Class 10: Distance and Section Formula

$\displaystyle \textbf{Question 1: }\text{Find the distance between the points }(3,6)\text{ and }(0,2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }(3,6)=(x_1,y_1)\text{ and }(0,2)=(x_2,y_2)$
$\displaystyle \text{Distance between the given points}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\displaystyle =\sqrt{(0-3)^2+(2-6)^2}$
$\displaystyle =\sqrt{9+16}=\sqrt{25}=5\text{ units}$
$\\$

$\displaystyle \textbf{Question 2: }\text{Find the distance between the origin and the point:}$
$\displaystyle \text{(i) }(-12,5)\qquad\text{(ii) }(15,-8)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Since the distance between the origin and }(x,y)\text{ is }\sqrt{x^2+y^2}$
$\displaystyle \text{Distance between the origin and }(-12,5)=\sqrt{(-12)^2+5^2}$
$\displaystyle =\sqrt{144+25}=\sqrt{169}=13\text{ units}$
$\displaystyle \text{(ii) Distance between the origin and }(15,-8)=\sqrt{15^2+(-8)^2}$
$\displaystyle =\sqrt{225+64}=\sqrt{289}=17\text{ units}$
$\\$

$\displaystyle \textbf{Question 3: }\text{Find the coordinates of points on the }x\text{-axis which are at a distance of }5\text{ units from the point }(6,-3).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the coordinates of the required point be }(x,0)$
$\displaystyle \text{Using the distance formula,}$
$\displaystyle 5=\sqrt{(x-6)^2+(0+3)^2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle 25=(x-6)^2+9$
$\displaystyle 25=x^2-12x+36+9$
$\displaystyle x^2-12x+20=0$
$\displaystyle (x-2)(x-10)=0$
$\displaystyle x=2\text{ or }x=10$
$\displaystyle \therefore \text{The required points are }(2,0)\text{ and }(10,0).$

$\displaystyle \textbf{Question 4: }\text{KM is a straight line of }13\text{ units. If K has the coordinate }(2,5)$
$\displaystyle \text{and M has the coordinates }(x,-7),\text{ find the value of }x.\hfill \text{[ICSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }K(2,5)=(x_1,y_1)\text{ and }M(x,-7)=(x_2,y_2)$
$\displaystyle KM=13\text{ units}$
$\displaystyle \Rightarrow \sqrt{(x-2)^2+(-7-5)^2}=13$
$\displaystyle \Rightarrow \sqrt{(x-2)^2+144}=13$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle (x-2)^2+144=169$
$\displaystyle x^2-4x+4+144=169$
$\displaystyle x^2-4x-21=0$
$\displaystyle (x-7)(x+3)=0$
$\displaystyle x=7\text{ or }x=-3$
$\\$

$\displaystyle \textbf{Question 5: }\text{Which point on the }y\text{-axis is equidistant from the points }(12,3)$
$\displaystyle \text{and }(-5,10)?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point on the }y\text{-axis be }(0,y)$
$\displaystyle \text{Given, }(0,y)\text{ is equidistant from }(12,3)\text{ and }(-5,10)$
$\displaystyle \Rightarrow \sqrt{(12-0)^2+(3-y)^2}=\sqrt{(-5-0)^2+(10-y)^2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle 144+(3-y)^2=25+(10-y)^2$
$\displaystyle 144+9-6y+y^2=25+100-20y+y^2$
$\displaystyle 153-6y=125-20y$
$\displaystyle 14y=-28$
$\displaystyle y=-2$
$\displaystyle \therefore \text{The required point on the }y\text{-axis is }(0,-2).$
$\\$

$\displaystyle \textbf{Question 6: }\text{Use the distance formula to show that the points }A(1,1),B(6,4)$
$\displaystyle \text{and }C(4,2)\text{ are collinear.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given solution is incorrect because the points are not collinear.}$
$\displaystyle AB=\sqrt{(6-1)^2+(4-1)^2}=\sqrt{25+9}=\sqrt{34}$
$\displaystyle BC=\sqrt{(4-6)^2+(2-4)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
$\displaystyle AC=\sqrt{(4-1)^2+(2-1)^2}=\sqrt{9+1}=\sqrt{10}$
$\displaystyle \text{Here, }AC+BC\neq AB$
$\displaystyle \therefore \text{The points }A(1,1),B(6,4)\text{ and }C(4,2)\text{ are not collinear.}$
$\displaystyle \text{Hence, the question appears to have an error.}$
$\\$

$\displaystyle \textbf{Question 7: }\text{Show that the points }A(8,3),B(0,9)\text{ and }C(14,11)\text{ are the vertices}$
$\displaystyle \text{of an isosceles right-angled triangle.}$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\sqrt{(0-8)^2+(9-3)^2}=\sqrt{64+36}=\sqrt{100}=10$
$\displaystyle BC=\sqrt{(14-0)^2+(11-9)^2}=\sqrt{196+4}=\sqrt{200}=10\sqrt{2}$
$\displaystyle CA=\sqrt{(8-14)^2+(3-11)^2}=\sqrt{36+64}=\sqrt{100}=10$
$\displaystyle AB^2+CA^2=10^2+10^2=100+100=200$
$\displaystyle BC^2=(10\sqrt{2})^2=200$
$\displaystyle \therefore BC^2=AB^2+CA^2$
$\displaystyle \therefore \triangle ABC\text{ is right-angled at }A.$
$\displaystyle \text{Also, }AB=CA=10$
$\displaystyle \therefore \triangle ABC\text{ is isosceles.}$
$\displaystyle \therefore \triangle ABC\text{ is an isosceles right-angled triangle.}$
$\\$

$\displaystyle \textbf{Question 8: }\text{Show that the quadrilateral }ABCD\text{ with }A(3,1),B(0,-2),C(1,1)\text{ and }D(4,4)$
$\displaystyle \text{is a parallelogram.}$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\sqrt{(0-3)^2+(-2-1)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
$\displaystyle BC=\sqrt{(1-0)^2+(1+2)^2}=\sqrt{1+9}=\sqrt{10}$
$\displaystyle CD=\sqrt{(4-1)^2+(4-1)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
$\displaystyle DA=\sqrt{(3-4)^2+(1-4)^2}=\sqrt{1+9}=\sqrt{10}$
$\displaystyle \text{Since }AB=CD\text{ and }BC=DA,$
$\displaystyle \therefore \text{the opposite sides of quadrilateral }ABCD\text{ are equal.}$
$\displaystyle \therefore \text{Quadrilateral }ABCD\text{ is a parallelogram.}$
$\\$

$\displaystyle \textbf{Question 9: }\text{Find the area of a circle whose centre is }(5,-3)\text{ and which passes through}$
$\displaystyle \text{the point }(-7,2).\text{ Take }\pi=3.14.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius }(r)=\text{distance between }(5,-3)\text{ and }(-7,2)$
$\displaystyle r=\sqrt{(-7-5)^2+(2+3)^2}=\sqrt{144+25}=\sqrt{169}=13$
$\displaystyle \text{Area of the circle}=\pi r^2$
$\displaystyle =3.14\times13^2$
$\displaystyle =3.14\times169$
$\displaystyle =530.66\text{ sq. units}$
$\displaystyle \therefore \text{The area of the circle is }530.66\text{ sq. units.}$
$\\$

$\displaystyle \textbf{Question 10: }\text{Find the points on the }x\text{-axis whose distances from the points }A(7,6)$
$\displaystyle \text{and }B(-3,4)\text{ are in the ratio }1:2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point on the }x\text{-axis be }P(x,0)$
$\displaystyle \text{Given, }\frac{PA}{PB}=\frac{1}{2}$
$\displaystyle \Rightarrow 2PA=PB$
$\displaystyle \Rightarrow 2\sqrt{(x-7)^2+(0-6)^2}=\sqrt{(x+3)^2+(0-4)^2}$
$\displaystyle \Rightarrow 4\{(x-7)^2+36\}=(x+3)^2+16$
$\displaystyle \Rightarrow 4(x^2-14x+49+36)=x^2+6x+9+16$
$\displaystyle \Rightarrow 4x^2-56x+340=x^2+6x+25$
$\displaystyle \Rightarrow 3x^2-62x+315=0$
$\displaystyle \Rightarrow 3x^2-27x-35x+315=0$
$\displaystyle \Rightarrow 3x(x-9)-35(x-9)=0$
$\displaystyle \Rightarrow (x-9)(3x-35)=0$
$\displaystyle \Rightarrow x=9\text{ or }x=\frac{35}{3}$
$\displaystyle \therefore \text{The required points are }(9,0)\text{ and }\left(\frac{35}{3},0\right).$
$\\$

$\displaystyle \textbf{Question 11: }\text{Point }P(x,y)\text{ is equidistant from the points }A(-2,0)\text{ and }B(3,-4).$
$\displaystyle \text{Prove that }10x-8y=21.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }PA=PB$
$\displaystyle \Rightarrow \sqrt{(x+2)^2+(y-0)^2}=\sqrt{(x-3)^2+(y+4)^2}$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle (x+2)^2+y^2=(x-3)^2+(y+4)^2$
$\displaystyle \Rightarrow x^2+4x+4+y^2=x^2-6x+9+y^2+8y+16$
$\displaystyle \Rightarrow 4x+4=-6x+8y+25$
$\displaystyle \Rightarrow 10x-8y=21$
$\displaystyle \therefore \text{Proved.}$
$\\$

$\displaystyle \textbf{Question 12: }\text{Find the coordinates of the circumcenter of the triangle }ABC,$
$\displaystyle \text{whose vertices }A,B\text{ and }C\text{ are }(4,6),(0,4)\text{ and }(6,2)\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the circumcenter be }P(x,y)$
$\displaystyle \therefore PA=PB$
$\displaystyle \Rightarrow \sqrt{(x-4)^2+(y-6)^2}=\sqrt{(x-0)^2+(y-4)^2}$
$\displaystyle \Rightarrow x^2-8x+16+y^2-12y+36=x^2+y^2-8y+16$
$\displaystyle \Rightarrow -8x-4y=-36$
$\displaystyle \Rightarrow 2x+y=9\qquad\text{... (i)}$
$\displaystyle \text{Also, }PA=PC$
$\displaystyle \Rightarrow \sqrt{(x-4)^2+(y-6)^2}=\sqrt{(x-6)^2+(y-2)^2}$
$\displaystyle \Rightarrow x^2-8x+16+y^2-12y+36=x^2-12x+36+y^2-4y+4$
$\displaystyle \Rightarrow 4x-8y=-12$
$\displaystyle \Rightarrow x-2y=-3\qquad\text{... (ii)}$
$\displaystyle \text{Solving (i) and (ii), we get }x=3\text{ and }y=3$
$\displaystyle \therefore \text{The circumcenter of the given triangle is }(3,3).$
$\\$

$\displaystyle \textbf{Question 13: }\text{Find the coordinates of point }P\text{ which divides the join of }A(4,-5)$
$\displaystyle \text{and }B(6,3)\text{ in the ratio }2:5.$
$\displaystyle \text{Answer:}$ x13 1
$\displaystyle \text{Let the coordinates of }P\text{ be }(x,y)$
$\displaystyle \text{Here, }m_1:m_2=2:5$
$\displaystyle x=\frac{m_1x_2+m_2x_1}{m_1+m_2}=\frac{2\times6+5\times4}{2+5}=\frac{32}{7}$
$\displaystyle y=\frac{m_1y_2+m_2y_1}{m_1+m_2}=\frac{2\times3+5\times(-5)}{2+5}=-\frac{19}{7}$
$\displaystyle \therefore P=\left(\frac{32}{7},-\frac{19}{7}\right)$
$\\$

$\displaystyle \textbf{Question 14: }\text{Find the ratio in which the point }(5,4)\text{ divides the line joining}$
$\displaystyle \text{the points }(2,1)\text{ and }(7,6).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }m_1:m_2$
$\displaystyle \text{Take }(2,1)=(x_1,y_1),\ (7,6)=(x_2,y_2)\text{ and }(5,4)=(x,y)$
$\displaystyle x=\frac{m_1x_2+m_2x_1}{m_1+m_2}$
$\displaystyle \Rightarrow 5=\frac{7m_1+2m_2}{m_1+m_2}$
$\displaystyle \Rightarrow 5m_1+5m_2=7m_1+2m_2$
$\displaystyle \Rightarrow 2m_1=3m_2$
$\displaystyle \Rightarrow \frac{m_1}{m_2}=\frac{3}{2}$
$\displaystyle \therefore \text{The required ratio is }3:2.$
$\\$

$\displaystyle \textbf{Question 15: }\text{In what ratio is the line joining the points }(4,2)\text{ and }(3,-5)$
$\displaystyle \text{divided by the }x\text{-axis? Also, find the coordinates of the point of intersection.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point of intersection be }(x,0)$
$\displaystyle \text{Take }(4,2)=(x_1,y_1)\text{ and }(3,-5)=(x_2,y_2)$ x15
$\displaystyle \text{Since }y=\frac{ky_2+y_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k(-5)+2}{k+1}$
$\displaystyle \Rightarrow -5k+2=0$
$\displaystyle \Rightarrow k=\frac{2}{5}$
$\displaystyle \therefore m_1:m_2=2:5$
$\displaystyle x=\frac{2\times3+5\times4}{2+5}=\frac{26}{7}$
$\displaystyle \therefore \text{The line is divided in the ratio }2:5$
$\displaystyle \text{and the point of intersection is }\left(\frac{26}{7},0\right).$
$\\$

$\displaystyle \textbf{Question 16: }\text{Calculate the ratio in which the line joining the points }(4,6)\text{ and }(-5,-4)$
$\displaystyle \text{is divided by the line }y=3.\text{ Also, find the coordinates of the point of intersection.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The coordinates of every point on the line }y=3\text{ are of the form }(x,3).$
$\displaystyle \text{Let the required ratio be }m_1:m_2$
$\displaystyle \text{Since }y=\frac{m_1y_2+m_2y_1}{m_1+m_2}$
$\displaystyle \Rightarrow 3=\frac{m_1(-4)+m_2(6)}{m_1+m_2}$
$\displaystyle \Rightarrow 3m_1+3m_2=-4m_1+6m_2$
$\displaystyle \Rightarrow 7m_1=3m_2$
$\displaystyle \Rightarrow \frac{m_1}{m_2}=\frac{3}{7}$
$\displaystyle \therefore \text{The required ratio is }3:7$
$\displaystyle \text{Now, }x=\frac{m_1x_2+m_2x_1}{m_1+m_2}$
$\displaystyle \Rightarrow x=\frac{3(-5)+7(4)}{3+7}=\frac{13}{10}$
$\displaystyle \therefore \text{The required point of intersection is }\left(\frac{13}{10},3\right).$
$\\$

$\displaystyle \textbf{Question 17: }\text{The origin }O,B(-6,9)\text{ and }C(12,-3)\text{ are vertices of }\triangle OBC.$
$\displaystyle \text{Point }P\text{ divides }OB\text{ in the ratio }1:2\text{ and point }Q\text{ divides }OC\text{ in the ratio }1:2.$
$\displaystyle \text{Find the coordinates of }P\text{ and }Q.\text{ Also, show that }PQ=\frac{1}{3}BC.$
$\displaystyle \text{Answer:}$ x17
$\displaystyle \text{For point }P,\ m_1:m_2=1:2,\ (x_1,y_1)=(0,0)\text{ and }(x_2,y_2)=(-6,9)$
$\displaystyle P=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)$
$\displaystyle =\left(\frac{1(-6)+2(0)}{1+2},\frac{1(9)+2(0)}{1+2}\right)=(-2,3)$
$\displaystyle \text{For point }Q,\ m_1:m_2=1:2,\ (x_1,y_1)=(0,0)\text{ and }(x_2,y_2)=(12,-3)$
$\displaystyle Q=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)$
$\displaystyle =\left(\frac{1(12)+2(0)}{1+2},\frac{1(-3)+2(0)}{1+2}\right)=(4,-1)$
$\displaystyle \text{Now, }PQ=\text{distance between }P(-2,3)\text{ and }Q(4,-1)$
$\displaystyle PQ=\sqrt{(4+2)^2+(-1-3)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$
$\displaystyle BC=\sqrt{(12+6)^2+(-3-9)^2}=\sqrt{324+144}=\sqrt{468}=6\sqrt{13}$
$\displaystyle \therefore PQ=2\sqrt{13}\text{ and }BC=6\sqrt{13}$
$\displaystyle \Rightarrow PQ=\frac{1}{3}BC$
$\displaystyle \therefore \text{Proved.}$
$\\$

$\displaystyle \textbf{Question 18: }\text{Find the coordinates of the points of trisection of the line segment}$
$\displaystyle \text{joining the points }A(6,-2)\text{ and }B(-8,10).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P\text{ and }Q\text{ be the points of trisection so that }AP=PQ=QB.$
$\displaystyle \text{For point }P:$
$\displaystyle m_1:m_2=AP:PB=1:2,\ (x_1,y_1)=(6,-2)\text{ and }(x_2,y_2)=(-8,10)$
$\displaystyle x=\frac{m_1x_2+m_2x_1}{m_1+m_2}=\frac{1(-8)+2(6)}{1+2}=\frac{4}{3}$
$\displaystyle y=\frac{m_1y_2+m_2y_1}{m_1+m_2}=\frac{1(10)+2(-2)}{1+2}=2$
$\displaystyle \therefore P=\left(\frac{4}{3},2\right)$
$\displaystyle \text{For point }Q:$
$\displaystyle m_1:m_2=AQ:QB=2:1,\ (x_1,y_1)=(6,-2)\text{ and }(x_2,y_2)=(-8,10)$
$\displaystyle x=\frac{m_1x_2+m_2x_1}{m_1+m_2}=\frac{2(-8)+1(6)}{2+1}=-\frac{10}{3}$
$\displaystyle y=\frac{m_1y_2+m_2y_1}{m_1+m_2}=\frac{2(10)+1(-2)}{2+1}=6$
$\displaystyle \therefore Q=\left(-\frac{10}{3},6\right)$
$\\$

$\displaystyle \textbf{Question 19: }\text{Show that }P(3,m-5)\text{ is a point of trisection of the line segment}$
$\displaystyle \text{joining the points }A(4,-2)\text{ and }B(1,4).\text{ Hence, find the value of }m.$
$\displaystyle \text{Answer:}$
$\displaystyle P\text{ will be a point of trisection of }AB\text{ if it divides }AB\text{ in the ratio }1:2\text{ or }2:1.$
$\displaystyle \text{Since }x=\frac{m_1x_2+m_2x_1}{m_1+m_2}$ x19 1
$\displaystyle \Rightarrow 3=\frac{m_1(1)+m_2(4)}{m_1+m_2}$
$\displaystyle \Rightarrow 3m_1+3m_2=m_1+4m_2$
$\displaystyle \Rightarrow 2m_1=m_2$
$\displaystyle \Rightarrow \frac{m_1}{m_2}=\frac{1}{2}$
$\displaystyle \Rightarrow m_1:m_2=1:2$
$\displaystyle \therefore P\text{ is a point of trisection of }AB.$
$\displaystyle \text{Now, }y=\frac{m_1y_2+m_2y_1}{m_1+m_2}$
$\displaystyle \Rightarrow m-5=\frac{1(4)+2(-2)}{1+2}$
$\displaystyle \Rightarrow m-5=0$
$\displaystyle \Rightarrow m=5$
$\\$

$\displaystyle \textbf{Question 20: }\text{Find the coordinates of the midpoint of the line segment joining}$
$\displaystyle \text{the points }P(4,-6)\text{ and }Q(-2,4).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
$\displaystyle =\left(\frac{4+(-2)}{2},\frac{-6+4}{2}\right)$
$\displaystyle =\left(\frac{2}{2},\frac{-2}{2}\right)=(1,-1)$
$\\$

$\displaystyle \textbf{Question 21: }\text{The midpoint of line segment }AB\text{ (shown in the diagram) is }(-3,5).$
$\displaystyle \text{Find the coordinates of }A\text{ and }B.$
$\displaystyle \text{Answer:}$ x21
$\displaystyle \text{Since point }A\text{ lies on the }x\text{-axis, let }A=(x,0).$
$\displaystyle \text{Since point }B\text{ lies on the }y\text{-axis, let }B=(0,y).$
$\displaystyle \text{Midpoint of }AB=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
$\displaystyle =\left(\frac{x+0}{2},\frac{0+y}{2}\right)=(-3,5)$
$\displaystyle \Rightarrow \frac{x}{2}=-3\Rightarrow x=-6$
$\displaystyle \Rightarrow \frac{y}{2}=5\Rightarrow y=10$
$\displaystyle \therefore A=(-6,0)\text{ and }B=(0,10).$
$\\$

$\displaystyle \textbf{Question 22: }\text{A}(14,-2),\ B(6,-2)\text{ and }D(8,2)\text{ are three vertices of a}$
$\displaystyle \text{parallelogram }ABCD.\text{ Find the coordinates of the fourth vertex }C.$
$\displaystyle \text{Answer:}$ x22
$\displaystyle \text{Let }C=(x,y).$
$\displaystyle \text{Since the diagonals of a parallelogram bisect each other,}$
$\displaystyle \text{Midpoint of }AC=\text{Midpoint of }BD$
$\displaystyle \Rightarrow \left(\frac{14+x}{2},\frac{-2+y}{2}\right)=\left(\frac{8+6}{2},\frac{2+(-2)}{2}\right)=(7,0)$
$\displaystyle \Rightarrow \frac{14+x}{2}=7\Rightarrow x=0$
$\displaystyle \Rightarrow \frac{-2+y}{2}=0\Rightarrow y=2$
$\displaystyle \therefore C=(0,2).$
$\\$

$\displaystyle \textbf{Question 23: }\text{In }\triangle ABC,\ P(-2,5)\text{ is midpoint of }AB,\ Q(2,4)\text{ is midpoint of }BC$
$\displaystyle \text{and }R(-1,2)\text{ is midpoint of }AC.\text{ Calculate the coordinates of vertices }A,B\text{ and }C.$
$\displaystyle \text{Answer:}$ x23
$\displaystyle \text{Let }A=(x_1,y_1),\ B=(x_2,y_2)\text{ and }C=(x_3,y_3).$
$\displaystyle \text{Since }P(-2,5)\text{ is midpoint of }AB,$
$\displaystyle \frac{x_1+x_2}{2}=-2\text{ and }\frac{y_1+y_2}{2}=5$
$\displaystyle \Rightarrow x_1+x_2=-4\qquad\text{... (i)}$
$\displaystyle \Rightarrow y_1+y_2=10\qquad\text{... (ii)}$
$\displaystyle \text{Since }Q(2,4)\text{ is midpoint of }BC,$
$\displaystyle \frac{x_2+x_3}{2}=2\text{ and }\frac{y_2+y_3}{2}=4$
$\displaystyle \Rightarrow x_2+x_3=4\qquad\text{... (iii)}$
$\displaystyle \Rightarrow y_2+y_3=8\qquad\text{... (iv)}$
$\displaystyle \text{Since }R(-1,2)\text{ is midpoint of }AC,$
$\displaystyle \frac{x_1+x_3}{2}=-1\text{ and }\frac{y_1+y_3}{2}=2$
$\displaystyle \Rightarrow x_1+x_3=-2\qquad\text{... (v)}$
$\displaystyle \Rightarrow y_1+y_3=4\qquad\text{... (vi)}$
$\displaystyle \text{Adding (i), (iii) and (v), we get}$
$\displaystyle x_1+x_2+x_2+x_3+x_1+x_3=-4+4-2$
$\displaystyle \Rightarrow 2(x_1+x_2+x_3)=-2$
$\displaystyle \Rightarrow x_1+x_2+x_3=-1\qquad\text{... (vii)}$
$\displaystyle \text{Subtracting (i) from (vii), }x_3=-1-(-4)=3$
$\displaystyle \text{Subtracting (iii) from (vii), }x_1=-1-4=-5$
$\displaystyle \text{Subtracting (v) from (vii), }x_2=-1-(-2)=1$
$\displaystyle \text{Adding (ii), (iv) and (vi), we get}$
$\displaystyle y_1+y_2+y_2+y_3+y_1+y_3=10+8+4$
$\displaystyle \Rightarrow 2(y_1+y_2+y_3)=22$
$\displaystyle \Rightarrow y_1+y_2+y_3=11\qquad\text{... (viii)}$
$\displaystyle \text{Subtracting (ii) from (viii), }y_3=11-10=1$
$\displaystyle \text{Subtracting (iv) from (viii), }y_1=11-8=3$
$\displaystyle \text{Subtracting (vi) from (viii), }y_2=11-4=7$
$\displaystyle \therefore A=(-5,3),\ B=(1,7)\text{ and }C=(3,1).$
$\\$

$\displaystyle \textbf{Question 24: }\text{The midpoint of the line segment joining }(3m,6)\text{ and }(-4,3n)\text{ is }(1,2m-1).$
$\displaystyle \text{Find the values of }m\text{ and }n.\hfill \text{[ICSE 2006]}$
$\displaystyle \text{Answer:}$ x24
$\displaystyle \text{Using the midpoint formula,}$
$\displaystyle \left(\frac{3m+(-4)}{2},\frac{6+3n}{2}\right)=(1,2m-1)$
$\displaystyle \Rightarrow \frac{3m-4}{2}=1$
$\displaystyle \Rightarrow 3m-4=2$
$\displaystyle \Rightarrow 3m=6$
$\displaystyle \Rightarrow m=2$
$\displaystyle \text{Also, }\frac{6+3n}{2}=2m-1$
$\displaystyle \Rightarrow 6+3n=4m-2$
$\displaystyle \Rightarrow 6+3n=4(2)-2$
$\displaystyle \Rightarrow 6+3n=6$
$\displaystyle \Rightarrow 3n=0$
$\displaystyle \Rightarrow n=0$
$\displaystyle \therefore m=2\text{ and }n=0.$
$\\$

$\displaystyle \textbf{Question 25: }\text{Find the coordinates of the point of intersection of the medians of }\triangle ABC,$
$\displaystyle \text{given }A=(-2,3),\ B=(6,7)\text{ and }C=(4,1).$ x25
$\displaystyle \text{Answer:}$
$\displaystyle \text{The point of intersection of medians of a triangle is called its centroid.}$
$\displaystyle \text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$
$\displaystyle =\left(\frac{-2+6+4}{3},\frac{3+7+1}{3}\right)$
$\displaystyle =\left(\frac{8}{3},\frac{11}{3}\right)$
$\displaystyle \therefore \text{The point of intersection of the medians is }\left(\frac{8}{3},\frac{11}{3}\right).$
$\\$

$\displaystyle \textbf{Question 26: }ABC\text{ is a triangle and }G(4,3)\text{ is the centroid of the triangle.}$
$\displaystyle \text{If }A=(1,3),\ B=(4,b)\text{ and }C=(a,1),\text{ find }a\text{ and }b.$
$\displaystyle \text{Find the length of side }BC.\hfill \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$ x26
$\displaystyle \text{Since }G(4,3)\text{ is the centroid of }\triangle ABC,$
$\displaystyle \left(\frac{1+4+a}{3},\frac{3+b+1}{3}\right)=(4,3)$
$\displaystyle \Rightarrow \frac{5+a}{3}=4$
$\displaystyle \Rightarrow 5+a=12$
$\displaystyle \Rightarrow a=7$
$\displaystyle \text{Also, }\frac{4+b}{3}=3$
$\displaystyle \Rightarrow 4+b=9$
$\displaystyle \Rightarrow b=5$
$\displaystyle \therefore B=(4,5)\text{ and }C=(7,1)$
$\displaystyle BC=\sqrt{(7-4)^2+(1-5)^2}$
$\displaystyle =\sqrt{9+16}=\sqrt{25}=5\text{ units}$
$\\$