Solution
1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. Given ;$(x_1,y_1)$ = (-1,7) $(x_2,y_2)$ = (4,-3) $m_1$ = 2; $m_2$=3 Therefore, coordinates of the points is given as, $(x,y)=({m_1 x_2 + m_2 x_1}/{m_1 + m_2},{m_1 y_2 + m_2 y_1}/{m_1 + m_2})$ $x={m_1 x_2 + m_2 x_1}/{m_1 + m_2}$ $x={2×4 + 3× (-1)}/{2 + 3}$ => ${8-3}/5$= $1$ $y={m_1 y_2 + m_2 y_1}/{m_1 + m_2}$ $y={2 ×(-3) + 3 × 7}/{2 + 3}$ = ${-6+21}/5$ = $3$ Ans: Coordinates of the points are (1,3)2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). Trisection means dividing a line segment in three equal parts or dividing a line segment in the ratio 1:2 and 2:1 Given $(x_1,y_1)= (4,-1)$ $(x_2,y_2) = (-2,-3)$ Case 1: Let the ratio be $(m_1,m_2)=(1,2)$ Therefore, coordinates of the points is given as, $(x,y)=({m_1 x_2 + m_2 x_1}/{m_1 + m_2},{m_1 y_2 + m_2 y_1}/{m_1 + m_2})$ $(x,y)=({1×(-2) + 2× 4}/{1 + 2},{1× (-3) + 2× (-1)}/{1 + 2})$ $(x,y)=({-2 + 8}/{3},{-3 -2}/{3})$ $(x,y)=(2,{-5}/{3})$ Case 2:Let the ratio be $(m_1,m_2)=(2,1)$ $(x,y)=({m_1 x_2 + m_2 x_1}/{m_1 + m_2},{m_1 y_2 + m_2 y_1}/{m_1 + m_2})$ $(x,y)=({2×(-2) + 1× 4}/{2+1},{2× (-3) + 1× (-1)}/{2+1})$ $(x,y)=({-4 +4}/{3},{-6 -1}/{3})$ $(x,y)=(0,{-7}/{3})$3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs $1/4$th the distance AD on the 2nd line and posts a green flag. Preet runs $1/5$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? 4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6). Let $m_1,m_2$ be the ratio which divide the line segment joining the point (– 3, 10) and (6, – 8) and (-1,6) be the coordinates. Therefore, coordinates of the points is given as, $(x,y)=({m_1 x_2 + m_2 x_1}/{m_1 + m_2},{m_1 y_2 + m_2 y_1}/{m_1 + m_2})$ $x$ = ${m_1 x_2 + m_2 x_1}/{m_1 + m_2}$ $-1$ = ${m_1 6 + m_2 (-3)}/{m_1 + m_2}$ $-m_1-m_2=6m_1-3m_2$ $6m_1+m_1=3m_2-m_2$ $7m_1=2m_2$ $y$= ${m_1 y_2 + m_2 y_1}/{m_1 + m_2}$ $6$= ${m_1 (-8) + m_2 (10)}/{m_1 + m_2}$ $6m_1+6m_2= -8m_1 +10m_2$ $6m_1+8m_1= 10m_2-6m_2$ $14m_1=4m_2$ $7m_1=2m_2$ Ans: Hence the required ratio is (2:7)5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division. Let $(m_1,m_2)$ be the ratio . Since the line is divided by the x-axis, the coordinates of point is $(x,0)$ Hence its $y$ coordinates will be 0. Since the $y$ coordinates is given by the eqn ${m_1 y_2 + m_2 y_1}/{m_1 + m_2}$, So, $0$= ${m_1× 5 + m_2 ×(-5)}/{m_1 + m_2}$ $5m_1-5m_2=0$ $m_1/m_2=1$ Hence $x$ coordinates is =${m_1 x_2 + m_2 x_1}/{m_1 + m_2}$ =${1×(-4) + 1× 1}/{1 + 1}$ =${-4 + 1}/{2}$ =${-3}/2$ Ans:Hence the x-axis divides the line segment joining A(1, – 5) and B(– 4, 5) in the ratio 1:1 and the required coordinate is ($-3/2,0$).6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. Let M be the midpoint of AC and BD. By using midpoint formula, coordinate of point M = $({1+x}/2,4)$ Also coordinate of point M = $({3+4}/2,{5+y}/2)$ Therefore ${1+x}/2=7/2$ and ${5+y}/2=4$ ${1+x}=7$ => $x=6$ and $5+y=8$ => $y=3$ Ans: $x$= 6 and $y$ = 37. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4). Let the coordinates of A be ($x_1,y_1$). Coordinates of B is (1,4) and the coordinate of center O is (2,-3) AB is the diameter of the circle with center O. Hence O is the midpoint of the line segment AB and divides AB in the ratio 1:1. By using midpoint formula, coordinates of the points are given as $x= {x_1+x_2}/2$ and $y={y_1+y_2}/2$ $2= {x_1+1}/2$ $x_1=3$ $y={y_1+y_2}/2$ $-3={y_1+4}/2$ $y_1= -10$ Ans: Coordinates of A = (3,-10)8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = $3/7$ AB and P lies on the line segment AB. Given: AP = $3/7$ AB ${AP}/{AB}=3/7$ ${AP}/{AP+BP}=3/7$ $7{AP}={3AP+3BP}$ ${4AP}=3BP$ ${AP}/{BP}=3/4$ Hence, P divides AB in the ratio 3:4. Therefore the coordinates of P are, $x$ = ${m_1x_2+m_2x_1}/{m_1+m_2}$ $x$ = ${3×2+4×(-2)}/{3+4}$ $x$ = ${6-8/{7}$ $x$ = $-2/7$ and $y$ = ${m_1y_2+m_2y_1}/{m_1+m_2}$ $y$ = ${3×(-4)+4×(-2)}/{3+4}$ $y$ = ${-12-8}/{7}$ $y$ = $-20/{7}$ Ans: Coordinates of P is ($-2/7,-20/7$)9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts. Let the point which divides the line segment into 4 equal parts be P,Q,R and the ratio of each part be $AP : PQ : QR : RB = 1 : 1 : 1 : 1$ P divides the line segment AB in the ratio $1:3$ Therefore Coordinates of P are $x$ = ${m_1x_2+m_2x_1}/{m_1+m_2}$ $x$ = ${1×2+3×(-2)}/{1+3}$ $x$ = ${2-6}/{4}$ $x$ = $-1$ and $y$ = ${m_1y_2+m_2y_1}/{m_1+m_2}$ $y$ = ${1×8+3×2}/{1+3}$ $y$ = ${8+6}/{4}$ $y$ = ${14}/{4}$ $y$ = $7/2$ P = $(-1,7/2)$ Q divides the line segment in the ratio $2:2$ i.e $1:1$ Therefore coordinates of Q are $(x_1+x_2)/2,(y_1+y_2)/2$ = $(-2+2)/2,(2+8)/2$ Q = $(0,5)$ R divides the line segment in the ratio $3:1$ Coordinates of R are $x$ = ${m_1x_2+m_2x_1}/{m_1+m_2}$ $x$ = ${3×2+1×(-2)}/{3+1}$ $x$ = ${6-2}/{4}$ $x$ = $4/4$ $x$ = $1$ and $y$ = ${m_1y_2+m_2y_1}/{m_1+m_2}$ $y$ = ${3×8+1×2}/{3+1}$ $y$ = ${24+2}/{4}$ $y$ = ${26}/{4}$ $y$ = $13/2$ R = $(1,13/2)$ Hence the coordinates are $(-1,7/2)$,$(0,5)$ ,$(1,13/2)$10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. Length of the diagonal of the Rhombus AC = $√{(x_1-x_2)^2 +(y_1-y_2)^2}$ =$√{(3-(-1))^2 +(0-4)^2}$ =$√{(4)^2 +(-4)^2}$ =$√{32}$ =$4√{2}$ unit Similarly , Length of the diagonal of the Rhombus BD = $√{(x_1-x_2)^2 +(y_1-y_2)^2}$ = $√{(4-(-2))^2 + (5-(-1))^2}$ = $√{6^2+6^2}$ = $√{36+36}$ = $√{72}$ = $6√{2}$ unit If $p$ and $q$ are the length of the diagonal of Rhombus Area of Rhombus = ${pq}/2$ = ${4√2× 6√2}/2$ = ${48}/2$ = $24$ sq.units Ans:Area of Rhombus =$24$ sq.units