Coordinate Geometry. Exercise 13.1

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1 3 Exercise 3. Question. Find the distance between the following pairs of points (i) ( 3) ( ) (ii) ( 5 7) ( 3) (iii) (a b) ( a b) Solution (i) Let A( 3 ) and B( ) be the given points. Here x y 3and x y AB ( x x ) + ( y y ) ( ) + ( 3) () + ( ) + 8 (ii) Let A( 57and ) B( 3 ) be the given points. Here x 5 y 7and x y 3 AB ( x x ) + ( y y ) ( + 5) + ( 3 7) () + ( ) (iii) Let Aa ( b)and B( a b) be the given points. Here x a y b and x a y b AB ( x x ) + ( y y ) ( a a) + ( b b) ( a) + ( b) a + b a + b Question. Find the distance between the points (0 0) and (36 5). Can you now find the distance between the two towns A and B. Solution Let M ( 0 0 ) and N ( 36 5) be the given points. Here x 0 y 0 and x 36 y 5 MN ( x x ) + ( y y ) ( 36 0) + ( 5 0) Since the position of towns A and B are given (0 0) and (36 5) respectively and so the distance between them is 39 km.

2 Question 3. Determine if the points ( 5) ( 3) and ( ) are collinear. Solution Let A( 5 ) B( 3 ) and C( ) be the given points. Then we have AB ( ) + ( 3 5) [ QDistance ( x x ) + ( y y ) + ( z z ) ] + ( ) + 5 BC ( ) + ( 3) ( ) + ( ) AC ( ) + ( 5) ( 3) + ( 6) Now AB + AC AB + AC BC Similarly AB BC + AC and AC AB + BC Hence AB and C are non-collinear. Question. Check whether ( 5 )( 6) and ( 7 ) are the vertices of an isosceles triangle. Solution Let P( 5 ) Q( 6) and R( 7 ) are the given points. Then PQ ( 6 5) + ( + ) ( x x ) + ( y y ) QR ( 7 6) + ( ) Since PQ QR PQR is an isosceles triangle.

3 Question 5. In a classroom friends are seated at the points A B C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli Don t you think ABCD is a square? Chameli disagrees. Using distance formula find which of them is correct. Rows Solution From the given figure the coordinates of points ABC and D are ( 3 ) ( 6 7) ( 9 ) and (6 ). AB ( 6 3) + ( 7 ) A B D Columns ] [QDistance ( x x ) + ( y y ) BC ( 9 6) + ( 7) 3 + ( 3) 8 C CD ( 6 9) + ( ) ( 3) + ( 3) and DA ( 3 6) + ( ) ( 3 ) AB BC CD DA Now AC ( 9 3) + ( )

4 + and BD ( 6 6) + ( 7) AC BD 6 Since the four sides and diagonals are equal. Hence ABCD is a square. So Champa is correct. Question 6. Name the type of quadrilateral formed if any by the following points and give reasons for your answer (i) ( ) ( 0 ) ( ) ( 30 ) (ii) ( 35 ) ( 3 ) ( 03 ) ( ) (iii) ( 5 )( 76 )( 3 )( ) Solution (i) Let A( ) B( 0 ) C( ) and D( 30be ) the given points. Then AB ( + ) + ( 0+ ) ] [Qdistance ( x x ) + ( y y ) BC ( ) + ( 0) ( ) CD ( 3 + ) + ( 0 ) ( ) + ( ) + 8 DA ( + 3) + ( 0) ( ) + ( ) + 8 AB BC CD DA Now AC ( + ) + ( + ) 0+ and BD ( 3 ) + ( 0 0) ( ) + 0 AC BD Since the four sides AB BC CD and DA are equal and also diagonals AC and BD are equal. The quadrilateral ABCD is a square.

5 (ii) Let A( 35 ) B( 3 ) C( 03and ) D( ) be the given points. From the figure the points AC and B are collinear. So no quadrilateral is formed by given points. y A( 3 5) x' C(0 3) (iii) Let A( 5 ) B( 76 ) C( 3 ) and D( ) be the given points. Then AB ( 7 ) + ( 6 5) ] [QDistance ( x x ) + ( y y ) BC ( 7) + ( 3 6) ( 3) + ( 3) CD ( ) + ( 3) ( 3) + ( ) DA ( ) + ( 5 ) 3 + ( 3) B(3) D( ) AC ( ) + ( 3 5) 0+ ( ) and BD ( 7) + ( 6) ( 6) + ( ) Since AB CD BC DA and AC BD The quadrilateral ABCD is a parallelogram. 5 3 y' Question 7. Find the point on the X-axis which is equidistant from ( 5 ) and ( 9). Solution Since the point on the X-axis. It s ordinate 0 So Ax ( 0 ) is any point on the X-axis. Since Ax ( 0 ) is equidistant from B( 5) and C( 9. ) AB AC AB AC ( x ) + ( 0+ 5) ( x + ) + ( 0 9) [QDistance ( x x ) + ( y y ) x ]

6 x x x + x x x 8 5 8x x 7 8 So the point equidistant from given points on the x-axis is ( 70. ) Question 8. Find the values of y for which the distance between the points P( 3) and Q( 0 y) is 0 units. Solution According to question Squaring both sides we get PQ 0 ( 0 ) + ( y + 3) 0 () 8 + ( y + 3) y y 0 y y + 6y y y + 6y 7 0 y + 9y 3y 7 0 y ( y + 9) 3 ( y + 9) 0 ( y + 9) ( y 3) 0 y 9 or y 3 y 9 or 3 Question 9. If Q( 0is ) equidistant from P( 5 3) values of x. Also find the distance QR and PR. ] [QDistance ( x x ) + ( y y ) and Rx ( 6 ) find the Solution Since the point Q( 0 ) is equidistant from P ( 5 3) and Rx ( 6 ). QP QR QP QR ( 5 0) + ( 3 ) ( x 0) + ( 6 ) 5 + x x + 5 x 6 x ± ] [QDistance ( x x ) + ( y y )

7 ± Thus R is ( 6) or ( 6. ) Now QR Distance between Q( 0 ) and R ( 6 ) ( 0) + ( 6 ) Also QR Distance between Q( 0 ) and R( 6 ) ( 0) + ( 6 ) ( ) and PR Distance between P ( 5 3) and R( 6 ) ( 5) + ( 6+ 3) ( ) Also PR Distance between P ( 5 3) and R( 6 ) ( 5) + ( 6+ 3) ( 9 ) Question 0. Find a relation between x and y such that the point ( x y) is equidistant from the points ( 36and ) ( 3. ) Solution C( 3. ) Let the point Ax ( y)be equidistant from the points B( 3 6 ) and AB AC AB AC ( x 3) + ( y 6) ( x + 3) + ( y ) ] [QDistance ( x x ) + ( y y ) x 6x + 9+ y y + 36 x + 6x + 9+ y 8y + 6 6x 6x y + 8y x y ( x + y 5) 0 3x + y 5 0 ( Q 0)

8 3 Exercise 3. Question. Find the coordinates of the point which divides the join of ( 7 ) and ( 3) in the ratio : 3. Solution Let the coordinates of a point are ( x y ). We have x y 7; x y 3 and m m 3 By using section formula mx m x x + () + 3( ) m + m my m y and y + ( 3) + 37 () m + m Hence coordinates of the point are( 3 ). Question. Find the coordinates of the points of trisection of the line segment joining ( ) and ( 3 ). Solution Let A( ) and B ( 3) be the line segments and points of trisection of the line segment be P and Q. Then AP PQ BQ k (Say) PB PQ + QB k and AQ AP + PQ k k k k ( ) ( 3) A P Q B AP : PB k : k : and AQ : QB k : k : Since P divides AB internally in the ratio :. So the coordinate of P are mx m x By using + my m y and + m + m m + m ( ) + ( 3) + ( ) and Q divides AB internally in the ratio :.

9 So the coordinates of Q are ( ) + ( 3) + ( ) + + mx By using m + m x my m y and + + m m + m So the two points of trisection are 5 and Question 3. To conduct Sports Day activities in your rectangular shaped school ground ABCD lines have been drawn with chalk powder at a distance of m each. 00 flower pots have been placed at a distance of m from each other along AD as shown in figure. Niharika runs th the distance AD on the nd line and posts a green flag. Preet runs th the 5 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? D C A B

10 Solution From the above figure the position of green flag posted by Niharika is M 00 ie..m ( 5) and red flag posted by Preet is N 8 00 ie.. 5 N ( 8 0 ). Now MN ( 8 ) + ( 0 5) () 6 + ( 5) Hence the distance between flags 6 m Let P be the position of the blue flag posted by Rashmi in the halfway of line segment MN. M ( 5) P N (8 0) + + P is given by x x y y ( 5. 5) Hence the blue flag is on the fifth line at a distance of.5 m above it. Question. Find the ratio in which the line segment joining the points ( 3 0) and (6 8) is divided by ( 6). Solution Let the point A( 6divide ) the line joining B ( 3 0) and C( 6 8) in the ratio k :. Then the coordinates of A are 6 k 3 8k + 0. k + k + QInternally ratio mx + m x my + m y m + m m + m But the coordinates of A are given by ( 6. ) B( 3 0) A( 6) k : C(6 8) On comparing coordinates we get 6k 3 and 8k k + k + 6k 3 k and 8k + 0 6k + 6 6k + k + 3and 8k 6k 6 0 7k and k k 7 So the point A divides BC in the ratio : 7.

11 Question 5. Find the ratio in which the line segment joining A( 5) and B( 5 ) is divided by the x-axis. Also find the coordinates of the point of division. Solution Let the required ratio be k :. So the coordinates of the point M of division A( 5) and B( 5are ) kx + x k + ky + y k + 5k 5 ie... k + k + k + But according to question line segment joining A( 5) and B( 5 ) is divided by the x-axis. So y-coordinates must be zero. 5k 5 0 k + 5k 5 0 5k 5 k So the required ratio is : and the point of division M is () + 5 () ie.. 0 ie Question 6. If ( ) ( y) ( x 6) and ( 35) are the vertices of a parallelogram taken in order find x and y. Solution Let A( ) B( y) Cx ( 6and ) D( 3 5 ) are the vertices of a parallelogram. D(3 5) Cx ( 6) H A( ) B( y) Since ABCD is a parallelogram. AC and BD will bisect each other. Hence mid-point of AC and mid-point of BD are same point. Mid-point of AC is + x Mid-point of BD is y QMid -point x + x y + y + x + 3 and y + x 7and y x 6 and y 3

12 Question 7. Find the coordinates of a point A where AB is the diameter of a circle whose centre is ( 3) and B is ( ). Solution Suppose AB be a diameter of the circle having its centre at C( 3) and coordinates of end point B are ( ). Ax ( y) C( 3) B( ) Let the coordinates of A be ( x y ). Since AB is diameter. C is the mid-point of AB. The coordinates of C are x + y +. QCoordinate of mid -point x + x y + y But it is given that the coordinates of C are ( 3). x + x + x 3 y + and 3 y + 6 y 0 So the required coordinates of A are (3 0). Question 8. If A and B are ( ) and ( ) respectively find the coordinates of P such that AP 3 AB and P lies on the line segment AB. 7 Solution According to the question A( ) Px ( y) 3: B( ) AP 3 AB 7 AB AP 7 3

13 AP + PB AP 3 PB + AP + PB AP 3 AP PB AP : PB 3: Suppose Pxy ( )be the point which divides the line segment joining the points A( ) and B( ) in the ratio 3 :. 3 + ( ) 6 8 mx + m x x Qx m + m 3 ( ) + ( ) my and y Qy 3 + m Hence the required coordinates of the point P are m y + m Question 9. Find the coordinates of the points which divide the line segment joining A( and ) B( 8into ) four equal parts. Solution Let P Q and R be the points on line segment AB such that AP PQ QR RB Let AP PQ QR RB k k k k k ( ) ( 8) A P Q R B Now AP k PB 3k 3 Therefore P divides AB internally in the ratio : 3. Q Internally ratio mx + m x my + my m + m m + m + 3 ( ) P

14 Again AR RB 3k 3 k Therefore R divides AB internally in the ratio 3 :. 3 + ( ) R AQ k Also QB k Q is the mid-point of AB. Q ie.. Q 0 0 ( 05) QMid -point x + x y + y So required points are 7 3 and ( 0 5 ). Question 0. Find the area of a rhombus if its vertices are (3 0) ( 5) ( ) and ( ) taken in order. [Hint Area of a rhombus (Product of its diagonals)] Solution Let A( 3 0) B( 5 ) C( ) and D( ) be the vertices of the rhombus ABCD. Diagonal AC ( 3) + ( 0) ( ) + ] [QDistance ( x x ) + ( y y ) Diagonal BD ( ) + ( 5) ( 6) + ( 6) Area of the rhombus ABCD AC BD 6 6 sq units

15 3 Exercise 3.3 Question. Find the area of the triangle whose vertices are (i) ( 3 ) ( 0 ) ( ) (ii) ( 5 ) ( 3 5) ( 5 ) Solution (i) Let A ( x y) ( 3) B ( x y) ( 0 ) C ( x3 y3) ( ) Area of ABC [ x ( y y3) + x ( y3 y) + x3 ( y y)] ) ) 3) 3 0)] ( ) sq units (ii) Let A ( x y) ( 5 ) B ( x y) ( 3 5) and C ( x3 y3) ( 5 ) Area of ABC [ x ( y y3) + x ( y3 y) + x3 ( y y)] [ 5 ( 5 ) + 3 ( + ) + 5 ( + 5)] [ 5( 7) + 33 () + 5()] sq units Question. In each of the following find the value of k for which the points are collinear (i) (7 ) (5 ) (3 k) (ii) (8 ) (k ) ( 5) Solution (i) Let A ( x y ) ( 7 ) B ( x y ) () 5and C ( x y ) ( k) Since the points are collinear. Area of ABC 0 [ x ( y y3) + x ( y3 y) + x3 ( y y)] ( k) + 5( k + ) + 3( ) 0 (Multiply by ) 7 7k + 5k k k 8 k (ii) Let A ( x y) ( 8 ) B ( x y) ( k ) and C ( x3 y3) ( 5) Since the points are collinear. Area of ABC 0 [ x ( y y3) + x ( y3 y) + x3 ( y y)] 0 8 ( + 5) + k ( 5 ) + ( + ) 0 (Multiply by )

16 8 () + k ( 6) + 5 ( ) k k 8 8 k 3 6 Question 3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0 ) ( ) and (0 3). Find the ratio of this area to the area of the given triangle. + Solution Mid-point of M QMid -point x + x y + y ( 0 ) + + Mid-point of N 0 ( 0) + + and Mid-point of P 0 3 ( ) Let N ( 0) ( x y) and P ( ) ( x y) and M ( 0) ( x3 y3) Area of NPM [ x( y y3) + x ( y3 y) + x3 ( y y)] + + [ ( ) ( 0) 0( 0 )] [ () ] A (0 ) (0 ) M N (0) C (03) P () B () Let A ( x y) ( 0 ) B ( x y) ( ) and C ( x3 y3) ( 03) Area of ABC [ x ( y y3) + x ( y3 y) + x3 ( y y)] [ 0 ( 3) ( 3 ) 0( )] + + ( 0 8 0) sq units Area of NPM Required ratio Area of ABC

17 Question. Find the area of the quadrilateral whose vertices taken in order are ( ) ( 3 5 ) ( 3 ) and ( 3. ) Solution Let A ( ) B ( 3 5 ) C ( 3 ) and D ( 3) be the vertices of the quadrilateral ABCD. Area of quadrilateral ABCD Area of ACD + Area of ABC [ ( 3) + 3 ( 3 + ) + ( + )] + [ ( 5 + ) 3 ( + ) + 3 ( + 5)] QArea of triangle [ x ( y y3) + x ( y3 y) + x3 ( y y)] [ ( 5) + 3 () 5 + ()] 0 + [ ( 3) 30 ( ) + 33 ( )] ( 0 5 0) ( 0 9) + 35 ( ) 56 8 sq units Question 5. You have studied in Class IX (Chapter 9 Example 3) that a median of a triangle divides it into two triangles of equal areas. Verify this result for ABC whose vertices are A ( 6) B ( 3 ) and C ( 5. ) Solution According to the question AD is the median of ABC therefore D is the mid-point of BC. A( 6) B(3 ) D( 0) C(5 ) Coordinates of D are i.e. ( 0 ) QMid -point x + x y + y Area of ADC [ ( 0 ) ( 6) 5 ( 6 0)] QArea [ x + + ( y y3) x ( y3 y) x3 ( y y)]

18 Here A ( ) ( x y ) 0 D ( 0) ( x y) and C ( 5) ( x3 y3) ( ) ( 6) 3 3 sq units (QArea of triangle is positive) and Area of ABD [ ( 0) 3 ( 0 6) ( 6 )] [Let A ( 6) ( x y) B ( 3 ) ( x y) and D ( 0) ( x3 y3) ] ( ) ( 6) 3 3 sq units (QArea of triangle is positive) Q Area of ADC Area of ABD Hence the median of the triangle divides it into two triangle of equal areas.

19 3 Exercise 3. (Optional) Question. Determine the ratio in which the line x + y 0 divides the line segment joining the points A ( ) and B ( 37. ) Solution Let the line x + y 0 divides the line segment joining the points A ( ) and B ( 3 7 ) in the ratio k : at the point P. The coordinates of P are 3 k + 7k. k + k + QInternally ratio mx + m x my m + m m But P lies on x + y 0 3 k + 7k + 0 k + k + 6k + + 7k k 0 9k 0 9k k 9 Point P divides the line in the ratio : 9. + m y + m Question. Find a relation between x and y if the points ( x y) ( ) and ( 70are ) collinear. Solution Since the points A ( x y) B ( ) and C ( 7 0 ) are collinear. Area of ABC 0 [ x ( y y3) + x ( y3 y) + x3 ( y y)] 0 [ x ( 0) + 0 ( y) + 7( y )] 0 x y + 7y 0 (Multiply by ) x + 6y 0 x + 3y 7 0 (Divide by )

20 Question 3. Find the centre of a circle passing through the points ( 6 6)( 3 7) and ( 33. ) Solution Let C ( x y) be the centre of the circle passing through the points P ( 6 6) Q ( 3 7) and R ( 3 3). Then PC QC CR (Radius of circle) Now PC QC PC QC ( x 6) + ( y + 6) ( x 3) + ( y + 7) ] [QDistance ( x x ) + ( y y ) x x y + y + 36 x 6x y + y + 9 x + 6x + y y x y + 0 and QC CR QC CR 3x + y 7 0 (Divide by ) (i) ( x 3) + ( y + 7) ( x 3) + ( y 3) P(6 6) R(33) Cxy ( ) Q(3 7) x 6x y + y + 9 x 6x + 9+ y 6y + 9 6x + 6x + y + 6y y y 0 (ii) Putting y in Eq. (i) we get 3x 7 0 3x 9 x 3 Hence centre is ( 3 ). Question. The two opposite vertices of a square are ( and ) ( 3. ) Find the coordinates of the other two vertices. Solution Let PQRM be a square and let P ( ) and R ( 3 ) be the vertices. Let the coordinates of Q be ( x y ). M R (3) P ( ) Q( x y) Q PQ MR PQ MR + + +

21 ( x + ) + ( y ) ( x 3) + ( y ) [QDistance ( x x ) + ( y y ) ] x + + x + y + y x + 9 6x + y + y x + 6x + 9 8x 8 x (i) In PQR we have PQ + QR PR ( x + ) + ( y ) + ( x 3) + ( y ) ( 3 + ) + ( ) x + + x + y + y + x + 9 6x + y + y + 0 x + y + x y 6x y x + y x 8y + 0 x + y x y + 0 (ii) Putting x from Eq. (i) in Eq. (ii) we get + y y + 0 y y 0 y ( y ) 0 y 0or Hence the required vertices of square are ( 0 ) and ( ). Question 5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance ofm from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot. B C P R Q A (i) Taking A as origin find the coordinates of the vertices of the triangle. (ii) What will be the coordinates of the vertices of PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe? Solution (i) When A is taken as origin AD and AB as coordinate axes i.e. X-axis and Y-axis respectively. Coordinates of P Q and R are respectively ( 6) (3 ) and (6 5). (ii) When C is taken as origin and CB as X-axis and CD asy-axis. D

22 Coordinates of P Q and R are respectively ( ) (3 6) and (0 3) Areas of triangles according to both conditions Condition (i) When A is taken as origin and AD and AB as coordinates axes. Area of PQR + + [ ( 5) 3 ( 5 6) 6( 6 )] QArea of triangle [ x ( y y 3 ) + x ( y 3 y ) + x 3 ( y y ) + + [ ( 3) 3 ( ) 6 ] 9 ( 3 + ) sq units Condition (ii) When C is taken as origin and CB and CD as axes Area of PQR [ ( ) ( ) ( )] [ ( )] [ ] 9 sq units Hence we observe that the areas of both triangles are same. Question 6. The vertices of a ABC are A ( 6 ) B ( 5 ) and C ( 7 ).A line is drawn to intersect sides AB and AC at D and E respectively such that AD AE. Calculate the area of the ADE and compare it with AB AC the area of ABC. AD Solution Given AB AD DB 3 A( 6) 3 3 D E 9 5 Hence D divides AB internally in the ratio : 3. The coordinates of D are ie or Again B ( 5) C (7 ) 3 QInternally ratio mx + m x my m + m m AE AC AE EC 3 + m y + m

23 Hence E divides AC internally in the ratio : 3. The coordinates of E are i.e. or y 9 5. QInternally ratio mx + m x my + m y m + m m + m Now area of ABC + + [ ( 5 ) ( 6) 7 ( 6 5)] QArea of triangle [ x ( y y 3 ) + x ( y 3 y ) + x 3 ( y y )] + [ 7] 5 sq units and area of ADE ( ) sq units 6 3 Area of ADE Area of ABC 5 / 3 5 : 6 / Question 7. Let A ( ) B ( 6 5) and C ( be ) the vertices of ABC. (i) The median from A meets BC at D. Find the coordinates of the point D. (ii) Find the coordinates of the point P on AD such that AP : PD :. (iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE : and CR : RF :. (iv) What do you observe? Note The point which is common to all the three medians is called the centroid and this point divides each median in the ratio :. (v) If A ( x y) B ( x y) and C ( x3 y3) are the vertices of ABC find the coordinates of the centroid of the triangle. Solution Let A ( ) B ( 65 ) and C ( ) be the vertices of ABC. A( ) 3 F R P Q E B (6 5) D C ( )

24 (i) Since AD is the median of ABC. D is the mid-point of BC. The coordinates of D are i.e. 7 9 QMid -point x + x y + y (ii) Since P divides AD is the ratio : so its coordinates are or i.e QInternally ratio mx + m x my m + m m + m y + m (iii) Since BE is the median of ABC so E is the mid-point of AC and its coordinates are E + + i.e. E 5 3. Since Q divides BE in the ratio : so its coordinates are Q or Q or Q Since CF is the median of ABC so F is the mid-point of AB. Therefore its coordinates are F i.e. F 5 7. Since R divides in the ratio : so its coordinates are 7 R or R or R (iv) We find that the points P Q and R coincide at the point. This 3 3 point is known as the centroid of the triangle. (v) Let A ( x y) B ( x y) and C( x3 y3) be the vertices of ABC whose medians are AD BE and CF respectively then DE and F are respectively the mid-points of BC CA and AB. Coordinates of D are x + x 3 y + y 3. A F E B D G C

25 Coordinates of a point G dividing AD intheratio:are ( x + x3) ( y ( ) + y ( ) 3) x + y x x x3 y y y The coordinates of E are x + x y + y 3. The coordinates of a point dividing BE intheratio:are ( x + x3) ( y ( ) + y ( ) 3) x + y x + x + x3 y + y + y Question 8. ABCD is a rectangle formed by the points A ( ) B ( ) C ( 5 ) and D ( 5 ). P Q R and S are the mid-points of AB BC CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer. Solution Given vertices of a rectangle are A ( ) B ( ) C ( 5 ) and D ( 5 ). Mid-point of AB is P + or P 3. QMid -point x + x y + y Mid-point of BC is Q 5 + or Q ( ). Mid-point of CD is R or R 5 3. Mid-point of AD is S + 5 or S ( ). Now PQ ( ) R 5 3 D(5 ) C (5) S( ) Q () A( ) P 3 B ( )

26 QR + 3 ( 5 ) RS ( 5) ( 3 ) SP + 3 ( ) ( 3 ) Q PQ QR RS SP Now PR ( 5 ) and SQ ( ) + ( + ) PR SQ Since all the sides are equal but diagonals are not equal. PQRS is a rhombus. 5 6

COORDINATE GEOMETRY BASIC CONCEPTS AND FORMULAE. To find the length of a line segment joining two points A(x 1, y 1 ) and B(x 2, y 2 ), use

COORDINATE GEOMETRY BASIC CONCEPTS AND FORMULAE I. Length of a Line Segment: The distance between two points A ( x1, 1 ) B ( x, ) is given b A B = ( x x1) ( 1) To find the length of a line segment joining