Information about coordinate Geometry straight line

Published on March 16, 2016

Author: SahilPuri14

Source: slideshare.net

2. Introduction Geometry, branch of mathematics that deals with shapes and sizes. Basic geometry allows us to determine properties such as the areas and perimeters of two-dimensional shapes and the surface areas and volumes of three-dimensional shapes.

3. Geometry used in daily Life People use formulas derived from geometry in everyday life for tasks such as figuring how much paint they will need to cover the walls of a house or calculating the amount of water a fish tank holds.

4. By the mean of coordinate system we are now able to find the dimensions between the two points : We can find the distance between the two points whose coordinates are given. we can find the coordinate of the point which divides a line segment joining two point in a given ratio. And to find the area of the triangle formed by three given points.

5. 1. distance between two point can be found by the formula called Distance Formula : distance between the point P(X1 , y1) and Q (x2 , y2) is PQ= (X2 –X1)² + (Y2 –Y1)²

6. this can be found out by Section Formula : the coordinate of a point dividing the line segment joining the point (X1 , Y1) and (X2, Y2) internally in the ratio m : n are : x-axis – m1x2+m2x1 y-axis – m1y2+m2y1 m1 + m2 m1 + m2 2. we can find the coordinate of the point which divides a line segment joining two point in a given ratio.

7. Area of triangle whose vertices are (x1, y1) , (x2, y2) and (x3, y2) is : ½ {x1(y2 –y3)+ x2(y3 –y1) +x3(y1 –y2)} 3. the area of the triangle formed by three given points

8. Straight line A straight line is a curve such that every point on the line segment joining any two points on it lies on it.

9. Despite its simplicity, the line is a vital concept of geometry and enters into our daily experience in numerous interesting and useful ways. Main focus is on representing the line algebraically, for which slope is most essential.

10. Slope of a line A line in a coordinate plane forms two angles with the x-axis, which are supplementary. The angle (say) made by line l with positive direction of x-axis and measured anticlockwise is called the inclination of the line. Obviously 0 180 180 - y x l o

11. definition 1:- If is the inclination of line l, then tan is called the slope or gradient of the line l. The slope of a line whose inclination is 90 is not defined The slope of a line is denoted by m thus, m = tan , 90 It may be observed that the slope of x-axis is zero and the slope of y-axis is not defined.

12. illustration 1: Find the slope of line whose inclination to the (+) ve direction of x-axis in anticlockwise sense is (1). 60 (2). 150 Sol:- (1) .slope = tan 60 =3 (2). slope = tan 150 = -cot 60 = -1 3

13. Slope of a line – coordinates are given we know that a line is completely determined when we are given two points on it. Hence, we proceed to find the slope of a line in terms of the coordinates of two points on the line. The inclination of the line l may be acute or obtuse . Lets us take each cases

14. Let P(x1, y1) and Q(x2, y2) be two point on non vertical line l whose inclination is . Draw perpendicular QR to x-axis and PM perpendicular to RQ as shown in fig,2. P(x1, y1) Q(x2, y2) R y O X M l Fig. 2.

15. CASE 1 :- when angle is acute in fig,2. MPQ = therefore, slope of line l = m = tan but in MPQ, we have tan = MQ = y2 – y1 MP x2 – x1 so, m = y2 – y1 x2 – x1 )(

16. CASE 2 :- when is obtuse in the fig,3. we have MPQ = 180 - therefore, = 180 - MPQ Y P (x1 , y1) R X O 180 - M Q (x2 , y2) l Fig.3

17. now, slope of line l m = tan = tan( 180 - MPQ) = -tan MPQ = - MQ = - y2 – y1 = y2 – y1 MP x1 – x2 x2 – x1

18. consequently, we see that in both the case the slope m of the line trough the point (x1 , y1) and (x2 , y2) is given by m = y2 – y1 x2 – x1

19. illustration 2 : find the slope of a line which passes through point (3, 2) and (-1, 5). Sol :- we known that the slope of a line passing through two point (x1, y1) and (x2, y2) is given by m = y2 – y1 x2- x1 here the line is passing through the point (3 ,2) and (-1 , 5). So, Its slope is given by m = 5 – 2 = - 3 -1 – 3 4

20. Condition for parallelism and perpendicularity of lines in terms of slopes 1. condition for parallelism of two lines if two lines m1 and m2 are perpendicular, then the angle between them is 90 In the coordinate plane , suppose that non – vertical line l1 and l2 have slope m1 and m2 respectively. Let their inclination be and , respectively.

21. if the line l1 is parallel to line l2, from the fig4,. Then their inclination are equal ,i.e., = , hence, tan = tan Therefore, m1 = m2, i.e., their slopes are equal l1 l2 O Y X Fig.,4

22. conversely, if the slope of two line l1 and l2 is same, i.e., m1 = m2 then, tan = tan . By the property of tangent function (between 0 and 180 ), = therefore, the line are parallel, Hence, two non vertical lines l1 and l2 are parallel if and only if there slopes are equal.

23. 2. condition for perpendicularity of two lines if two lines of slope m1 and m2 are perpendicular, then the angle between them is of 90 Y X l1 l2 O Fig.,5

24. if the line l1 and l2 are perpendicular from the fig .5, then = + 90 Therefore, tan = tan( + 90) =- cot = - 1 tan i.e., m2 = - 1 or m1 m2 = - 1 m1

25. Conversely, if m1 m2 = -1, i.e., tan tan = -1 Then tan = - cot = tan ( + 90) or tan ( - 90) therefore, and differ by 90. Thus, line l1 and l2 are perpendicular to each other. hence, two non–vertical lines are perpendicular to each other if and only if their slope are negative reciprocal of each other, m2 = -1 or, m1 m2 = -1 m1

26. illustration 3 : find the slope of line: 1. passing through (3, -2 )and (-1, 4) 2. passing through (3, -2) and (7, -2) Sol :- 1. slope of line through (3, -2) and (-1, 4) m = 4 –(-2) = 6 = 3 -1 -3 -4 2 2. The slope of line through (3, -2) and (7, -2) m = -2 – (- 2) = 0 = 0 7 – 3 4

27. angle between two lines When we think about more then one line in a plane then we find that these lines are either parallel or intersecting. Here we will discus the angle between two line in terms of there slopes.

28. let l1 and l2 be two vertical lines with slope m1 and m2, respectively . If and are the inclination of lines l1 and l2, respectively. Then m1 = tan and m2 = tan Y X l1 l2 Fig.,6

29. We know that two lines intersect each other, they make two pairs of vertically opposite angle such that sum of any two adjacent angle is 180. let and be the adjacent angle between the line l1 and l2 (fig., 4) then = - and , 90

30. Therefore, tan = tan( - ) = tan - tan = m2 – m1 1 + tan tan 1+ m1 m2 1+ m1 m2 And, = tan (180 - ) = - tan = - m2 – m1 1+m1 m2 Now, there arises two cases: 1. when , m2 – m1 ( is positive) 1+ m1 m2 2. When , m2 – m1 ( is negative) 1+ m1 m2

31. Case 1 : if, m2 – m1 ( is positive ) 1+ m1 m2 Here tan will be positive and will be negative which means will be acute and will be obtuse

32. Case 2: if, m2 - m1 (is negative) 1+ m1 m2 here tan will be negative and will be positive which means that will be obtuse and will be acute

33. thus ,acute angle between line l1 and l2 which slope m1 and m2 respectively, is given by tan = m2 – m1 1+ m1 m2 the obtuse angle can be found out by = ( 180 - )

34. illustration 4 : If A (-2, 1) , B (2, 3) and C (-2, -4) are three points, find the angle between BA and BC Sol :- let m1 and m2 be the slope of BA and BC respectively. then, m1 = 3 – 1 = 2 = 1 m2 = -4 -3 = 7 2- (- 2) 4 2 -2 -2 4 7 - 1 Tan = m2 – m1 = 4 2 = 10/8 = ± 2/3 1+ m1 m2 1+ 7 1 15/8 4 2 = tan ( 2/3 )-1

35. By :- Sahil Puri Class :- XI ‘B’ Roll no :- 21 School :- Kendriya Vidyalaya NO.2 Ambala Cantt

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