Title 7. COORDINATE GEOMETRY.pdf ✅

CLASS - X MATHEMATICS 137 FOUNDATION 7. COORDINATE GEOMETRY VOL - I Cartesian co-ordinate System Quadrants Plotting of points Collinearity of Three Points Section Formula Area of a Polygon Straight Lines For learning this lesson effectively, From grade 9 co-ordinate geometry topic will be useful. Cartesian co-ordinate axes On a graph paper two mutually perpendicular straight lines X’OX and YOY’, intersecting each other at point O. The horizontal line X’OX is called X-axis and the vertical line YOY’ is called Y-axis and the two lines X’OX and YOY’ taken together are known as coordinate axes or rectangular axes or axes or reference. The point ‘O’ is called origin. SYNOPSIS-1 1.1 ORDERED PAIR A pair of real number a and b listed in a specific order, where ‘a’ always occupies the first position and ‘b’ the second position, is called an ordered pair. (a, b) (b, a) Example: Consider an equation y = 3x + 1 where x = 0, 1, 2. Write the ordered pairs specifying (x, y) Solution: We have y = 3x + 1 When x = 0, y = 3 × 0 + 1 = 1 When x = 1, y = 3 × 1 + 1 = 4 When x = 2, y = 3 × 2 + 1 = 7 The ordered pairs are (0, 1), (1, 4) and (2, 7). 1.2 CARTESIAN CO-ORDINATE SYSTEM Quadrants : x and y axes divide the plane into four quadrants and each axis has two directions. CHAPTER 7 COORDINATE GEOMETRY
138 MATHEMATICS VOL - I CLASS - X 7. COORDINATE GEOMETRY FOUNDATION Signs of x and y-axes The coordinate axis divide the coordinate in to four regions called quadrants. i) In the first quadrant, both coordinate i.e. abscissa and ordinate are positive. ii) In the second quadrant, for a point, abscissa is negative are ordinate is positive. iii) In the third quadrant, for a point, both abscissa and ordinate are negative. iv) In the fourth quadrant, for a point, the abscissa is positive and the ordinate is negative. Quadrant x-Coordinate y-Coordinate Sign of Coordinates I + + (+, +) II - + (-, +) III - - (-, -) IV + - (+, -) Cartesian co-ordinate of a point : The horizontal line is labelled as XOX’ and called the x-axis. The Vertical line is labelled as YOY’ and called the y -axis. The point of intersection of the two axis O is the origin. This set of axis forms a cartesian co-ordinate system. On the x-axis, to the right of O is positive, to the left is negative. On the y-axis, up is positive and down is negative. To locate a point (x, y), we go x units along the right on x-axis and then y units up in the vertical direction. Example: To locate (-2,1), we go 2 units horizontally in the negative direction along the x-axis and then 1 unit up parallel to y-axis. Note: The co-ordinates of the origin are (0,0) Abscissa : The distance of the point from y-axis is called its Abscissa. Ordinate : The distance of the point P from x-axis is called its ordinate. Quadrant : A Quadrant is 1/4 part of plane divided by co-ordinate axes. Points on axes : If a point P lies on x-axis, its coordinate is in the form of (x, 0) and if a point Q lies on y-axis its coordinate is in the form of (0, y) i.e., if a point P lies on x-axis its distance from x-axis will be zero therefore we can say that its y-coordinate will be zero. Similarly if a point Q lies on y-axis its distance from y-axis will be zero therefore we can say that its x-coordinate will be zero.
CLASS - X MATHEMATICS 139 FOUNDATION 7. COORDINATE GEOMETRY VOL - I Plotting of points : In order to plot the points in a plane, we may use the following algorithm. Step-I: Draw two mutually perpendicular lines on the graph paper. One horizontal and other vertical. Step-II: Mark their intersection point as O (origin) Step-III: Choose a suitable scale on x-axis and mark the points on both the axes. Step-IV: Obtain the coordinates of the point which is to be plotted. Let the point be P(a, b). To plot this point start from the origin and |a| units move along ox or ox’ according as a is positive or negative respectively. Suppose we arrive at point M. From point M move vertically upward or downward |b| through units according as b is positive or negative. The point where we arrive finally is the required point P(a, b). SOLVED EXAMPLES: 1. Plot the points (1,2), (-1,3), (-2,-4), (3,-2), (2,0) and (0,3) in a rectangular coordinate system. Solution: As shown in the figure, XOX′ and YOY′ is rectangular system of coordinate axes. The axes are labelled with positive numbers along OX and OY and with negative numbers along OX′ and OY′. The first point (1, 2) is represented by P, the second point (–1, 3) is represented by Q, the third point (–2, –4) is represented by R, the fourth point (3, –2) is represented by S, the fifth point (2, 0) is represented by T and the sixth point (0, 3) is represented by L. 2. The base AB of an equilateral triangle ABC with side 2a lies along the x-axis such that the mid-point of AB is at the origin and the vertex C is above the x-axis. Find the coordinates of the vertices of the triangle ABC. Solution: As shown in the figure, Length of OA = length of OB = ( ) = = 1 2 1 2 length of AB a ( ) 2 a The point A lies on OX, i.e., on the positive side of the x-axis and therefore, the coordinates of A are (a, 0). The point B lies on OX′, i.e., on the negative side of the x-axis and therefore, the coordinates of B are (–a, 0). The ∆ABC is equilateral, therefore, the point C lies on OY. Now, in right ∆AOC, we have OC2 = AC2 – OA2 = (2a)2 – (a)2 = 3a2 ⇒ OC = 3a . Therefore, the coordinates of the point C are ( , 0 3a) . 1.3 DISTANCE BETWEEN TWO POINTS The distance between any two points in the plane is the length of the line segment joining them. The distance between two points P(x1 , y1 ) and Q(x2 , y2 ) is given by PQ = − ( ) x x + − ( ) y y 2 1 2 2 1 2 Proof: We have to find the distance between two points P(x1 , y1 ) and Q(x2 , y2 ). PR and QS perpendiculars have been drawn to the x-axis.
140 MATHEMATICS VOL - I CLASS - X 7. COORDINATE GEOMETRY FOUNDATION A perpendicular from the point P on QS is also drawn which meets at T as shown in figure. OR = x1 , OS = x2 ⸫ RS = x2 – x1 ⇒ PT = x2 – x1 [RS = PT] Again, PR = y1 , QS = y2 ⸫ QT = y2 – y1. . Now, in right ∆PTQ, we have, PQ2 = PT2 + QT2 [By Pythagoras Theorem] ⇒ PQ2 = (x2 – x1 )2 + (y2 – y1 )2 Hence, PQ = ( ) x x ( ) y y 2 1 2 2 1 2 − + − .AB = (Difference of abscissa) +(Difference of ordinates) 2 2 It is called distance formula. The distance is always non-negative, so we take only the positive square root. Because the value of x1 – x2 will always be negative and its square will be positive. Similarly, the value of y1 – y2 will always be negative and its square will be positive. Hence, there will not be any change in the value of both the formulae. Distance of a Point from Origin: The distance of a point (x, y) from origin is x + y 2 2 . Proof: Let us take a point P (x, y) in the given plane of axes X’OX and Y’OY as shown in the fig. Here, the point P (x, y) is in the first quadrant but it can be taken anywhere in all the four quadrants. We have to find the distance OP, i.e., the distance of the point P from the origin O. From the point P, draw PM ⊥ OX and PL ⊥ OY. Then we have OM = x and MP = OL = y OP2 = OM2 + MP2 = x2 + y2 . Therefore, OP = x +y 2 2 Proof For Geometrical Figures: (a) For an isosceles triangle : Prove that two sides are equal. (b) For an equilateral triangle : Prove that three sides are equal. (c) For a right-angled triangle : Prove that the sum of the squares of two sides is equal to the square of the third side. (d) For a square : Prove that all sides are equal and diagonals are equal. (e) For a rhombus : Prove that all sides are equal and diagonals are not equal. (f) For a rectangle : Prove that the opposite sides are equal and diagonals are also equal. (g) For a parallelogram : Prove that the opposite sides are equal in length and diagonals are not equal. SOLVED EXAMPLES: 1. Find the distance between the points (3, 4) and (6, –3). Solution: Let the given points are A(3, 4) and B(6, –3). Here, x1 = 3, y1 = 4 and x2 = 6, y2 = –3 AB = − ( ) x x + − ( ) y y 1 2 2 1 2 2