![Area of a Triangle
Area of ABCΔ = Area of trapezium ABQP + Area of
trapezium BQRC– Area of trapezium APRC.
We also know that ,
Area of trapezium =
Therefore,
Area of ABCΔ =
( )( )embetween thdistancesidesparallelofsum
2
1
( ) ( ) ( )PRCR+AP
2
1
CR+BQ
2
1
QPAP+BQ
2
1
−+ QR
( )( ) ( )( ) ( )( )133123321212
2
1
2
1
2
1
xxyyxxyyxxyy −+−−++−+=
( ) ( ) ( )[ ]133311312333223211211222
2
1
xyxyxyxyxyxyxyxyxyxyxyxy −+−−−+−+−+−=
( ) ( ) ( )[ ]123312231
2
1
yyxyyxyyx −+−+−=
Area of Δ ABC](https://image.slidesharecdn.com/coordinategeometry-180912070032/85/CLASS-X-MATHS-Coordinate-geometry-28-320.jpg)
CLASS X MATHS Coordinate geometry
- 2. 1. Cartesian Coordinate system and Quadrants 2. Distance formula 3. Area of a triangle 4. Section formula Objectives
- 3. • The use of algebra to study geometric properties i.e. operates on symbols is defined as the coordinate system. What is co-ordinate geometry ? IntroductIon
- 4. A system of geometry where the position of points on the plane is described using an ordered pair of numbers. The method of describing the location of points in this way was proposed by the French mathematician René Descartes . He proposed further that curves and lines could be described by equations using this technique, thus being the first to link algebra and geometry. In honor of his work, the coordinates of a point are often referred to as its Cartesian coordinates, and the coordinate plane as the Cartesian Coordinate Plane. Coordinate Geometry
- 5. SoME BASIc PoIntS To locate the position of a point on a plane,we require a pair of coordinate axes. The distance of a point from the y-axis is called its x-coordinate,OR abscissa. The distance of a point from the x-axis is called its y-coordinate,OR ordinate. The coordinates of a point on the x-axis are of the form (x, 0) and of a point on the y-axis are of the form (0, y).
- 6. RECAP Coordinate Plane XX’ Y Y’ O Origin 1 2 3 4 +ve direction -1-2-3-4 -ve direction -1 -2 -3 -vedirection 1 2 3 +vedirection X-axis : X’OX Y-axis : Y’OY
- 8. Coordinates XX’ Y Y’ O 1 2 3 4-1-2-3-4 -1 -2 -3 1 2 3 (2,1) (-3,-2) Ordinate Abcissa (?,?)
- 9. XX’ O Y Y’ III III IV (+,+)(-,+) (-,-) (+,-) Q : (1,0) lies in which Quadrant? Ist? IInd? A : None. Points which lie on the axes do not lie in any quadrant.
- 11. Let us now find the distance between any two points P(x1, y1) and Q(x1, y2) Draw PR and QS ⊥ x-axis. A perpendicular from the point P on QS is draw to meet it at the point T So, OR = x1 , OS = x2 , PR = PS = y1 , QS = y2 Then , PT = x2 – x1 , QT = y2 – y1 x Y P (x1 , y1) Q(x2 , y2) T R SO Distance Formula
- 12. Now, applying the Pythagoras theorem in ΔPTQ, we get Therefore 222 QTPTPQ += ( ) ( )2 12 2 12 yyxxPQ −+−= which is called the distance formula.
- 13. Example 1: Find the distance between P(1,-3) and Q(5,7). The exact distance between A(1, -3) and B(5, 7) is
- 14. Distance From Origin Distance of P(x, y) from the origin is ( ) ( )2 2 x 0 y 0= − + − 2 2 x y= +
- 15. Applications of Distance Formula To check which type of triangle is formed by given 3 coordinates. and To check which type of quadrilateral is formed by given 4 coordinates.
- 16. Applications of Distance Formula Parallelogram Prove opposite sides are equal or diagonals bisect each other
- 17. Applications of Distance Formula Rhombus Prove all 4 sides are equal
- 18. Applications of Distance Formula Rectangle Prove opposite sides are equal and diagonals are equal.
- 19. Applications of Distance Formula Square Prove all 4 sides are equal and diagonals are equal.
- 20. Collinearity of Three Points Use Distance Formula a b c Show that a+b = c
- 21. Section Formula 2 1 m m PB PA = Consider any two points A(x1 , y1) and B(x1 ,y2) and assume that P (x, y) divides AB internally in the ratio m1: m2 i.e. Draw AR, PS and BT ⊥ x-axis. Draw AQ and PC parallel to the x-axis. Then, by the AA similarity criterion, x Y A (x1 , y1) B(x2 , y2) P (x , y) R SO T m1 m2 Q C
- 22. Section Formula ΔPAQ ~ ΔBPC ---------------- (1) Now, AQ = RS = OS – OR = x– x1 PC = ST = OT – OS = x2– x PQ = PS – QS = PS – AR = y– y1 BC = BT– CT = BT – PS = y2– y Substituting these values in (1), we get BC PQ PC AQ BP PA == ( ) ( ) ( ) ( )yy yy xx xx m m − − = − − = 2 1 2 1 2 1 ( ) ( ) ( ) ( )yy yy xx xx m m − − = − − = 2 1 2 1 2 1
- 23. Section Formula For x - coordinate Taking or ( ) ( )xx xx m m − − = 2 1 2 1 ( ) ( )1221 xxmxxm −=− 122121 xmxmxmxm −=−or or ( )121221 mmxxmxm +=+ 12 1221 mm xmxm x + + =
- 24. Section Formula For y – coordinate Taking ( ) ( )yy yy m m − − = 2 1 2 1 ( ) ( )1221 yymyym −=− 122121 ymymymym −=− ( )121221 mmyymym +=+ 12 1221 mm ymym y + + = or or or
- 25. Midpoint Midpoint of A(x1, y1) and B(x2,y2) m:n ≡ 1:1 1 2 1 2x x y y P , 2 2 + + ∴ ≡ Find the Mid-Point of P(1,-3) and Q(5,7).
- 26. Area of a Triangle Let ABC be any triangle whose vertices are A(x1, y1), B(x2 , y2) and C(x3 , y3). Draw AP, BQ and CR perpendiculars from A,B and C, respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapezium, Now, from figure QP = (x2 – x1) PR = (x3 – x1) QR = (x3 – x2) x Y A (x1 , y1) B(x2 , y2) C (x3 , y3) P QO R
- 27. Area of a Triangle XX’ Y’ O Y A(x1, y1) C(x3, y3) B(x2,y2) M L N Area of ∆ ABC = Area of trapezium ABML + Area of trapezium ALNC - Area of trapezium BMNC
- 28. Area of a Triangle Area of ABCΔ = Area of trapezium ABQP + Area of trapezium BQRC– Area of trapezium APRC. We also know that , Area of trapezium = Therefore, Area of ABCΔ = ( )( )embetween thdistancesidesparallelofsum 2 1 ( ) ( ) ( )PRCR+AP 2 1 CR+BQ 2 1 QPAP+BQ 2 1 −+ QR ( )( ) ( )( ) ( )( )133123321212 2 1 2 1 2 1 xxyyxxyyxxyy −+−−++−+= ( ) ( ) ( )[ ]133311312333223211211222 2 1 xyxyxyxyxyxyxyxyxyxyxyxy −+−−−+−+−+−= ( ) ( ) ( )[ ]123312231 2 1 yyxyyxyyx −+−+−= Area of Δ ABC
- 29. PROJECT Mark coordinate axes on your city map and find distances between important landmarks- bus stand, railway station, airport, hospital, school, Your house, Any river etc.
- 31. THANK YOU
Editor's Notes
- #7: Description of axes. The axes are arbitrarily chosen mutually perpendicular lines passing through an arbitrarily chosen point identified as the origin. Left to right and bottom to top are positive directions.
- #8: The cartesian plane is divided into four quadrants. Quadrants are numbered in anticlockwise direction. All abcissae in a given quadrant will have the same sign and all ordinates in a given quadrant will have the same sign.
- #9: Any point on the plane can be uniquely identified by its x-coordinate and y-coordinate. The x-coordinate is also called the abcissa, the y coordinate the ordinate of the point. The point is identified as an ordered pair (abcissa, ordinate) or (x,y).
- #11: Explain the derivation step by step according to animation.
- #17: To show that a figure is a parallelogram, prove that opposite sides are equal.
- #18: To show that a figure is a rhombus, prove all sides are equal.
- #19: To show that a figure is a rectangle, prove that opposite sides are equal and the diagonals are equal. Corollary : Parellelogram but not rectangle diagonals are unequal
- #20: To show that a figure is a square, prove all sides are equal and the diagonals are equal. Corollary : Rhombus but not square diagonals are unequal
- #21: To show that three points are collinear, show that a+b = c