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- 1. CHAPTER 5The Straight Line
- 2. Learning Objectives5.1 Understand the concept of gradient of a straight line.5.2 Understand the concept of gradient of a straight line in Cartesian coordinates.5.3 Understand the concept of intercept.5.4 Understand and use equation of a straight line.5.5 Understand and use the concept of parallel lines.
- 3. y2 − y1m= x2 − x1 y = mx + c
- 4. 5.1 graDient OF a straigHt Line(A) Determine the vertical and horizontal distances between two given points on a straight line F E GExample of application: AN ESCALATOR.EG - horizontal distance(how far a person goes)GF - vertical distances(height changed)
- 5. Example 1State the horizontal and verticaldistances for the following case. 10 m 16 mSolution:The horizontal distance = 16 mThe vertical distance = 10 m
- 6. (B)Determine the ratio of the verticaldistance to the horizontal distance 10 m 16 m Let us look at the ratio of the vertical distance to the horizontal distances of the slope as shown in figure.
- 7. Vertical distance = 10 mHorizontal distance = 16 mTherefore,Solution: vertical distance 16 = horizontal distance 10 = 1.6
- 8. 5.2 GRADIENT OF THE STRAIGHT LINE IN CARTESIAN COORDINATESy • Coordinate T = (X2,Y1) • horizontal distance R(x2,y2) = PT = Difference in x-coordinates y 2 – y1 = x2 – x 1 x2 – x 1 • Vertical distance P(x1,y1) T(x2,y1) = RT x = Difference in y-coordinates 0 = y2 – y 1
- 9. Solution: vertical distance gradient of PR = horizontal distance RT = PT y 2 − y1 = x2 − x1 REMEMBER!!! For a line passing through two points (x1,y1) and (x2,y2), y2 − y1 m= x2 − x1 where m is the gradient of a straight line
- 10. Example 2• Determine the gradient of the straight line passing through the following pairs of pointsi) P(0,7) , Q(6,10)ii)L(6,1) , N(9,7)Solution: 10 − 7 7 −1 Gradient PQ = Gradient LN = 6−0 9−6 3 units 6 units = = 6 units 3 units 1 =2 = 2
- 11. (C) Determine the relationship between the value of the gradient and the(i)Steepness(ii)Direction of inclination of a straight line• What does gradient represents?? Steepness of a line with respect to the x- axis.
- 12. B • a right-angled triangle. Line AB is a slope, making an angle θ with the horizontal line AC θA C vertical distance tan θ = horizontal distance = gradient of AB
- 13. y y B B θ θ x x 0 0 A AWhen gradient of AB is When gradient of AB ispositive: negative:• inclined upwards • inclined downwards• acute angle • obtuse angle.• tan θ is positive • tan θ is negative
- 14. Activity:Determine the gradient of the given lines in figureand measure the angle between the line and the x-axis (measured in anti-clocwise direction) y Line Gradient Sign θ V(1,4) N(3,3) Q(-2,4) MN S(-3,1) 0 x PQ M(-2,-2) R(3,-1) RS U(-1,-4) P(2,-4) UV
- 15. REMEMBER!!!The value of the gradient of a line:• Increases as the steepness increases• Is positive if it makes an acute angle• Is negative if it makes an obtuse angle
- 16. Lines Gradient y AB 0A B 0 x
- 17. Lines Gradient y D CD Undefined C0 x
- 18. Lines Gradient y F E EF Positive0 x
- 19. Lines Gradient y H G0 x GH Negative
- 20. Lines Gradient y AB D H F 0A B G CD Undefined E EF C Positive 0 x GH Negative
- 21. 5.3 Intercepts y-intercept x-intercept• Another way finding m, the gradient: y - intercept m=− x - intercept
- 22. 5.4 Equation of a straight line• Slope intercept form y = mx + c• Point-slope form given 1 point and gradient: y − y1 = m( x − x1 ) given 2 point: y − y1 y 2 − y1 = x − x1 x2 − x1
- 23. 5.5 Parallel lines • When the gradient of two straight lines are equal, it can be concluded that the two straight lines are parallel.Example:Is the line 2x-y=6 parallel to line 2y=4x+3?Solution:2x-y=6y→ → is 2. y=2x-6 gradient 32y=4x+3 → + 2 → y = 2x gradient is 2.Since their gradient is same hence they are parallel.

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