\displaystyle \textbf{Question 1: }\text{Find the coordinates of the point which divides the line segment joining }(-1,3)
\displaystyle \text{and }(4,-7)\text{ internally in the ratio }3:4.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ divide the line segment joining }A(-1,3)\text{ and }B(4,-7)
\displaystyle \text{internally in the ratio }3:4.
\displaystyle \text{Using the section formula,}
\displaystyle x=\frac{3(4)+4(-1)}{3+4}=\frac{12-4}{7}=\frac{8}{7}
\displaystyle y=\frac{3(-7)+4(3)}{3+4}=\frac{-21+12}{7}=-\frac{9}{7}
\displaystyle \therefore \text{The required point is }\left(\frac{8}{7},-\frac{9}{7}\right).
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the coordinates of the point where the diagonals of the parallelogram}
\displaystyle \text{formed by joining the points }(-2,-1),\,(1,0),\,(4,3)\text{ and }(1,2)\text{ meet.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,-1),\ B(1,0),\ C(4,3)\text{ and }D(1,2)\text{ be the vertices.}
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \text{Mid-point of }AC=\left(\frac{-2+4}{2},\frac{-1+3}{2}\right)=(1,1).
\displaystyle \text{Also, mid-point of }BD=\left(\frac{1+1}{2},\frac{0+2}{2}\right)=(1,1).
\displaystyle \therefore \text{The diagonals meet at }(1,1).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Points }A(3,1),\,B(5,1),\,C(a,b)\text{ and }D(4,3)\text{ are vertices of a}
\displaystyle \text{parallelogram }ABCD.\text{ Find the values of }a\text{ and }b.\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore \text{Mid-point of }AC=\text{mid-point of }BD.
\displaystyle \left(\frac{3+a}{2},\frac{1+b}{2}\right)=\left(\frac{5+4}{2},\frac{1+3}{2}\right)
\displaystyle \therefore \frac{3+a}{2}=\frac{9}{2}\quad\text{and}\quad\frac{1+b}{2}=2
\displaystyle 3+a=9\quad\text{and}\quad1+b=4
\displaystyle \therefore a=6,\quad b=3.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the ratio in which the point }(2,y)\text{ divides the line segment joining}
\displaystyle \text{the points }A(-2,2)\text{ and }B(3,7).\text{ Also, find the value of }y.\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(2,y)\text{ divide }AB\text{ internally in the ratio }m:n.
\displaystyle \text{Using the section formula for the }x\text{-coordinate,}
\displaystyle 2=\frac{3m-2n}{m+n}
\displaystyle 2m+2n=3m-2n
\displaystyle m=4n
\displaystyle \therefore m:n=4:1.
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle y=\frac{4(7)+1(2)}{4+1}=\frac{30}{5}=6.
\displaystyle \therefore \text{The required ratio is }4:1\text{ and }y=6.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A(-1,3),\,B(1,-1)\text{ and }C(5,1)\text{ are the vertices of a triangle }ABC,
\displaystyle \text{find the length of the median through }A.
\displaystyle \text{Answer:}
\displaystyle \text{Let }D\text{ be the mid-point of }BC.
\displaystyle D=\left(\frac{1+5}{2},\frac{-1+1}{2}\right)=(3,0).
\displaystyle \text{Therefore, }AD=\sqrt{(3+1)^2+(0-3)^2}
\displaystyle =\sqrt{16+9}=\sqrt{25}=5\text{ units.}
\displaystyle \therefore \text{The length of the median through }A\text{ is }5\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The points }(3,-4)\text{ and }(-6,2)\text{ are the extremities of a diagonal of a}
\displaystyle \text{parallelogram. If the third vertex is }(-1,-3),\text{ find the coordinates of the fourth vertex.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,-4)\text{ and }C(-6,2)\text{ be the extremities of a diagonal.}
\displaystyle \text{Let }B(-1,-3)\text{ and }D(x,y)\text{ be the other two vertices.}
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \text{Mid-point of }AC=\left(\frac{3-6}{2},\frac{-4+2}{2}\right)=\left(-\frac{3}{2},-1\right).
\displaystyle \therefore \left(\frac{-1+x}{2},\frac{-3+y}{2}\right)=\left(-\frac{3}{2},-1\right)
\displaystyle \frac{-1+x}{2}=-\frac{3}{2}\quad\text{and}\quad\frac{-3+y}{2}=-1
\displaystyle x=-2,\quad y=1.
\displaystyle \therefore \text{The fourth vertex is }(-2,1).
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the coordinates of the points }C\text{ which divide the line segment joining}
\displaystyle A(-2,2)\text{ and }B(2,8)\text{ into four equal parts.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three points of division be }C_1,\ C_2\text{ and }C_3.
\displaystyle C_1\text{ divides }AB\text{ internally in the ratio }1:3.
\displaystyle C_1=\left(\frac{1(2)+3(-2)}{1+3},\frac{1(8)+3(2)}{1+3}\right) =\left(-1,\frac{7}{2}\right).

\displaystyle C_2\text{ divides }AB\text{ internally in the ratio }1:1.
\displaystyle C_2=\left(\frac{-2+2}{2},\frac{2+8}{2}\right)=(0,5).

\displaystyle C_3\text{ divides }AB\text{ internally in the ratio }3:1.
\displaystyle C_3=\left(\frac{3(2)+1(-2)}{3+1},\frac{3(8)+1(2)}{3+1}\right)  =\left(1,\frac{13}{2}\right).
\displaystyle \therefore \text{The required points are }\left(-1,\frac{7}{2}\right),\ (0,5)\text{ and }\left(1,\frac{13}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Points }P,Q,R\text{ and }S\text{ divide the line segment joining the points }A(1,2)
\displaystyle \text{and }B(6,7)\text{ in }5\text{ equal parts. Find the coordinates of the points }P,Q\text{ and }R.
\displaystyle \hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle P\text{ divides }AB\text{ internally in the ratio }1:4.
\displaystyle P=\left(\frac{1(6)+4(1)}{1+4},\frac{1(7)+4(2)}{1+4}\right)=(2,3).

\displaystyle Q\text{ divides }AB\text{ internally in the ratio }2:3.
\displaystyle Q=\left(\frac{2(6)+3(1)}{2+3},\frac{2(7)+3(2)}{2+3}\right)=(3,4).

\displaystyle R\text{ divides }AB\text{ internally in the ratio }3:2.
\displaystyle R=\left(\frac{3(6)+2(1)}{3+2},\frac{3(7)+2(2)}{3+2}\right)=(4,5).
\displaystyle \therefore P=(2,3),\quad Q=(3,4),\quad R=(4,5).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the ratio in which the line segment joining }(-2,-3)\text{ and }(5,6)\text{ is}
\displaystyle \text{divided by (i) }x\text{-axis (ii) }y\text{-axis. Also, find the coordinates of the point of division}
\displaystyle \text{in each case.}\hfill\text{[CBSE 2013, 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,-3)\text{ and }B(5,6).

\displaystyle \text{(i) Let the }x\text{-axis divide }AB\text{ internally in the ratio }m:n.
\displaystyle \text{The }y\text{-coordinate of the point of division is }0.
\displaystyle \therefore \frac{6m-3n}{m+n}=0
\displaystyle 6m-3n=0
\displaystyle \therefore m:n=1:2.
\displaystyle x=\frac{1(5)+2(-2)}{1+2}=\frac{1}{3}
\displaystyle \therefore \text{The ratio is }1:2\text{ and the point of division is }\left(\frac{1}{3},0\right).

\displaystyle \text{(ii) Let the }y\text{-axis divide }AB\text{ internally in the ratio }m:n.
\displaystyle \text{The }x\text{-coordinate of the point of division is }0.
\displaystyle \therefore \frac{5m-2n}{m+n}=0
\displaystyle 5m-2n=0
\displaystyle \therefore m:n=2:5.
\displaystyle y=\frac{2(6)+5(-3)}{2+5}=-\frac{3}{7}
\displaystyle \therefore \text{The ratio is }2:5\text{ and the point of division is }\left(0,-\frac{3}{7}\right).
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Prove that }(4,3),\,(6,4),\,(5,6)\text{ and }(3,5)\text{ are the angular points}
\displaystyle \text{of a square.}

\displaystyle \text{Answer:}
\displaystyle \text{Let }A(4,3),\ B(6,4),\ C(5,6)\text{ and }D(3,5)\text{ be the given points.}
\displaystyle AB=\sqrt{(6-4)^2+(4-3)^2}=\sqrt{5}
\displaystyle BC=\sqrt{(5-6)^2+(6-4)^2}=\sqrt{5}
\displaystyle CD=\sqrt{(3-5)^2+(5-6)^2}=\sqrt{5}
\displaystyle DA=\sqrt{(4-3)^2+(3-5)^2}=\sqrt{5}
\displaystyle \therefore AB=BC=CD=DA.
\displaystyle \text{Hence, }ABCD\text{ is a rhombus.}
\displaystyle AC=\sqrt{(5-4)^2+(6-3)^2}=\sqrt{10}
\displaystyle BD=\sqrt{(3-6)^2+(5-4)^2}=\sqrt{10}
\displaystyle \therefore AC=BD.
\displaystyle \text{A rhombus with equal diagonals is a square.}
\displaystyle \therefore \text{The given points are the angular points of a square.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove that the points }(-4,-1),\,(-2,-4),\,(4,0)\text{ and }(2,3)\text{ are the}
\displaystyle \text{vertices of a rectangle.}
\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{13}
\displaystyle BC=\sqrt{(4+2)^2+(0+4)^2}=2\sqrt{13}
\displaystyle CD=\sqrt{(2-4)^2+(3-0)^2}=\sqrt{13}
\displaystyle DA=\sqrt{(-4-2)^2+(-1-3)^2}=2\sqrt{13}
\displaystyle \therefore AB=CD\quad\text{and}\quad BC=DA.
\displaystyle \text{Hence, }ABCD\text{ is a parallelogram.}
\displaystyle AC=\sqrt{(4+4)^2+(0+1)^2}=\sqrt{65}
\displaystyle BD=\sqrt{(2+2)^2+(3+4)^2}=\sqrt{65}
\displaystyle \therefore AC=BD.
\displaystyle \text{A parallelogram with equal diagonals is a rectangle.}
\displaystyle \therefore \text{The given points are the vertices of a rectangle.}
\displaystyle \\

\displaystyle \textbf{Question 12: }ABCD\text{ is a rectangle formed by joining the points }A(-1,-1),
\displaystyle B(-1,4),\,C(5,4)\text{ and }D(5,-1).\,P,Q,R\text{ and }S\text{ are the mid-points of sides}
\displaystyle AB,BC,CD\text{ and }DA\text{ respectively.}
\displaystyle \text{(i) Is the quadrilateral }PQRS\text{ a square? a rectangle? or a rhombus? Justify your answer.}
\displaystyle \text{Answer:} \displaystyle P=\left(-1,\frac{3}{2}\right),\quad Q=(2,4),\quad R=\left(5,\frac{3}{2}\right),\quad S=(2,-1).
\displaystyle PQ=\sqrt{(2+1)^2+\left(4-\frac{3}{2}\right)^2}=\frac{\sqrt{61}}{2}
\displaystyle QR=\sqrt{(5-2)^2+\left(\frac{3}{2}-4\right)^2}=\frac{\sqrt{61}}{2}
\displaystyle RS=\sqrt{(2-5)^2+\left(-1-\frac{3}{2}\right)^2}=\frac{\sqrt{61}}{2}
\displaystyle SP=\sqrt{(-1-2)^2+\left(\frac{3}{2}+1\right)^2}=\frac{\sqrt{61}}{2}
\displaystyle \therefore PQ=QR=RS=SP.
\displaystyle \therefore PQRS\text{ is a rhombus.}
\displaystyle PR=6\quad\text{and}\quad QS=5.
\displaystyle \text{Since }PR\ne QS,\ PQRS\text{ is neither a square nor a rectangle.}

\displaystyle \text{(ii) Show that the diagonals of the quadrilateral }PQRS\text{ bisect each other.}
\displaystyle \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Mid-point of }PR=\left(\frac{-1+5}{2},\frac{\frac{3}{2}+\frac{3}{2}}{2}\right)=\left(2,\frac{3}{2}\right).
\displaystyle \text{Mid-point of }QS=\left(\frac{2+2}{2},\frac{4-1}{2}\right)=\left(2,\frac{3}{2}\right).
\displaystyle \therefore PR\text{ and }QS\text{ have the same mid-point.}
\displaystyle \therefore \text{The diagonals of }PQRS\text{ bisect each other.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the ratio in which the point }P(-1,y)\text{ lying on the line segment joining}
\displaystyle A(-3,10)\text{ and }B(6,-8)\text{ divides it. Also, find the value of }y.\hfill\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(-1,y)\text{ divide }AB\text{ internally in the ratio }m:n.
\displaystyle \text{Using the section formula for the }x\text{-coordinate,}
\displaystyle -1=\frac{6m-3n}{m+n}
\displaystyle -m-n=6m-3n
\displaystyle 7m=2n
\displaystyle \therefore m:n=2:7.
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle y=\frac{2(-8)+7(10)}{2+7}=\frac{54}{9}=6.
\displaystyle \therefore \text{The required ratio is }2:7\text{ and }y=6.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the coordinates of a point }A,\text{ where }AB\text{ is a diameter of the circle}
\displaystyle \text{whose centre is }(2,-3)\text{ and }B\text{ is }(1,4).\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(x,y)\text{ and }B(1,4)\text{ be the endpoints of the diameter.}
\displaystyle \text{The centre }(2,-3)\text{ is the mid-point of }AB.
\displaystyle \therefore \left(\frac{x+1}{2},\frac{y+4}{2}\right)=(2,-3)
\displaystyle \frac{x+1}{2}=2\quad\text{and}\quad\frac{y+4}{2}=-3
\displaystyle x+1=4\quad\text{and}\quad y+4=-6
\displaystyle x=3,\quad y=-10.
\displaystyle \therefore \text{The coordinates of }A\text{ are }(3,-10).
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In what ratio does the point }(-4,6)\text{ divide the line segment joining}
\displaystyle \text{the points }A(-6,10)\text{ and }B(3,-8)?\hfill\text{[CBSE 2015, 17]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(-4,6)\text{ divide }AB\text{ internally in the ratio }m:n.
\displaystyle \text{Using the section formula for the }x\text{-coordinate,}
\displaystyle -4=\frac{3m-6n}{m+n}
\displaystyle -4m-4n=3m-6n
\displaystyle 7m=2n
\displaystyle \therefore m:n=2:7.
\displaystyle \text{Also, }\frac{2(-8)+7(10)}{2+7}=\frac{54}{9}=6.
\displaystyle \therefore \text{The point }(-4,6)\text{ divides }AB\text{ internally in the ratio }2:7.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{(i) Find the ratio in which the }y\text{-axis divides the line segment joining}
\displaystyle \text{the points }(5,-6)\text{ and }(-1,-4).\text{ Also, find the coordinates of the point of division.}
\displaystyle \hfill\text{[CBSE 2006, 10, 16, 19, 2023, 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the }y\text{-axis divide the line segment internally in the ratio }m:n.
\displaystyle \text{Since the point of division lies on the }y\text{-axis, its }x\text{-coordinate is }0.
\displaystyle \therefore 0=\frac{m(-1)+n(5)}{m+n}
\displaystyle -m+5n=0
\displaystyle \therefore m:n=5:1.
\displaystyle y=\frac{5(-4)+1(-6)}{5+1}=-\frac{26}{6}=-\frac{13}{3}
\displaystyle \therefore \text{The required ratio is }5:1\text{ and the point of division is }\left(0,-\frac{13}{3}\right).

\displaystyle \text{(ii) Find the ratio in which the line segment joining the points }A(6,3)\text{ and }B(-2,-5)
\displaystyle \text{is divided by }x\text{-axis.}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the }x\text{-axis divide }AB\text{ internally in the ratio }m:n.
\displaystyle \text{Since the point of division lies on the }x\text{-axis, its }y\text{-coordinate is }0.
\displaystyle \therefore 0=\frac{m(-5)+n(3)}{m+n}
\displaystyle -5m+3n=0
\displaystyle \therefore m:n=3:5.
\displaystyle \therefore \text{The }x\text{-axis divides the line segment internally in the ratio }3:5.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{(i) If }A\text{ and }B\text{ are }(1,4)\text{ and }(5,2)\text{ respectively, find the}
\displaystyle \text{coordinates of }P\text{ when }\frac{AP}{BP}=\frac{3}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\frac{AP}{BP}=\frac{3}{4},\ P\text{ divides }AB\text{ internally in the ratio }3:4.
\displaystyle \text{Using the section formula,}
\displaystyle P=\left(\frac{3(5)+4(1)}{3+4},\frac{3(2)+4(4)}{3+4}\right)
\displaystyle =\left(\frac{19}{7},\frac{22}{7}\right).
\displaystyle \therefore \text{The coordinates of }P\text{ are }\left(\frac{19}{7},\frac{22}{7}\right).

\displaystyle \text{(ii) }P(-2,5)\text{ and }Q(3,2)\text{ are two points. Find the coordinates of the point }R\text{ on the}
\displaystyle \text{line segment }PQ\text{ such that }PR=2QR.\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }PR=2QR,\ R\text{ divides }PQ\text{ internally in the ratio }2:1.
\displaystyle \text{Using the section formula,}
\displaystyle R=\left(\frac{2(3)+1(-2)}{2+1},\frac{2(2)+1(5)}{2+1}\right)
\displaystyle =\left(\frac{4}{3},3\right).
\displaystyle \therefore \text{The coordinates of }R\text{ are }\left(\frac{4}{3},3\right).

\displaystyle \text{(iii) Find the coordinates of the point }C\text{ which lies on the line }AB\text{ produced such that}
\displaystyle AC=2BC,\text{ where the coordinates of }A\text{ and }B\text{ are }(-1,7)\text{ and }(4,-3)
\displaystyle \text{respectively.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }C\text{ lies on }AB\text{ produced, }AC=AB+BC.
\displaystyle \text{Given }AC=2BC.
\displaystyle \therefore AB+BC=2BC
\displaystyle \therefore AB=BC.
\displaystyle \therefore B(4,-3)\text{ is the mid-point of }AC.
\displaystyle \text{Let }C=(x,y).
\displaystyle \therefore \left(\frac{-1+x}{2},\frac{7+y}{2}\right)=(4,-3)
\displaystyle \frac{-1+x}{2}=4\quad\text{and}\quad\frac{7+y}{2}=-3
\displaystyle x=9,\quad y=-13.
\displaystyle \therefore \text{The coordinates of }C\text{ are }(9,-13).
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{(i) Find the ratio in which }P(4,m)\text{ divides the line segment joining the}
\displaystyle \text{points }A(2,3)\text{ and }B(6,-3).\text{ Hence, find }m.\hfill\text{[CBSE 2004, 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(4,m)\text{ divide }AB\text{ internally in the ratio }p:q.
\displaystyle \text{Using the section formula for the }x\text{-coordinate,}
\displaystyle 4=\frac{6p+2q}{p+q}
\displaystyle 4p+4q=6p+2q
\displaystyle 2p=2q
\displaystyle \therefore p:q=1:1.
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle m=\frac{1(-3)+1(3)}{1+1}=0.
\displaystyle \therefore \text{The required ratio is }1:1\text{ and }m=0.

\displaystyle \text{(ii) Find the ratio in which the point }\left(\frac{8}{5},y\right)\text{ divides the line segment joining}
\displaystyle \text{the points }(1,2)\text{ and }(2,3).\text{ Also, find the value of }y.\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\left(\frac{8}{5},y\right)\text{ divide the line segment internally in the ratio }p:q.
\displaystyle \text{Using the section formula for the }x\text{-coordinate,}
\displaystyle \frac{8}{5}=\frac{2p+q}{p+q}
\displaystyle 8p+8q=10p+5q
\displaystyle 2p=3q
\displaystyle \therefore p:q=3:2.
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle y=\frac{3(3)+2(2)}{3+2}=\frac{13}{5}.
\displaystyle \therefore \text{The required ratio is }3:2\text{ and }y=\frac{13}{5}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find the coordinate of the point }R\text{ on the line segment joining the points}
\displaystyle P(-1,3)\text{ and }Q(2,5)\text{ such that }PR=\frac{3}{5}PQ.
\displaystyle \text{Answer:}
\displaystyle PR=\frac{3}{5}PQ
\displaystyle \therefore RQ=PQ-PR=PQ-\frac{3}{5}PQ=\frac{2}{5}PQ
\displaystyle \therefore PR:RQ=3:2.
\displaystyle \text{Using the section formula,}
\displaystyle R=\left(\frac{3(2)+2(-1)}{3+2},\frac{3(5)+2(3)}{3+2}\right)
\displaystyle =\left(\frac{4}{5},\frac{21}{5}\right).
\displaystyle \therefore \text{The coordinates of }R\text{ are }\left(\frac{4}{5},\frac{21}{5}\right).
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }(a,b)\text{ is the mid-point of the line segment joining the points }A(10,-6),
\displaystyle B(k,4)\text{ and }a-2b=18,\text{ find the value of }k\text{ and the distance }AB.
\displaystyle \text{Answer:}
\displaystyle \text{Since }(a,b)\text{ is the mid-point of }AB,
\displaystyle a=\frac{10+k}{2}\quad\text{and}\quad b=\frac{-6+4}{2}=-1.
\displaystyle \text{Given, }a-2b=18
\displaystyle \frac{10+k}{2}-2(-1)=18
\displaystyle \frac{10+k}{2}=16
\displaystyle 10+k=32
\displaystyle \therefore k=22.
\displaystyle AB=\sqrt{(22-10)^2+(4+6)^2}
\displaystyle =\sqrt{144+100}=\sqrt{244}=2\sqrt{61}\text{ units.}
\displaystyle \therefore k=22\text{ and }AB=2\sqrt{61}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If the points }P,Q(x,7),R,S(6,y)\text{ in this order divide the line segment}
\displaystyle \text{joining }A(2,p)\text{ and }B(7,10)\text{ in }5\text{ equal parts, find }x,y\text{ and }p.
\displaystyle \hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle Q\text{ divides }AB\text{ internally in the ratio }2:3.
\displaystyle \therefore x=\frac{2(7)+3(2)}{2+3}=\frac{20}{5}=4
\displaystyle 7=\frac{2(10)+3p}{2+3}
\displaystyle 35=20+3p
\displaystyle \therefore p=5.
\displaystyle S\text{ divides }AB\text{ internally in the ratio }4:1.
\displaystyle \therefore y=\frac{4(10)+1(5)}{4+1}=\frac{45}{5}=9.
\displaystyle \therefore x=4,\quad y=9,\quad p=5.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If a vertex of a triangle is }(1,1)\text{ and the middle points of the sides through}
\displaystyle \text{it are }(-2,3)\text{ and }(5,2),\text{ find the other vertices.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(1,1)\text{ be the given vertex and }B(x_1,y_1),\,C(x_2,y_2)\text{ be the other vertices.}
\displaystyle \text{Since }(-2,3)\text{ is the mid-point of }AB,
\displaystyle \left(\frac{1+x_1}{2},\frac{1+y_1}{2}\right)=(-2,3)
\displaystyle 1+x_1=-4\quad\text{and}\quad1+y_1=6
\displaystyle \therefore x_1=-5,\quad y_1=5.
\displaystyle \text{Since }(5,2)\text{ is the mid-point of }AC,
\displaystyle \left(\frac{1+x_2}{2},\frac{1+y_2}{2}\right)=(5,2)
\displaystyle 1+x_2=10\quad\text{and}\quad1+y_2=4
\displaystyle \therefore x_2=9,\quad y_2=3.
\displaystyle \therefore \text{The other vertices are }(-5,5)\text{ and }(9,3).
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If the mid-point of the line joining }(3,4)\text{ and }(k,7)\text{ is }(x,y)
\displaystyle \text{and }2x+2y+1=0,\text{ find the value of }k.
\displaystyle \text{Answer:}
\displaystyle \text{Since }(x,y)\text{ is the mid-point,}
\displaystyle x=\frac{3+k}{2}\quad\text{and}\quad y=\frac{4+7}{2}=\frac{11}{2}.
\displaystyle \text{Given, }2x+2y+1=0
\displaystyle 2\left(\frac{3+k}{2}\right)+2\left(\frac{11}{2}\right)+1=0
\displaystyle 3+k+11+1=0
\displaystyle \therefore k=-15.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }A\text{ and }B\text{ are two points having coordinates }(-2,-2)\text{ and }(2,-4)
\displaystyle \text{respectively, find the coordinates of }P\text{ such that }AP=\frac{3}{7}AB.\text{ Also, find the}
\displaystyle \text{coordinates of }P\text{ on }AB\text{ such that }BP=\frac{4}{7}AB.\quad\text{[CBSE 2008, 2009, 2025]}
\displaystyle \text{Answer:}
\displaystyle AP=\frac{3}{7}AB
\displaystyle \therefore PB=AB-AP=AB-\frac{3}{7}AB=\frac{4}{7}AB
\displaystyle \therefore AP:PB=3:4.
\displaystyle \text{Thus, }P\text{ divides }AB\text{ internally in the ratio }3:4.
\displaystyle P=\left(\frac{3(2)+4(-2)}{3+4},\frac{3(-4)+4(-2)}{3+4}\right)
\displaystyle =\left(-\frac{2}{7},-\frac{20}{7}\right).
\displaystyle \text{Also, }BP=\frac{4}{7}AB\text{ gives the same point }P.
\displaystyle \therefore \text{In both cases, }P=\left(-\frac{2}{7},-\frac{20}{7}\right).
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If two vertices of a parallelogram are }(3,2),\,(-1,0)\text{ and the diagonals}
\displaystyle \text{cut at }(2,-5),\text{ find the other vertices of the parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }O(2,-5)\text{ be the point of intersection of the diagonals.}
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \text{Let }A(3,2)\text{ and }C(x,y)\text{ be opposite vertices.}
\displaystyle \therefore \left(\frac{3+x}{2},\frac{2+y}{2}\right)=(2,-5)
\displaystyle 3+x=4\quad\text{and}\quad2+y=-10
\displaystyle \therefore x=1,\quad y=-12.
\displaystyle \text{Let }B(-1,0)\text{ and }D(p,q)\text{ be the other pair of opposite vertices.}
\displaystyle \therefore \left(\frac{-1+p}{2},\frac{q}{2}\right)=(2,-5)
\displaystyle -1+p=4\quad\text{and}\quad q=-10
\displaystyle \therefore p=5,\quad q=-10.
\displaystyle \therefore \text{The other vertices are }(1,-12)\text{ and }(5,-10).
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If the coordinates of the mid-points of the sides of a triangle are }(3,4),
\displaystyle (4,6)\text{ and }(5,7),\text{ find its vertices.}\hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(x_1,y_1),\,B(x_2,y_2)\text{ and }C(x_3,y_3)\text{ be the vertices of the triangle.}
\displaystyle \text{Let }D(3,4),\,E(4,6)\text{ and }F(5,7)\text{ be the mid-points of }BC,CA\text{ and }AB.
\displaystyle x_2+x_3=6,\quad x_3+x_1=8,\quad x_1+x_2=10
\displaystyle y_2+y_3=8,\quad y_3+y_1=12,\quad y_1+y_2=14
\displaystyle \text{Adding the three equations for }x,\quad 2(x_1+x_2+x_3)=24
\displaystyle \therefore x_1+x_2+x_3=12.
\displaystyle \therefore x_1=12-6=6,\quad x_2=12-8=4,\quad x_3=12-10=2.
\displaystyle \text{Adding the three equations for }y,\quad 2(y_1+y_2+y_3)=34
\displaystyle \therefore y_1+y_2+y_3=17.
\displaystyle \therefore y_1=17-8=9,\quad y_2=17-12=5,\quad y_3=17-14=3.
\displaystyle \therefore \text{The vertices of the triangle are }(6,9),\ (4,5)\text{ and }(2,3).
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{The line segment joining the points }P(3,3)\text{ and }Q(6,-6)\text{ is trisected}
\displaystyle \text{at the points }A\text{ and }B\text{ such that }A\text{ is nearer to }P.\text{ If }A\text{ also lies on the line given by}
\displaystyle 2x+y+k=0,\text{ find the value of }k.\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ is the trisection point nearer to }P,\ PA:AQ=1:2.
\displaystyle \text{Using the section formula,}
\displaystyle A=\left(\frac{1(6)+2(3)}{1+2},\frac{1(-6)+2(3)}{1+2}\right)=(4,0).
\displaystyle \text{Since }A(4,0)\text{ lies on }2x+y+k=0,
\displaystyle 2(4)+0+k=0
\displaystyle \therefore k=-8.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{The line segment joining the points }(3,-4)\text{ and }(1,2)\text{ is trisected at}
\displaystyle \text{the points }P\text{ and }Q.\text{ If the coordinates of }P\text{ and }Q\text{ are }(p,-2)\text{ and}
\displaystyle \left(\frac{5}{3},q\right)\text{ respectively, find the values of }p\text{ and }q.\hfill\text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,-4)\text{ and }B(1,2)\text{ be the given points.}
\displaystyle P\text{ divides }AB\text{ internally in the ratio }1:2.
\displaystyle P=\left(\frac{1(1)+2(3)}{1+2},\frac{1(2)+2(-4)}{1+2}\right)  =\left(\frac{7}{3},-2\right).
\displaystyle \therefore p=\frac{7}{3}.
\displaystyle Q\text{ divides }AB\text{ internally in the ratio }2:1.
\displaystyle Q=\left(\frac{2(1)+1(3)}{2+1},\frac{2(2)+1(-4)}{2+1}\right)  =\left(\frac{5}{3},0\right).
\displaystyle \therefore q=0.
\displaystyle \therefore p=\frac{7}{3}\quad\text{and}\quad q=0.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{The line joining the points }(2,1)\text{ and }(5,-8)\text{ is trisected at the points }P
\displaystyle \text{and }Q.\text{ If point }P\text{ lies on the line }2x-y+k=0,\text{ find the value of }k.
\displaystyle \hfill\text{[CBSE 2005, 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,1)\text{ and }B(5,-8)\text{ be the given points.}
\displaystyle P\text{ divides }AB\text{ internally in the ratio }1:2.
\displaystyle P=\left(\frac{1(5)+2(2)}{1+2},\frac{1(-8)+2(1)}{1+2}\right)
\displaystyle =(3,-2).
\displaystyle \text{Since }P(3,-2)\text{ lies on }2x-y+k=0,
\displaystyle 2(3)-(-2)+k=0
\displaystyle 8+k=0
\displaystyle \therefore k=-8.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Find the ratio in which the line }2x+3y-5=0\text{ divides the line segment}
\displaystyle \text{joining the points }(8,-9)\text{ and }(2,1).\text{ Also, find the coordinates of the point of division.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P\text{ divide the line segment joining }A(8,-9)\text{ and }B(2,1)
\displaystyle \text{internally in the ratio }m:n.
\displaystyle P=\left(\frac{2m+8n}{m+n},\frac{m-9n}{m+n}\right).
\displaystyle \text{Since }P\text{ lies on }2x+3y-5=0,
\displaystyle 2\left(\frac{2m+8n}{m+n}\right)+3\left(\frac{m-9n}{m+n}\right)-5=0
\displaystyle 4m+16n+3m-27n-5m-5n=0
\displaystyle 2m-16n=0
\displaystyle \therefore m:n=8:1.
\displaystyle P=\left(\frac{8(2)+1(8)}{8+1},\frac{8(1)+1(-9)}{8+1}\right)
\displaystyle =\left(\frac{8}{3},-\frac{1}{9}\right).
\displaystyle \therefore \text{The required ratio is }8:1\text{ and the point of division is }\left(\frac{8}{3},-\frac{1}{9}\right).
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{A point }P\text{ divides the line segment joining the points }A(3,-5)\text{ and}
\displaystyle B(-4,8)\text{ such that }\frac{AP}{PB}=\frac{k}{1}.\text{ If }P\text{ lies on the line }x+y=0,\text{ find the value}
\displaystyle \text{of }k.\hfill\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\frac{AP}{PB}=\frac{k}{1},\ P\text{ divides }AB\text{ internally in the ratio }k:1.
\displaystyle \text{Using the section formula,}
\displaystyle P=\left(\frac{-4k+3}{k+1},\frac{8k-5}{k+1}\right).
\displaystyle \text{Since }P\text{ lies on }x+y=0,
\displaystyle \frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0
\displaystyle 4k-2=0
\displaystyle \therefore k=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{The mid-point }P\text{ of the line segment joining the points }A(-10,4)\text{ and}
\displaystyle B(-2,0)\text{ lies on the line segment joining the points }C(-9,-4)\text{ and }D(-4,y).\text{ Find}
\displaystyle \text{the ratio in which }P\text{ divides }CD.\text{ Also, find the value of }y.\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle P=\left(\frac{-10-2}{2},\frac{4+0}{2}\right)=(-6,2).
\displaystyle \text{Let }P\text{ divide }CD\text{ internally in the ratio }m:n.
\displaystyle \text{Using the section formula for the }x\text{-coordinate,}
\displaystyle -6=\frac{m(-4)+n(-9)}{m+n}
\displaystyle -6m-6n=-4m-9n
\displaystyle 2m=3n
\displaystyle \therefore m:n=3:2.
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle 2=\frac{3y+2(-4)}{3+2}
\displaystyle 10=3y-8
\displaystyle 3y=18
\displaystyle \therefore y=6.
\displaystyle \therefore P\text{ divides }CD\text{ in the ratio }3:2\text{ and }y=6.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If the point }C(-1,2)\text{ divides internally the line segment joining the points}
\displaystyle A(2,5)\text{ and }B(x,y)\text{ in the ratio }3:4,\text{ find the value of }x^2+y^2.\hfill\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }C(-1,2)\text{ divides }AB\text{ internally in the ratio }3:4,
\displaystyle -1=\frac{3x+4(2)}{3+4}\quad\text{and}\quad2=\frac{3y+4(5)}{3+4}.
\displaystyle -7=3x+8\quad\text{and}\quad14=3y+20
\displaystyle 3x=-15\quad\text{and}\quad3y=-6
\displaystyle \therefore x=-5,\quad y=-2.
\displaystyle \therefore x^2+y^2=(-5)^2+(-2)^2=25+4=29.
\displaystyle \\

\displaystyle \textbf{Question 34: }ABCD\text{ is a parallelogram with vertices }A(x_1,y_1),\,B(x_2,y_2)\text{ and}
\displaystyle C(x_3,y_3).\text{ Find the coordinates of the fourth vertex }D\text{ in terms of }x_1,x_2,x_3,
\displaystyle y_1,y_2\text{ and }y_3.
\displaystyle \text{Answer:}
\displaystyle \text{Let the coordinates of }D\text{ be }(x,y).
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore \text{Mid-point of }AC=\text{mid-point of }BD.
\displaystyle \left(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\right)=\left(\frac{x_2+x}{2},\frac{y_2+y}{2}\right)
\displaystyle x_1+x_3=x_2+x\quad\text{and}\quad y_1+y_3=y_2+y
\displaystyle x=x_1+x_3-x_2\quad\text{and}\quad y=y_1+y_3-y_2.
\displaystyle \therefore D=(x_1+x_3-x_2,\ y_1+y_3-y_2).
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{If the points }A(6,1),\,B(p,2),\,C(9,4)\text{ and }D(7,q)\text{ are the vertices}
\displaystyle \text{of a parallelogram }ABCD,\text{ find the values of }p\text{ and }q.\text{ Hence, check whether }ABCD
\displaystyle \text{is a rectangle or not.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore \text{Mid-point of }AC=\text{mid-point of }BD.
\displaystyle \left(\frac{6+9}{2},\frac{1+4}{2}\right)=\left(\frac{p+7}{2},\frac{2+q}{2}\right)
\displaystyle \frac{15}{2}=\frac{p+7}{2}\quad\text{and}\quad\frac{5}{2}=\frac{2+q}{2}
\displaystyle \therefore p=8,\quad q=3.
\displaystyle \text{Thus, }A(6,1),\,B(8,2),\,C(9,4)\text{ and }D(7,3).
\displaystyle AC=\sqrt{(9-6)^2+(4-1)^2}=3\sqrt{2}
\displaystyle BD=\sqrt{(7-8)^2+(3-2)^2}=\sqrt{2}
\displaystyle \therefore AC\ne BD.
\displaystyle \text{Since the diagonals of the parallelogram are not equal, }ABCD\text{ is not a rectangle.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{The points }A(x_1,y_1),\,B(x_2,y_2)\text{ and }C(x_3,y_3)\text{ are the vertices of }\triangle ABC.
\displaystyle \text{(i) The median from }A\text{ meets }BC\text{ at }D.\text{ Find the coordinates of the point }D.
\displaystyle \text{(ii) Find the coordinates of the point }P\text{ on }AD\text{ such that }AP:PD=2:1.
\displaystyle \text{(iii) Find the coordinates of }Q\text{ and }R\text{ on medians }BE\text{ and }CF\text{ respectively such that}
\displaystyle BQ:QE=2:1\text{ and }CR:RF=2:1.
\displaystyle \text{(iv) What are the coordinates of the centroid of the triangle }ABC?
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }AD\text{ is a median, }D\text{ is the mid-point of }BC.
\displaystyle \therefore D=\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right).

\displaystyle \text{(ii) }P\text{ divides }AD\text{ internally in the ratio }2:1.
\displaystyle P=\left(\frac{2\left(\frac{x_2+x_3}{2}\right)+x_1}{3},\frac{2\left(\frac{y_2+y_3}{2}\right)+y_1}{3}\right)
\displaystyle \therefore P=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).

\displaystyle \text{(iii) Since }E\text{ is the mid-point of }AC,
\displaystyle E=\left(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\right).
\displaystyle \text{Since }BQ:QE=2:1,
\displaystyle Q=\left(\frac{2\left(\frac{x_1+x_3}{2}\right)+x_2}{3},\frac{2\left(\frac{y_1+y_3}{2}\right)+y_2}{3}\right)
\displaystyle =\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).
\displaystyle \text{Similarly, since }F\text{ is the mid-point of }AB\text{ and }CR:RF=2:1,
\displaystyle R=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).

\displaystyle \text{(iv) Since }P,Q\text{ and }R\text{ have the same coordinates, the three medians meet at this point.}
\displaystyle \therefore \text{The centroid of }\triangle ABC\text{ is }\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).
\displaystyle \\

\displaystyle \textbf{Question 37: }A(4,2),\,B(6,5)\text{ and }C(1,4)\text{ are the vertices of }\triangle ABC.
\displaystyle \text{(i) The median from }A\text{ meets }BC\text{ in }D.\text{ Find the coordinates of the point }D.
\displaystyle \text{(ii) Find the coordinates of point }P\text{ on }AD\text{ such that }AP:PD=2:1.
\displaystyle \text{(iii) Find the coordinates of points }Q\text{ and }R\text{ on medians }BE\text{ and }CF\text{ respectively}
\displaystyle \text{such that }BQ:QE=2:1\text{ and }CR:RF=2:1.
\displaystyle \text{(iv) What do you observe?}\hfill\text{[CBSE 2009, 10]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }AD\text{ is a median, }D\text{ is the mid-point of }BC.
\displaystyle D=\left(\frac{6+1}{2},\frac{5+4}{2}\right)=\left(\frac{7}{2},\frac{9}{2}\right).

\displaystyle \text{(ii) }P\text{ divides }AD\text{ internally in the ratio }2:1.
\displaystyle P=\left(\frac{2\left(\frac{7}{2}\right)+1(4)}{2+1},\frac{2\left(\frac{9}{2}\right)+1(2)}{2+1}\right)
\displaystyle =\left(\frac{11}{3},\frac{11}{3}\right).

\displaystyle \text{(iii) Since }E\text{ is the mid-point of }AC,
\displaystyle E=\left(\frac{4+1}{2},\frac{2+4}{2}\right)=\left(\frac{5}{2},3\right).
\displaystyle \text{Since }BQ:QE=2:1,
\displaystyle Q=\left(\frac{2\left(\frac{5}{2}\right)+1(6)}{3},\frac{2(3)+1(5)}{3}\right)=\left(\frac{11}{3},\frac{11}{3}\right).
\displaystyle \text{Since }F\text{ is the mid-point of }AB,
\displaystyle F=\left(\frac{4+6}{2},\frac{2+5}{2}\right)=\left(5,\frac{7}{2}\right).
\displaystyle \text{Since }CR:RF=2:1,
\displaystyle R=\left(\frac{2(5)+1(1)}{3},\frac{2\left(\frac{7}{2}\right)+1(4)}{3}\right)=\left(\frac{11}{3},\frac{11}{3}\right).

\displaystyle \text{(iv) We observe that }P,Q\text{ and }R\text{ have the same coordinates.}
\displaystyle \therefore \text{The three medians are concurrent at }\left(\frac{11}{3},\frac{11}{3}\right),\text{ the centroid of }\triangle ABC.
\displaystyle \text{The centroid divides each median internally in the ratio }2:1.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Find the length of the median through the vertex }B\text{ of }\triangle ABC\text{ with}
\displaystyle \text{vertices }A(9,-2),\,B(-3,7)\text{ and }C(-1,10).\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }D\text{ be the mid-point of }AC.
\displaystyle D=\left(\frac{9-1}{2},\frac{-2+10}{2}\right)=(4,4).
\displaystyle BD=\sqrt{(4+3)^2+(4-7)^2}
\displaystyle =\sqrt{49+9}=\sqrt{58}\text{ units.}
\displaystyle \therefore \text{The length of the median through }B\text{ is }\sqrt{58}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If the mid-point of the line segment joining the points }A(3,4)\text{ and }B(k,6)
\displaystyle \text{is }P(x,y)\text{ and }x+y-10=0,\text{ then find the value of }k.\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }P(x,y)\text{ is the mid-point of }AB,
\displaystyle x=\frac{3+k}{2}\quad\text{and}\quad y=\frac{4+6}{2}=5.
\displaystyle \text{Given, }x+y-10=0
\displaystyle \frac{3+k}{2}+5-10=0
\displaystyle \frac{3+k}{2}=5
\displaystyle 3+k=10
\displaystyle \therefore k=7.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{If }(a,b)\text{ is the mid-point of the line segment joining the points }A(10,-6)
\displaystyle \text{and }B(k,4)\text{ and }a-2b=18,\text{ then find the value of }k.\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }(a,b)\text{ is the mid-point of }AB,
\displaystyle a=\frac{10+k}{2}\quad\text{and}\quad b=\frac{-6+4}{2}=-1.
\displaystyle \text{Given, }a-2b=18
\displaystyle \frac{10+k}{2}-2(-1)=18
\displaystyle \frac{10+k}{2}=16
\displaystyle 10+k=32
\displaystyle \therefore k=22.
\displaystyle \\

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