \therefore \frac{2x+9}{5}=1\text{ and }\frac{2y-18}{5}=-2
\displaystyle \Rightarrow 2x+9=5\Rightarrow 2x=-4\Rightarrow x=-2
\displaystyle \Rightarrow 2y-18=-10\Rightarrow 2y=8\Rightarrow y=4
\displaystyle \text{Thus, coordinates of }B\text{ are }(-2,4).

\displaystyle \textbf{Question 31.}~M\text{ and }N\text{ are two points on the X-axis and Y-axis, respectively. } \\ P(3,2)\text{ divides the line segment }MN\text{ in the ratio }2:3.\text{ Find (i) the coordinates of } \\ M\text{ and }N,\text{ (ii) slope of the line }MN.~\text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the coordinates of }M\text{ and }N\text{ be }(x,0)\text{ and }(0,y)\text{ respectively.}

\displaystyle \text{(i) Using internal section formula, }
\displaystyle P(3,2)=\left(\frac{3x+0\cdot2}{2+3},\frac{2y+0\cdot3}{2+3}\right)
\displaystyle \Rightarrow (3,2)=\left(\frac{3x}{5},\frac{2y}{5}\right)
\displaystyle \Rightarrow \frac{3x}{5}=3\text{ and }\frac{2y}{5}=2
\displaystyle \Rightarrow x=5\text{ and }y=5
\displaystyle \text{Hence, the coordinates of }M\text{ and }N\text{ are }(5,0)\text{ and }(0,5)\text{ respectively.}
\displaystyle \text{(ii) Slope of line }MN=\frac{y_2-y_1}{x_2-x_1}=\frac{5-0}{0-5}=\frac{5}{-5}=-1

\displaystyle \textbf{Question 32.}~\text{In what ratio is the line joining }P(5,3)\text{ and }Q(-5,3)\text{ divided} \\ \text{by the Y-axis? Also, find the coordinates of the point of intersection.}~\text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let line joining }P(5,3)\text{ and }Q(-5,3)\text{ be divided by the Y-axis in the ratio }k:1\text{ at point }A.
\displaystyle \therefore \text{Coordinates of }A=\left(\frac{-5k+5}{k+1},\frac{3k+3}{k+1}\right)
\displaystyle \text{Since, }A\text{ lies on the Y-axis, therefore x-coordinate of }A\text{ is }0.
\displaystyle \Rightarrow \frac{-5k+5}{k+1}=0\Rightarrow -5k+5=0
\displaystyle \Rightarrow -5k=-5\Rightarrow k=1
\displaystyle \therefore \text{Required ratio }=k:1=1:1.
\displaystyle \text{The coordinates of point of intersection }A=\left(\frac{-5\cdot1+5}{1+1},\frac{3\cdot1+3}{1+1}\right)=(0,3)

\displaystyle \textbf{Question 33.}~\text{Find the ratio in which the line joining }(-2,5)\text{ and }(-5,-6) \\ \text{ is divided by the line }y=-3.\text{ Hence, find the point of intersection.}~\text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line }y=-3\text{ divides the line segment joining }A(-2,5)\text{ and }B(-5,-6)\text{ in the ratio }k:1\text{ at the point }C.
\displaystyle \text{Then, the coordinates of }C\text{ are }\left(\frac{-5k-2}{k+1},\frac{-6k+5}{k+1}\right).

\displaystyle \text{But }C\text{ lies on }y=-3,\text{ therefore }
\displaystyle \frac{-6k+5}{k+1}=-3\Rightarrow -6k+5=-3k-3
\displaystyle \Rightarrow -6k+3k=-3-5
\displaystyle \Rightarrow -3k=-8\Rightarrow k=\frac{8}{3}
\displaystyle \text{Hence, the required ratio is }8:3\text{ internally.}
\displaystyle \text{On putting }k=\frac{8}{3}\text{ in the coordinates of }C,\text{ we get }
\displaystyle C=\left(\frac{-5\cdot\frac{8}{3}-2}{\frac{8}{3}+1},\frac{-6\cdot\frac{8}{3}+5}{\frac{8}{3}+1}\right)
\displaystyle =\left(\frac{-\frac{40}{3}-\frac{6}{3}}{\frac{8}{3}+\frac{3}{3}},\frac{-\frac{48}{3}+\frac{15}{3}}{\frac{8}{3}+\frac{3}{3}}\right)
\displaystyle =\left(\frac{-\frac{46}{3}}{\frac{11}{3}},\frac{-\frac{33}{3}}{\frac{11}{3}}\right)=\left(-\frac{46}{11},-3\right)

\displaystyle \textbf{Question 34.}~A(1,4),~B(4,1)\text{ and }C(x,4)\text{ are the vertices of }\triangle ABC. \\ \text{ If the centroid of the triangle is }G(4,3),\text{ then }x\text{ is equal to }~\text{[ICSE Semester-II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }A(x_1,y_1)=A(1,4),~B(x_2,y_2)=B(4,1)\text{ and }C(x_3,y_3)=C(x,4) \\ \text{ and centroid }G(x,y)=G(4,3).
\displaystyle \therefore G(x,y)=\left(\frac{x_1+x_2+x_3}{3},~\frac{y_1+y_2+y_3}{3}\right)
\displaystyle (4,3)=\left(\frac{1+4+x}{3},~\frac{4+1+4}{3}\right)
\displaystyle \text{On comparing, we get }4=\frac{5+x}{3}
\displaystyle \Rightarrow 12=5+x
\displaystyle \Rightarrow x=7

\displaystyle \textbf{Question 18.}~\text{Find a point }P\text{ which divides internally the line segment joining} \\ \text{the points }A(-3,9)\text{ and }B(1,-3)\text{ in the ratio }1:3.~\text{[ICSE Semester-II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Let coordinates of the point }P\text{ be }(x,y).
\displaystyle \text{Given }A(-3,9)\text{ and }B(1,-3)\text{ and }AP:PB=1:3.
\displaystyle \therefore P(x,y)=\left(\frac{1\cdot1+3(-3)}{1+3},~\frac{1(-3)+3\cdot9}{1+3}\right)
\displaystyle =\left(\frac{1-9}{4},~\frac{-3+27}{4}\right)=(-2,6)

\displaystyle \textbf{Question 35.}~\text{If the vertices of a triangle are }(1,3),~(2,-4)\text{ and }(-3,1),\text{ then} \\ \text{the coordinate of its centroid is }~\text{[ICSE Semester-II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let the given coordinates be }A(1,3),~B(2,-4)\text{ and }C(-3,1).
\displaystyle \text{Now, we know that coordinates of centroid of a triangle are}
\displaystyle \left(\frac{x_1+x_2+x_3}{3},~\frac{y_1+y_2+y_3}{3}\right)
\displaystyle \text{Here, }x_1=1,~x_2=2,~x_3=-3\text{ and }y_1=3,~y_2=-4,~y_3=1.
\displaystyle \text{Coordinates of centroid }=\left(\frac{1+2+(-3)}{3},~\frac{3+(-4)+1}{3}\right)=(0,0).

\displaystyle \textbf{Question 36.}~\text{Find the ratio, in which the X-axis divides internally the line joining} \\ \text{points }A(6,-4)\text{ and }B(-3,8).~\text{[ICSE Semester-II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }K\text{ divides the line segment joining the points }A(6,-4)\text{ and }B(-3,8)\text{ in the ratio } \\ m_1:m_2.
\displaystyle \text{By section formula, we get}
\displaystyle \therefore \text{Coordinates of }K\text{ are }\left(\frac{-3m_1+6m_2}{m_1+m_2},\frac{8m_1-4m_2}{m_1+m_2}\right)
\displaystyle \text{Here }(x_1,y_1)=(6,-4)\text{ and }(x_2,y_2)=(-3,8).
\displaystyle \text{But line segment }AB\text{ is intersected at point }K\text{ by the X-axis, so coordinates of }K\text{ will be }(x,0).
\displaystyle \therefore (x,0)=\left(\frac{-3m_1+6m_2}{m_1+m_2},\frac{8m_1-4m_2}{m_1+m_2}\right)
\displaystyle \text{On equating y-coordinates both sides, we get}
\displaystyle \frac{8m_1-4m_2}{m_1+m_2}=0
\displaystyle \Rightarrow 8m_1-4m_2=0
\displaystyle \Rightarrow 8m_1=4m_2
\displaystyle \Rightarrow \frac{m_1}{m_2}=\frac{4}{8}=\frac{1}{2}
\displaystyle \text{Hence, the required ratio is }1:2.

\displaystyle \textbf{Question 37.}~\text{Point }M(2,b)\text{ is the mid-point of the line segment joining} \\ \text{points }P(a,7)\text{ and }Q(6,5).\text{ Find the values of }a\text{ and }b.~\text{[ICSE Semester-II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{The point }M(2,b)\text{ be the mid-point of the line segment joining points }P(a,7)\text{ and }Q(6,5).
\displaystyle \text{Coordinates of mid-point of }PQ\text{ are}
\displaystyle M(2,b)=\left(\frac{a+6}{2},\frac{7+5}{2}\right)
\displaystyle \text{On comparing, }\frac{a+6}{2}=2\text{ and }\frac{12}{2}=b
\displaystyle \Rightarrow a+6=4\Rightarrow a=4-6=-2\text{ and }b=6
\displaystyle \text{Hence, the value of }a=-2\text{ and }b=6.

\displaystyle \textbf{Question 38.}~A\text{ and }B\text{ are two points on the X-axis and Y-axis respectively.} \\ \text{[ICSE 2023]}

\displaystyle \text{(i) Write down the coordinates of }A\text{ and }B.
\displaystyle \text{(ii) }P\text{ is a point on }AB\text{ such that }AP:PB=3:1. \text{ Using section formula find}\text{ the coordinates of }P.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Coordinates of }A=(4,0)\text{ and coordinates of }B=(0,4).
\displaystyle \text{(ii) Given that }P\text{ is a point on line }AB\text{ and }AP:PB=3:1.
\displaystyle \text{We have by section formula, Coordinates of }P=\left(\frac{m x_2+n x_1}{m+n},\frac{m y_2+n y_1}{m+n}\right)
\displaystyle \text{Let point }P(x,y)\text{ divides the line segment joining points }A(x_1,y_1)\text{ and }B(x_2,y_2) \\ \text{internally in ratio }m:n.
\displaystyle \text{Here, }m=3,n=1,x_1=4,x_2=0,y_1=0,y_2=4.
\displaystyle \therefore \text{Coordinates of }P(x,y)=\left(\frac{3\cdot0+1\cdot4}{3+1},\frac{3\cdot4+1\cdot0}{3+1}\right)
\displaystyle =\left(\frac{4}{4},\frac{12}{4}\right)=(1,3)
\displaystyle \text{Hence, the coordinates of }P\text{ are }(1,3).

\displaystyle \textbf{Question 39.}~\text{From the given figure:}~\text{[ICSE 2023]}

\displaystyle \text{(i) Write down the coordinates of }A\text{ and }B.
\displaystyle \text{(ii) If }P\text{ divides }AB\text{ in the ratio }2:3,\text{ then find the coordinates of }P.
\displaystyle \text{Answer:}

\displaystyle \text{(i) From figure we can see that line }AB\text{ intersects X-axis at }(5,0)\text{ and Y-axis at }(0,3).
\displaystyle \text{Hence, coordinates of }A(5,0)\text{ and coordinates of }B(0,3).
\displaystyle \text{(ii) }P\text{ divides }AB\text{ in the ratio }2:3.
\displaystyle \text{Let }m_1=2\text{ and }m_2=3.
\displaystyle \text{Then, coordinates of }P\text{ are }\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)
\displaystyle \text{Here, }(x_1,y_1)=(5,0)\text{ and }(x_2,y_2)=(0,3).
\displaystyle \Rightarrow \left(\frac{2\cdot0+3\cdot5}{2+3},\frac{2\cdot3+3\cdot0}{2+3}\right)
\displaystyle =\left(\frac{15}{5},\frac{6}{5}\right)=\left(3,\frac{6}{5}\right)
\displaystyle \text{Hence, the coordinates of }P\text{ are }\left(3,\frac{6}{5}\right).

\displaystyle \textbf{Question 40.}~\text{The coordinates of the vertices of }\triangle ABC\text{ are }(-4,-2),~(6,2) \\ \text{ and }(4,6).\text{ The centroid }G\text{ of }\triangle ABC\text{ is }~\text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, coordinates of vertices of }\triangle ABC\text{ are }A(-4,-2),~B(6,2)\text{ and }C(4,6).
\displaystyle \text{Let }(x_1,y_1)=(-4,-2),~(x_2,y_2)=(6,2)\text{ and }(x_3,y_3)=(4,6).
\displaystyle \text{We have by centroid formula,}
\displaystyle \text{Centroid }G\text{ of }\triangle ABC
\displaystyle =\left(\frac{x_1+x_2+x_3}{3},~\frac{y_1+y_2+y_3}{3}\right)
\displaystyle =\left(\frac{-4+6+4}{3},~\frac{-2+2+6}{3}\right)=(2,2)
\displaystyle \text{Hence, the coordinates of centroid are }G(2,2).

\displaystyle \textbf{Question 41.}~\text{The centroid of }\triangle ABC\text{ is }G(6,7).\text{ If the coordinates} \\ \text{of vertices are }A(a,5),~B(7,9)\text{ and }C(5,7),\text{ respectively, find }a.~\text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }G(6,7)\text{ is centroid of }\triangle ABC\text{ and coordinates of vertices }A,B\text{ and }C \\ \text{ are }(a,5),(7,9)\text{ and }(5,7)\text{ respectively.}
\displaystyle \text{We know that coordinates of centroid of a triangle are}

\displaystyle G=\left(\frac{x_1+x_2+x_3}{3},~\frac{y_1+y_2+y_3}{3}\right)
\displaystyle \text{Here, }(x_1,y_1)=(a,5),~(x_2,y_2)=(7,9)\text{ and }(x_3,y_3)=(5,7).
\displaystyle \therefore \text{Centroid of triangle }G=\left(\frac{a+7+5}{3},~\frac{5+9+7}{3}\right)=\left(\frac{a+12}{3},~7\right)
\displaystyle \text{But centroid is given by }G(6,7).
\displaystyle \therefore (6,7)=\left(\frac{a+12}{3},~7\right)
\displaystyle \text{On equating x-coordinate of both sides, we get}
\displaystyle 6=\frac{a+12}{3}\Rightarrow 18=a+12
\displaystyle \Rightarrow a=18-12=6

\displaystyle \textbf{Question 42.}~\text{Points }A(x,y),~B(3,-2)\text{ and }C(4,-5)\text{ are collinear. The value} \\ \text{of }y\text{ in terms of }x\text{ is }~\text{[ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given points are }A(x,y),~B(3,-2)\text{ and }C(4,-5).
\displaystyle \text{Since, the points are collinear, therefore the area of triangle formed by these points will be zero.}
\displaystyle \therefore \frac{1}{2}\left|x(-2+5)+3(-5-y)+4(y+2)\right|=0
\displaystyle \text{On squaring both sides, we get}
\displaystyle 3x-15-3y+4y+8=0
\displaystyle \Rightarrow 3x+y-7=0
\displaystyle \Rightarrow y=7-3x

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