\displaystyle \textbf{Question 1: }\text{Find the mid-point of the line segment joining the points:}
\displaystyle \text{(i) }(-6,7)\text{ and }(3,5)
\displaystyle \text{(ii) }(5,-3)\text{ and }(-1,7)
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }P\text{ is the midpoint, }m_1:m_2=1:1.
\displaystyle \text{Let }P=(x,y).
\displaystyle x=\frac{1\times3+1\times(-6)}{1+1}=-\frac{3}{2}.
\displaystyle y=\frac{1\times5+1\times7}{1+1}=6.
\displaystyle \therefore P=\left(-\frac{3}{2},6\right).
\displaystyle \text{(ii) Since }P\text{ is the midpoint, }m_1:m_2=1:1.
\displaystyle \text{Let }P=(x,y).
\displaystyle x=\frac{1\times(-1)+1\times5}{1+1}=2.
\displaystyle y=\frac{1\times7+1\times(-3)}{1+1}=2.
\displaystyle \therefore P=(2,2).
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\displaystyle \textbf{Question 2: }\text{Points }A\text{ and }B\text{ have coordinates }(3,5)\text{ and }(x,y)\text{ respectively.}
\displaystyle \text{The midpoint of }AB\text{ is }(2,3).\text{ Find the values of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{Given midpoint of }AB=(2,3).
\displaystyle 2=\frac{x+3}{2}\Rightarrow x=1.
\displaystyle 3=\frac{y+5}{2}\Rightarrow y=1.
\displaystyle \therefore B=(1,1).
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\displaystyle \textbf{Question 3: }A(5,3),B(-1,1)\text{ and }C(7,-3)\text{ are the vertices of triangle }ABC.
\displaystyle \text{If }L\text{ is the midpoint of }AB\text{ and }M\text{ is the midpoint of }AC,\text{ show that }LM=\frac{1}{2}BC.
\displaystyle \text{Answer:}
\displaystyle \text{Since }L\text{ is the midpoint of }AB,\text{ the ratio is }1:1.
\displaystyle \text{Let }L=(x_1,y_1).
\displaystyle x_1=\frac{1\times(-1)+1\times5}{1+1}=2.
\displaystyle y_1=\frac{1\times1+1\times3}{1+1}=2.
\displaystyle \therefore L=(2,2).
\displaystyle \text{Since }M\text{ is the midpoint of }AC,\text{ the ratio is }1:1.
\displaystyle \text{Let }M=(x_2,y_2).
\displaystyle x_2=\frac{1\times7+1\times5}{1+1}=6.
\displaystyle y_2=\frac{1\times(-3)+1\times3}{1+1}=0.
\displaystyle \therefore M=(6,0).
\displaystyle LM=\sqrt{(6-2)^2+(0-2)^2}=\sqrt{20}.
\displaystyle BC=\sqrt{(7+1)^2+(-3-1)^2}
\displaystyle =\sqrt{8^2+(-4)^2}=\sqrt{80}=2\sqrt{20}.
\displaystyle \therefore LM=\frac{1}{2}BC.
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\displaystyle \textbf{Question 4: }\text{Given }M\text{ is the midpoint of }AB,\text{ find the coordinates of:}
\displaystyle \text{(i) }A;\text{ if }M=(1,7)\text{ and }B=(-5,10)
\displaystyle \text{(ii) }B;\text{ if }A=(3,-1)\text{ and }M=(-1,3)
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given midpoint of }AB=(1,7)\text{ and }B=(-5,10).
\displaystyle \text{Let }A=(x,y).
\displaystyle 1=\frac{x+(-5)}{2}\Rightarrow 2=x-5\Rightarrow x=7.
\displaystyle 7=\frac{y+10}{2}\Rightarrow 14=y+10\Rightarrow y=4.
\displaystyle \therefore A=(7,4).
\displaystyle \text{(ii) Given midpoint of }AB=(-1,3)\text{ and }A=(3,-1).
\displaystyle \text{Let }B=(x,y).
\displaystyle -1=\frac{3+x}{2}\Rightarrow -2=3+x\Rightarrow x=-5.
\displaystyle 3=\frac{-1+y}{2}\Rightarrow 6=-1+y\Rightarrow y=7.
\displaystyle \therefore B=(-5,7).
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\displaystyle \textbf{Question 5: }P(-3,2)\text{ is the midpoint of line segment }AB\text{ as shown in the figure.}
\displaystyle \text{Find the coordinates of points }A\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle \text{Given midpoint of }AB=(-3,2).
\displaystyle \text{Let }A=(0,y)\text{ and }B=(x,0).
\displaystyle -3=\frac{x+0}{2}\Rightarrow x=-6.
\displaystyle 2=\frac{y+0}{2}\Rightarrow y=4.
\displaystyle \therefore A=(0,4)\text{ and }B=(-6,0).
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\displaystyle \textbf{Question 6: }\text{In the given figure, }P(4,2)\text{ is the midpoint of line segment }AB.
\displaystyle \text{Find the coordinates of }A(x,0)\text{ and }B(0,y).
\displaystyle \text{Answer:}
\displaystyle \text{Given midpoint of }AB=(4,2).
\displaystyle \text{Let }A=(x,0)\text{ and }B=(0,y).
\displaystyle 4=\frac{x+0}{2}\Rightarrow x=8.
\displaystyle 2=\frac{0+y}{2}\Rightarrow y=4.
\displaystyle \therefore A=(8,0)\text{ and }B=(0,4).
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\displaystyle \textbf{Question 7: }(-5,2),(3,-6)\text{ and }(7,4)\text{ are the vertices of a triangle.}
\displaystyle \text{Find the length of its median through the vertex }(3,-6).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be the midpoint of the line joining }(-5,2)\text{ and }(7,4).
\displaystyle x=\frac{7+(-5)}{2}=1.
\displaystyle y=\frac{4+2}{2}=3.
\displaystyle \therefore P=(1,3).
\displaystyle \text{Length of the median}=\sqrt{(3-1)^2+(-6-3)^2}.
\displaystyle =\sqrt{2^2+(-9)^2}=\sqrt{4+81}=\sqrt{85}.
\displaystyle \therefore \text{The length of the median is }\sqrt{85}.
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\displaystyle \textbf{Question 8: }\text{Given a line }ABCD\text{ in which }AB=BC=CD,\ B=(0,3)\text{ and }C=(1,8).
\displaystyle \text{Find the coordinates of }A\text{ and }D.
\displaystyle \text{Answer:}
\displaystyle \text{Since }AB=BC,\ B\text{ is the midpoint of }AC.
\displaystyle \text{Let }A=(x_1,y_1).
\displaystyle 0=\frac{1+x_1}{2}\Rightarrow x_1=-1.
\displaystyle 3=\frac{8+y_1}{2}\Rightarrow y_1=-2.
\displaystyle \therefore A=(-1,-2).
\displaystyle \text{Since }BC=CD,\ C\text{ is the midpoint of }BD.
\displaystyle \text{Let }D=(x_2,y_2).
\displaystyle 1=\frac{x_2+0}{2}\Rightarrow x_2=2.
\displaystyle 8=\frac{y_2+3}{2}\Rightarrow y_2=13.
\displaystyle \therefore D=(2,13).
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\displaystyle \textbf{Question 9: }\text{One end of the diameter of a circle is }(-2,5).
\displaystyle \text{Find the coordinates of the other end if the centre of the circle is }(2,-1).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P=(x,y)\text{ be the other end of the diameter.}
\displaystyle \text{Since the centre is the midpoint of the diameter,}
\displaystyle 2=\frac{-2+x}{2}\Rightarrow x=6.
\displaystyle -1=\frac{5+y}{2}\Rightarrow y=-7.
\displaystyle \therefore \text{The coordinates of the other end of the diameter are }(6,-7).
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\displaystyle \textbf{Question 10: }A(2,5),B(1,0),C(-4,3)\text{ and }D(-3,8)\text{ are the vertices of quadrilateral }ABCD.
\displaystyle \text{Find the coordinates of the midpoints of }AC\text{ and }BD.\text{ Give a special name to the quadrilateral.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x_1,y_1)\text{ be the midpoint of }AC.
\displaystyle x_1=\frac{-4+2}{2}=-1.
\displaystyle y_1=\frac{3+5}{2}=4.
\displaystyle \therefore P=(-1,4).
\displaystyle \text{Let }M(x_2,y_2)\text{ be the midpoint of }BD.
\displaystyle x_2=\frac{1+(-3)}{2}=-1.
\displaystyle y_2=\frac{0+8}{2}=4.
\displaystyle \therefore M=(-1,4).
\displaystyle \text{Since the diagonals bisect each other, }ABCD\text{ is a parallelogram.}
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\displaystyle \textbf{Question 11: }P(4,2)\text{ and }Q(-1,5)\text{ are the vertices of parallelogram }PQRS.
\displaystyle \text{The diagonals intersect at }(-3,2).\text{ Find the coordinates of }R\text{ and }S.
\displaystyle \text{Answer:}
\displaystyle \text{Since the diagonals bisect each other, }(-3,2)\text{ is the midpoint of }PR.
\displaystyle \text{Let }R=(x_1,y_1).
\displaystyle -3=\frac{4+x_1}{2}\Rightarrow x_1=-10.
\displaystyle 2=\frac{2+y_1}{2}\Rightarrow y_1=2.
\displaystyle \therefore R=(-10,2).
\displaystyle \text{Also, }(-3,2)\text{ is the midpoint of }QS.
\displaystyle \text{Let }S=(x_2,y_2).
\displaystyle -3=\frac{-1+x_2}{2}\Rightarrow x_2=-5.
\displaystyle 2=\frac{5+y_2}{2}\Rightarrow y_2=-1.
\displaystyle \therefore S=(-5,-1).
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\displaystyle \textbf{Question 12: }A(-1,0),B(1,3)\text{ and }D(3,5)\text{ are the vertices of a parallelogram }ABCD.
\displaystyle \text{Find the coordinates of vertex }C.
\displaystyle \text{Answer:}
\displaystyle \text{Let }M(x,y)\text{ be the midpoint of }BD.
\displaystyle x=\frac{1+3}{2}=2.
\displaystyle y=\frac{3+5}{2}=4.
\displaystyle \therefore M=(2,4).
\displaystyle \text{Since the diagonals of a parallelogram bisect each other,}
\displaystyle \text{the midpoint of }AC\text{ is also }M(2,4).
\displaystyle \text{Let }C=(x,y).
\displaystyle 2=\frac{x+(-1)}{2}\Rightarrow x=5.
\displaystyle 4=\frac{y+0}{2}\Rightarrow y=8.
\displaystyle \therefore C=(5,8).
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\displaystyle \textbf{Question 13: }\text{The points }(2,-1),(-1,4)\text{ and }(-2,2)\text{ are midpoints of the sides of a triangle.}
\displaystyle \text{Find its vertices.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the vertices of the triangle be }A(x_1,y_1),B(x_2,y_2)\text{ and }C(x_3,y_3).
\displaystyle \text{Given midpoint of }AB=(2,-1).
\displaystyle 2=\frac{x_1+x_2}{2}\Rightarrow x_1+x_2=4\qquad\cdots\text{(i)}
\displaystyle -1=\frac{y_1+y_2}{2}\Rightarrow y_1+y_2=-2\qquad\cdots\text{(ii)}
\displaystyle \text{Given midpoint of }BC=(-1,4).
\displaystyle -1=\frac{x_2+x_3}{2}\Rightarrow x_2+x_3=-2\qquad\cdots\text{(iii)}
\displaystyle 4=\frac{y_2+y_3}{2}\Rightarrow y_2+y_3=8\qquad\cdots\text{(iv)}
\displaystyle \text{Given midpoint of }CA=(-2,2).
\displaystyle -2=\frac{x_3+x_1}{2}\Rightarrow x_3+x_1=-4\qquad\cdots\text{(v)}
\displaystyle 2=\frac{y_3+y_1}{2}\Rightarrow y_3+y_1=4\qquad\cdots\text{(vi)}
\displaystyle \text{Adding (i), (iii) and (v), we get}
\displaystyle 2(x_1+x_2+x_3)=-2
\displaystyle x_1+x_2+x_3=-1\qquad\cdots\text{(vii)}
\displaystyle \text{Adding (ii), (iv) and (vi), we get}
\displaystyle 2(y_1+y_2+y_3)=10
\displaystyle y_1+y_2+y_3=5\qquad\cdots\text{(viii)}
\displaystyle \text{Using (iii) and (vii), }x_1=-1-(-2)=1.
\displaystyle \text{Using (v) and (vii), }x_2=-1-(-4)=3.
\displaystyle \text{Using (i) and (vii), }x_3=-1-4=-5.
\displaystyle \text{Using (iv) and (viii), }y_1=5-8=-3.
\displaystyle \text{Using (vi) and (viii), }y_2=5-4=1.
\displaystyle \text{Using (ii) and (viii), }y_3=5-(-2)=7.
\displaystyle \therefore \text{The vertices are }(1,-3),(3,1)\text{ and }(-5,7).
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\displaystyle \textbf{Question 14: }\text{Points }A(-5,x),B(y,7)\text{ and }C(1,-3)\text{ are collinear such that }AB=BC.
\displaystyle \text{Calculate the values of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{Since }AB=BC,\ B\text{ is the midpoint of }AC.
\displaystyle y=\frac{1+(-5)}{2}=-2.
\displaystyle 7=\frac{-3+x}{2}\Rightarrow x=17.
\displaystyle \therefore A=(-5,17)\text{ and }B=(-2,7).
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\displaystyle \textbf{Question 15: }\text{Points }P(a,-4),Q(2,b)\text{ and }R(0,2)\text{ are collinear.}
\displaystyle \text{If }Q\text{ lies between }P\text{ and }R\text{ such that }PR=2QR,\text{ calculate }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{Since }PR=2QR,\text{ we have }PQ=QR.
\displaystyle \therefore Q\text{ is the midpoint of }PR.
\displaystyle 2=\frac{a+0}{2}\Rightarrow a=4.
\displaystyle b=\frac{-4+2}{2}=-1.
\displaystyle \therefore a=4\text{ and }b=-1.
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\displaystyle \textbf{Question 16: }\text{Calculate the coordinates of the centroid of triangle }ABC,
\displaystyle \text{if }A=(7,-2),\ B=(0,1)\text{ and }C=(-1,4).
\displaystyle \text{Answer:}
\displaystyle \text{Let }G\text{ be the centroid of triangle }ABC.
\displaystyle x=\frac{7+0+(-1)}{3}=2.
\displaystyle y=\frac{-2+1+4}{3}=1.
\displaystyle \therefore \text{The coordinates of the centroid are }(2,1).
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\displaystyle \textbf{Question 17: }\text{The coordinates of the centroid of triangle }PQR\text{ are }(2,-5).
\displaystyle \text{If }Q=(-6,5)\text{ and }R=(11,8),\text{ calculate the coordinates of }P.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P=(x,y).
\displaystyle \text{Since the centroid is }(2,-5),
\displaystyle 2=\frac{x+(-6)+11}{3}\Rightarrow x=1.
\displaystyle -5=\frac{y+5+8}{3}\Rightarrow y=-28.
\displaystyle \therefore P=(1,-28).
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\displaystyle \textbf{Question 18: }A(5,x),B(-4,3)\text{ and }C(y,-2)\text{ are the vertices of triangle }ABC
\displaystyle \text{whose centroid is the origin. Calculate the values of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{Since the centroid of triangle }ABC\text{ is }(0,0),
\displaystyle 0=\frac{5+(-4)+y}{3}\Rightarrow y=-1.
\displaystyle 0=\frac{x+3+(-2)}{3}\Rightarrow x=-1.
\displaystyle \therefore x=-1\text{ and }y=-1.
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