\displaystyle \textbf{Question 1: }\text{Calculate the distance between the points }(6,-4)\text{ and }(3,2),
\displaystyle \text{correct to }2\text{ decimal places.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance}=\sqrt{(3-6)^2+(2-(-4))^2}.
\displaystyle =\sqrt{9+36}=\sqrt{45}=6.71.
\\

\displaystyle \textbf{Question 2: }\text{Find the distance between the points }(-2,-2)\text{ and }(1,0),
\displaystyle \text{correct to }3\text{ significant figures.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance}=\sqrt{(1-(-2))^2+(0-(-2))^2}.
\displaystyle =\sqrt{9+4}=\sqrt{13}=3.61.
\\

\displaystyle \textbf{Question 3: }\text{Show that the points }P(7,3),Q(6,3+\sqrt{3})\text{ and }R(5,3)
\displaystyle \text{form an equilateral triangle.}
\displaystyle \text{Answer:}
\displaystyle PQ=\sqrt{(6-7)^2+\left((3+\sqrt{3})-3\right)^2}.
\displaystyle =\sqrt{1+3}=\sqrt{4}=2.
\displaystyle PR=\sqrt{(5-7)^2+(3-3)^2}.
\displaystyle =\sqrt{4+0}=\sqrt{4}=2.
\displaystyle QR=\sqrt{(5-6)^2+\left(3-(3+\sqrt{3})\right)^2}.
\displaystyle =\sqrt{1+(-\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=2.
\displaystyle \therefore PQ=PR=QR=2.
\displaystyle \therefore \triangle PQR\text{ is an equilateral triangle.}
\\

\displaystyle \textbf{Question 4: }\text{The circle with centre }(x,y)\text{ passes through the points }(3,11),(14,0)\text{ and }(12,8).
\displaystyle \text{Find the values of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{Since the distances of all points from the centre are equal,}
\displaystyle \sqrt{(3-x)^2+(11-y)^2}=\sqrt{(12-x)^2+(8-y)^2}.
\displaystyle (3-x)^2+(11-y)^2=(12-x)^2+(8-y)^2.
\displaystyle x^2-6x+9+y^2-22y+121=x^2-24x+144+y^2-16y+64.
\displaystyle 18x-6y=78.
\displaystyle 3x-y=13\qquad\cdots\text{(i)}
\displaystyle \sqrt{(3-x)^2+(11-y)^2}=\sqrt{(14-x)^2+(0-y)^2}.
\displaystyle (3-x)^2+(11-y)^2=(14-x)^2+(0-y)^2.
\displaystyle x^2-6x+9+y^2-22y+121=x^2-28x+196+y^2.
\displaystyle 22x-22y=66.
\displaystyle x-y=3\qquad\cdots\text{(ii)}
\displaystyle \text{Solving (i) and (ii), we get }x=5\text{ and }y=2.
\displaystyle \therefore \text{The centre is }(5,2).
\\

\displaystyle \textbf{Question 5: }\text{The points }A(-1,2),B(x,y)\text{ and }C(4,5)\text{ are such that }BA=BC.
\displaystyle \text{Find a linear relation between }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle BA=BC.
\displaystyle \sqrt{(x+1)^2+(y-2)^2}=\sqrt{(4-x)^2+(5-y)^2}.
\displaystyle (x+1)^2+(y-2)^2=(4-x)^2+(5-y)^2.
\displaystyle x^2+2x+1+y^2-4y+4=x^2-8x+16+y^2-10y+25.
\displaystyle 2x-4y+5=-8x-10y+41.
\displaystyle 10x+6y=36.
\displaystyle \therefore 5x+3y=18.
\\

\displaystyle \textbf{Question 6: }\text{Given a triangle }ABC\text{ in which }A=(4,4),B=(0,5)\text{ and }C=(5,10).
\displaystyle \text{A point }P\text{ lies on }BC\text{ such that }BP:PC=3:2.\text{ Find the length of }AP.
\displaystyle \text{Answer:}
\displaystyle \text{Point }P(x,y)\text{ divides }BC\text{ in the ratio }3:2.
\displaystyle x=\frac{2\times0+3\times5}{3+2}=3.
\displaystyle y=\frac{2\times5+3\times10}{3+2}=8.
\displaystyle \therefore P=(3,8).
\displaystyle AP=\sqrt{(3-4)^2+(8-4)^2}.
\displaystyle =\sqrt{1+16}=\sqrt{17}.
\displaystyle \therefore AP=\sqrt{17}.
\\

\displaystyle \textbf{Question 7: }A(20,0)\text{ and }B(10,-20)\text{ are two fixed points.}
\displaystyle \text{Find }P\text{ in }AB\text{ such that }3PB=AB.\text{ Also, find }Q\text{ in }AB\text{ such that }AB=6AQ.
\displaystyle \text{Answer:}
\displaystyle 3PB=AB\Rightarrow PB=\frac{1}{3}AB.
\displaystyle \therefore AP:PB=2:1.
\displaystyle P=\left(\frac{1\times20+2\times10}{2+1},\frac{1\times0+2\times(-20)}{2+1}\right).
\displaystyle \therefore P=\left(\frac{40}{3},-\frac{40}{3}\right).
\displaystyle AB=6AQ\Rightarrow AQ=\frac{1}{6}AB.
\displaystyle \therefore AQ:QB=1:5.
\displaystyle Q=\left(\frac{5\times20+1\times10}{1+5},\frac{5\times0+1\times(-20)}{1+5}\right).
\displaystyle \therefore Q=\left(\frac{55}{3},-\frac{10}{3}\right).
\\

\displaystyle \textbf{Question 8: }A(-8,0),B(0,16)\text{ and }C(0,0)\text{ are the vertices of triangle }ABC.
\displaystyle \text{Point }P\text{ lies on }AB\text{ and }Q\text{ lies on }AC\text{ such that }AP:PB=3:5\text{ and }AQ:QC=3:5.
\displaystyle \text{Show that }PQ=\frac{3}{8}BC.
\displaystyle \text{Answer:}
\displaystyle \text{Point }P(x_1,y_1)\text{ divides }AB\text{ in the ratio }3:5.
\displaystyle x_1=\frac{5\times(-8)+3\times0}{3+5}=-5.
\displaystyle y_1=\frac{5\times0+3\times16}{3+5}=6.
\displaystyle \therefore P=(-5,6).
\displaystyle \text{Point }Q(x_2,y_2)\text{ divides }AC\text{ in the ratio }3:5.
\displaystyle x_2=\frac{5\times(-8)+3\times0}{3+5}=-5.
\displaystyle y_2=\frac{5\times0+3\times0}{3+5}=0.
\displaystyle \therefore Q=(-5,0).
\displaystyle PQ=\sqrt{(-5+5)^2+(0-6)^2}=6.
\displaystyle BC=\sqrt{(0-0)^2+(16-0)^2}=16.
\displaystyle \frac{3}{8}BC=\frac{3}{8}\times16=6.
\displaystyle \therefore PQ=\frac{3}{8}BC.
\\

\displaystyle \textbf{Question 9: }\text{Find the coordinates of points of trisection of the line segment joining }(6,9)\text{ and the origin.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x_1,y_1)\text{ and }Q(x_2,y_2)\text{ be the points of trisection.}
\displaystyle \text{Point }P\text{ divides }(6,9)\text{ and }(0,0)\text{ in the ratio }1:2.
\displaystyle x_1=\frac{2\times6+1\times0}{1+2}=4.
\displaystyle y_1=\frac{2\times9+1\times0}{1+2}=6.
\displaystyle \therefore P=(4,6).
\displaystyle \text{Point }Q\text{ divides }(6,9)\text{ and }(0,0)\text{ in the ratio }2:1.
\displaystyle x_2=\frac{1\times6+2\times0}{2+1}=2.
\displaystyle y_2=\frac{1\times9+2\times0}{2+1}=3.
\displaystyle \therefore Q=(2,3).
\\

\displaystyle \textbf{Question 10: }\text{A line segment joining }A\left(-1,\frac{5}{3}\right)\text{ and }B(a,5)
\displaystyle \text{is divided in the ratio }1:3\text{ at }P,\text{ where }AB\text{ intersects the }y\text{-axis. }[\text{ICSE }1994]
\displaystyle \text{(i) Calculate }a.\qquad\text{(ii) Calculate the coordinates of }P.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P=(0,y).
\displaystyle 0=\frac{1\cdot a+3\cdot(-1)}{1+3}\Rightarrow a=3.
\displaystyle y=\frac{1\cdot5+3\cdot\frac{5}{3}}{1+3}=\frac{10}{4}=\frac{5}{2}.
\displaystyle \therefore a=3\text{ and }P\left(0,\frac{5}{2}\right).
\\

\displaystyle \textbf{Question 11: }\text{In what ratio is the line joining }A(0,3)\text{ and }B(4,-1)
\displaystyle \text{divided by the }x\text{-axis? Write the coordinates of the point where }AB\text{ intersects the }x\text{-axis. }[\text{ICSE }1993]
\displaystyle \text{Answer:}
\displaystyle \text{Let the required ratio be }k:1\text{ and the point of intersection be }P(x,0).
\displaystyle 0=\frac{k(-1)+3}{k+1}\Rightarrow k=3.
\displaystyle \therefore AP:PB=3:1.
\displaystyle x=\frac{3\cdot4+1\cdot0}{3+1}=3.
\displaystyle \therefore P=(3,0).
\\

\displaystyle \textbf{Question 12: }\text{The midpoint 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{Given midpoint of }AB=(4,-3).
\displaystyle \text{Let }A=(8,0)\text{ and }B=(0,y).
\displaystyle 4=\frac{8+0}{2}.
\displaystyle -3=\frac{0+y}{2}\Rightarrow y=-6.
\displaystyle \therefore A=(8,0)\text{ and }B=(0,-6).
\\

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

\displaystyle \textbf{Question 14: }\text{Find the coordinates of the centroid of triangle }ABC\text{ whose vertices are}
\displaystyle A(-1,3),\ B(1,-1)\text{ and }C(5,1). \hfill [\text{ICSE }2006]
\displaystyle \text{Answer:}
\displaystyle \text{Let }G(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 15: }\text{The midpoint of the line segment joining }(4a,2b-3)\text{ and }(-4,3b)
\displaystyle \text{is }(2,-2a).\text{ Find the values of }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{Given midpoint }=(2,-2a).
\displaystyle 2=\frac{4a+(-4)}{2}\Rightarrow a=2.
\displaystyle -2(2)=\frac{(2b-3)+3b}{2}.
\displaystyle -8=5b-3\Rightarrow 5b=-5\Rightarrow b=-1.
\displaystyle \therefore a=2\text{ and }b=-1.
\\

\displaystyle \textbf{Question 16: }\text{The midpoint of the line segment joining }(2a,4)\text{ and }(-2,2b)
\displaystyle \text{is }(1,2a+1).\text{ Find the values of }a\text{ and }b. \hfill [\text{ICSE }2007]
\displaystyle \text{Answer:}
\displaystyle \text{Given midpoint }=(1,2a+1).
\displaystyle 1=\frac{2a+(-2)}{2}\Rightarrow a=2.
\displaystyle 2a+1=\frac{4+2b}{2}.
\displaystyle 2(2)+1=\frac{4+2b}{2}\Rightarrow b=3.
\displaystyle \therefore a=2\text{ and }b=3.
\\

\displaystyle \textbf{Question 17: }\text{(i) Write down the coordinates of 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 x=\frac{1\times17+2\times(-4)}{1+2}=3.
\displaystyle y=\frac{1\times10+2\times1}{1+2}=4.
\displaystyle \therefore P=(3,4).
\displaystyle \text{(ii) }OP=\sqrt{(3-0)^2+(4-0)^2}=\sqrt{25}=5.
\displaystyle \text{(iii) Let the }y\text{-axis divide }AB\text{ in the ratio }k:1.
\displaystyle \text{Let the point of division be }Q(0,y).
\displaystyle 0=\frac{k\times17+1\times(-4)}{k+1}.
\displaystyle 17k-4=0\Rightarrow k=\frac{4}{17}.
\displaystyle \therefore \text{The required ratio is }4:17.
\\

\displaystyle \textbf{Question 18: }\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 }D\text{ so that }ABCD\text{ is a square. }[\text{ICSE }1992]
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(-1+5)^2+(-2-4)^2}=\sqrt{4^2+(-6)^2}=\sqrt{52}.
\displaystyle BC=\sqrt{(5+1)^2+(2+2)^2}=\sqrt{6^2+4^2}=\sqrt{52}.
\displaystyle AC=\sqrt{(5+5)^2+(2-4)^2}=\sqrt{10^2+(-2)^2}=\sqrt{104}.
\displaystyle \therefore AB=BC.
\displaystyle \text{Also, }AB^2+BC^2=52+52=104=AC^2.
\displaystyle \therefore \triangle ABC\text{ is an isosceles right-angled triangle, right-angled at }B.
\displaystyle \text{For square }ABCD,\text{ we have }D=A+C-B.
\displaystyle D=(-5,4)+(5,2)-(-1,-2)=(1,8).
\displaystyle \therefore D=(1,8).
\\

\displaystyle \textbf{Question 19: }M\text{ is the midpoint of the line segment joining }A(-3,7)\text{ and }B(9,-1).
\displaystyle \text{Find }M.\text{ If }R(2,2)\text{ divides }MO\text{ in the ratio }p:q,\text{ find }p:q.
\displaystyle \text{Answer:}
\displaystyle M=\left(\frac{-3+9}{2},\frac{7+(-1)}{2}\right)=(3,3).
\displaystyle \text{Let }R(2,2)\text{ divide }MO\text{ in the ratio }p:q.
\displaystyle \text{Here }M=(3,3)\text{ and }O=(0,0).
\displaystyle 2=\frac{p\cdot0+q\cdot3}{p+q}.
\displaystyle 2p+2q=3q\Rightarrow q=2p.
\displaystyle \therefore p:q=1:2.
\\

\displaystyle \textbf{Question 20: }\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 3=\frac{k\cdot6+1\cdot2}{k+1}.
\displaystyle 3k+3=6k+2\Rightarrow k=\frac{1}{3}.
\displaystyle \therefore \text{The required ratio is }1:3.
\displaystyle x=\frac{1\cdot3+3\cdot(-4)}{1+3}=-\frac{9}{4}.
\displaystyle \therefore P=\left(-\frac{9}{4},3\right).
\displaystyle AP=\sqrt{\left(-\frac{9}{4}+4\right)^2+(3-2)^2}.
\displaystyle =\sqrt{\left(\frac{7}{4}\right)^2+1}=\sqrt{\frac{49}{16}+\frac{16}{16}}.
\displaystyle =\sqrt{\frac{65}{16}}=\frac{\sqrt{65}}{4}.
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.