Transcript for:
Understanding Coordinate Geometry Basics

R JAMUNAR Syllabus Topic: Coordinate Geometry Title: Plotting Points, Gradient, Parallel and Perpendicular Lines, Distance, Midpoint, Equation of a Line Plotting Points R JAMUNAR A point is defined as an ordered pair or coordinate, (��, ��), such that �� is determined by its horizontal distance from the origin and �� is determined by its vertical distance from the origin. Consider the arrow diagram below: �� �� Rule: �� = 5�� 1 2 3 V V V 5 10 15 input output To show this information on a graph, we will plot the following points: • (1,5) • (2,10) • (3,15) R JAMUNAR Title: Graph showing the relation �� = 5��. output R JAMU NAR input Worked Example Plot the following points on the same grid: (a) ��(2,3) (e) ��(−3, −4) (b)��(0,5) (f) ��(4, −3) (c) ��(������������) (g) ��(1, −2) (d)��(−5,0) (h) ��(0, −1) R JAMUNAR Consider the point ��(2,3) ��-coordinate ��-coordinate Note: Points are always written of the form (��, ��). R JAMUNAR The number 2 is called the ��-coordinate. The number 3 is called the ��-coordinate. �� �� �� �� �� �� �� �� �� �� R JAMUNAR Points to Note: • Plan Ahead: Notice that on the ��-axis, you need to include values from −5 to 4 and on the ��-axis, you need to include values from −4 to 5. Essentially, you should have 5 units on each side of the origin for this graph. JAMUNAR • Ensure to occupy 80% of the graph page when drawing graphs. • Use a small “×” to plot the points. • Points are normally represented by capital letters and lines are represented by common letters. • The origin has coordinates (0,0). Gradient ➢ Another word for gradient is “slope” or “steepness”. ➢ The gradient of a straight line tells us how steep a line is, therefore, the bigger the gradient, the steeper the line. Exercise: Which line represents each gradient below? Gradient Representation of Line Positive R Negative Zero Infinite (undefined) R JAMUNAR Solution: Gradient Representation of Line Positive Negative Zero R Infinite (undefined) A Parallel Lines R JAMUN Parallel lines are a fixed distance apart and will never meet, no matter how long they are extended. ��1 ��2 Lines that are parallel have the same gradient. Therefore, for parallel lines ���� = ����. R JAMUNAR Perpendicular Lines Two lines are perpendicular if one is at right angles to each other. In other words, if the two lines cross and the angle between them is 90°. ��2 R JAMU NAR ��1 If two lines are perpendicular, then their gradients will multiply together to give -1. Therefore, for perpendicular lines, ���� = −������. Gradient Formula The gradient can be calculated as the change in �� over the change in ��. Gradient = �������� ������ Gradient = ���� ���� Gradient = ����−���� ����−���� R JAMUNAR Worked Example �� �� R JAMUNAR �� �� (a) What are the coordinates of �� and ��? (b)What is the gradient of the line ����? Solution: (a) The coordinates of point �� is ��(10,20). The coordinates of point �� is ��(45,35). R JAMUNAR (b)Title: Graph showing the line ����. �� Note: Line ���� is the same as line ����. R JAMUNAR �� �� �� Consider ��(10,20) and ��(45,35) (��1, ��1) (��2, ��2) Note: It does not matter which point is labelled as (��1, ��1) and (��2, ��2). R JAMUNAR Gradient, m = ��2−��1 ��2−��1 = 35−20 45−10 = 15 35 Note: Leave answer as a fraction. = 37 R JAMUNAR Distance/Length of a line To calculate the distance between two points in a straight line, we use the distance formula below which is an application of Pythagoras’ Theorem. Distance Formula: Distance = √(���� − ����)�� + (���� − ����)�� Relation of the distance formula to Pythagoras’ Theorem Recall: �� �� �� Pythagoras’ Theorem states that: ���� = ���� + ���� R JAMUNAR Consider the gradient of the line ����. �� �� ��2 R JAMUNAR ���� �� ��1 �� �� ��1 ��2 Using Pythagoras’ Theorem to find the distance of the line ����, we get: ��2 = ��2 + ��2 ��2 = (��2 − ��1)2 + (��2 − ��1)2 �� = √(��2 − ��1)2 + (��2 − ��1)2 R JAMUNAR Worked Example �� ��(3,8) R JAMUNAR ��(12,3) ��(0,0) �� (a) Find the gradient of the line ����. (b)State which line has a larger gradient between ���� and ����? (c) Find the length of the line ����. (d)Find the midpoint of ����. R JAMUNAR Solution: (a) The gradient of the line ���� is calculated as follows: Consider ��(3 , 8) and ��(12 , 3) R JAMUNAR (��1, ��1) (��2, ��2) Gradient of ����, �� = ��2−��1 ��2−��1 = 3−8 12−3 = −59 OR Consider ��(12, 3) and ��(3 , 8) (��1, ��1) (��2, ��2) Gradient of ����, m = ��2−��1 ��2−��1 = 8−3 3−12 = 5−9 Note: −�� ��=��−��= −���� Note: It does not matter which point is labelled as (��1, ��1) and (��2, ��2). R JAMUNAR (b)Title: Graph showing line ���� and ���� �� ��(3,8) R JAMUNAR ��(12,3) ��(0,0) �� Looking at the graph, it can be deduced that the line ���� is steeper than ����. Hence, line ���� has a larger gradient. (c) Consider ��(3 , 8) and ��(12 , 3) (��1, ��1) (��2, ��2) R JAMUNAR The distance of line ���� is calculated as follows: Distance of line ���� = √(��2 − ��1)2 + (��2 − ��1)2 = √(12 − 3)2 + (3 − 8)2 Note: Ensure to put your R JAMUNAR = √92 + (−5)2 = √81 + 25 = √106 = 10.3 units (to 3 significant figures) (d)To calculate the midpoint, the following formula is used: Midpoint Formula: Midpoint = (����+���� ��,����+���� ��) Consider ��(0 , 0) and ��(12 , 3) (��1, ��1) (��2, ��2) Midpoint = (��1+��2 2,��1+��2 2) = (0+12 2,0+3 2) = (122,32) = (6,32) units when calculating distance. If units are not given, just write “units” after your answer. R JAMUNAR Worked Example 2 �� �� R JAMUNAR �� �� �� (a) Find the gradient of the line ����. (b)Find the distance ����. (c) Find the midpoint, ��, of ����. (d)Find the gradient of ����. R JAMUNAR Solution: �� ��(6, 18) R JAMUNAR × ��(2, 4) �� ��(4, 11) ��(14, 10) (a) Points are ��(2, 4) and ��(6, 18). The gradient of ���� is as follows: R JAMUNAR Gradient = ��2−��1 ��2−��1 Gradient = 18−4 6−2 Gradient = 144 Gradient = 3.5 R JAMUNAR (b)Points are ��(2, 4) and ��(6, 18). Distance ���� = √(��2 − ��1)2 + (��2 − ��1)2 Distance ���� = √(6 − 2)2 + (18 − 4)2 Distance ���� = √(4)2 + (14)2 Distance ���� = √16 + 196 Distance ���� = √212 Distance ���� = 14.6 units (c) Midpoint, �� = (��1+��2 2,��1+��2 2) Midpoint, �� = (2+6 2,4+18 2) Midpoint, �� = (82,222) Midpoint, �� = (4, 11) (d)Points are ��(4, 11) and ��(14, 10). Gradient of ���� =��2−��1 ��2−��1 Gradient of ���� =10−11 14−4 Gradient of ���� =−1 10 R JAMUNAR Equation of a line The general form for the equation of a line is �� = ���� + �� where �� = gradient of the line R JAMUNAR �� = ��-intercept Worked Example �� ��(4,11) ��(0,3) �� Find the equation of the line ����. R JAMUNAR Solution: Consider ��(0 , 3) and ��(4, 11) (��1, ��1) (��2, ��2) R JAMUNAR Gradient of ����, �� = ��2−��1 ��2−��1 = 11−3 4−0 = 84 = 2 From the graph, it can be seen that the line cuts the ��-axis at 2. Hence, �� = 3. ∴ The equation of the line ���� is �� = 2�� + 3. R JAMUNAR Exercise �� �� R JAMUNAR �� �� �� Which of the following equations corresponds to the lines ��, �� and ��? �� =12�� + 4 �� = −1�� + 12 �� =12�� + 8 R JAMUNAR Solution: Note �� �� R JAMUNAR �� �� �� Equation of line: �� = ���� + �� If the constants �� and �� change, the line replicates this change. This can be seen in the following table: R JAMUNAR Equation of line Representation of line �� = 2�� + 4 UNAR R�� = −2�� + 4 JAM R JAMUNAR �� = 3�� + 10 UNAR R�� = 3�� + 20 JAM R JAMUNAR �� = 3�� − 1 UNAR R�� = 3�� − 15 JAM R JAMUNAR Worked Example 1 A line has a gradient of 12and passes through the point (2,3). Find the equation of the line. Solution: METHOD 1: R JAMUNAR • Substitute into �� = ���� + �� • Find ��. • Write equation. Substitute point (2,3) and gradient = 12into �� = ���� + ��, 3 =12(2) + �� 3 = 1 + �� �� = 3 − 1 �� = 2 So, �� = 2. ∴ The equation of the line is �� =12�� + 2, in the form �� = ���� + ��, where �� =12and �� = 2. R JAMUNAR METHOD 2: • Use �� − ���� = ��(�� − ����) �� =��2−��1 ��2−��1 R JAMUNAR Rearranging this equation and eliminating one of the points gives, �� − ��1 = ��(�� − ��1) Substitute point (2,3) and gradient = 12into �� − ��1 = ��(�� − ��1), �� − 3 =12(�� − 2) �� − 3 =12�� − 1 �� =12�� − 1 + 3 �� =12�� + 2 ∴ The equation of the line is �� =12�� + 2, in the form �� = ���� + ��, where �� =12and �� = 2. R JAMUNAR Worked Example 2 Find the equation of the line passing through the points ��(1, 10) and ��(5, −4). Solution: The points are ��(1, 10) and ��(5, −4). �� =��2−��1 ��2−��1 �� =−4−10 5−1 �� =−14 4 �� = −72 R JAMUNAR Substituting �� = −72and point (1, 10) into �� = ���� + �� gives, 10 = −72(1) + �� 10 = −72+ �� �� = 10 +72 �� =272 ∴ The equation of the line is �� = −72�� +272, in the form �� = ���� + ��, where �� = −72and �� =272. R JAMUNAR Alternatively, Substituting �� = −72and point (1, 10) into �� − ��1 = ��(�� − ��1) gives, �� − 10 = −72(�� − 1) R JAMUNAR �� − 10 = −72�� +72 �� = −72�� +72+ 10 �� = −72�� +272 ∴ The equation of the line is �� = −72�� +272, in the form �� = ���� + ��, where �� = −72and �� =272. R JAMUNAR Finding the equation of a perpendicular line Question: �� ��(3, 12) R JAMUNAR ��(1, 2) ��(4, 0) �� Find: (a) Find the gradient of the line ����. (b)Find the equation of the line perpendicular to ���� but passing through point ��. R JAMUNAR Solution: (a) Points are ��(1, 2) and ��(3, 12). Gradient of ���� =��2−��1 ��2−��1 Gradient of ���� =12−2 3−1 R JAMUNAR Gradient of ���� =102 Gradient of ���� = 5 (b)��1 = 5 So, the gradient of the perpendicular line, ��2 = −1��1 So, the gradient of the perpendicular line, ��2 = −15 Substituting point (4, 0) and gradient = −15into �� − ��1 = ��(�� − ��1) �� − 0 = −15(�� − 4) �� = −15�� +45 ∴ The equation of the line is �� = −15�� +45, in the form �� = ���� + ��, where �� = −15and �� =45.