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Equation of the Line Straight Lines Mr. J McCarthy
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xy 1 2 3 4
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xy 13 2 3 4
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xy 13 23 3 4
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xy 13 23 33 4
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xy 13 23 33 43 Equation of the Line y=3
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xy 1 2 3 4
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xy 21 2 3 4
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xy 21 22 3 4
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xy 21 22 23 4
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xy 21 22 23 24 Equation of the Line x=2
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xy 0 1 2
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xy -2 0 1 2
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xy -2 0 1 2
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xy -2 0 10 2
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xy -2 0 10 21 Equation of the Line y = x – 1 What is the relationship between x and y? x is always 1 more than y. y is always 1 less than x. What is 1 less than x? x – 1
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xy -3 0 1 2
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xy 0 0 1 2
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xy 0 03 1 2
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xy 0 03 14 2
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xy 0 03 14 25 Equation of the Line y = x + 3 What is the relationship between x and y? x is always 3 less than y. y is always 3 more than x. What is 3 more than x? x + 3
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How to get the equation of a line when the line is on a Graph? xY 00 12 24 36 xY 03 14 25 36 xY 00 21 42 63
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Equation of the Line
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y = mx + c y = x + 3 For any line the format of y = mx + c m is the slope of the line the point where it cuts the y axis is (0,c) m = 1 (0,3)
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y = mx + c y = 2x – 2 For any line the format of y = mx + c m is the slope of the line the point where it cuts the y axis is (0,c) m = 2 (0, – 2) Rise Run 2 1 2 2 1
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Equation of the Line formula If we know a point on a line and the lines slope then we can use a formula to figure out the equation of the line: y – y 1 = m(x – x 1 ) Example: There is a line with a Point (4,2) and a slope of 3. What is the equation of the line? y – y 1 = m(x – x 1 ) y – 2 = 3(x – 4) y – 2 = 3x – 12 y = 3x – 12 + 2 y = 3x – 10 (x 1, y 1 ) m
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Equation of the Line formula There is a line with a Point (4,2) and a slope of 3. What is the equation of the line? y – y 1 = m(x – x 1 ) y – 2 = 3(x – 4) y – 2 = 3x – 12 y = 3x – 12 + 2 y = 3x – 10 (x 1, y 1 )m
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Junior Cert Question
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Equation of the Line Question 1. The line K contains the point (2,4). K has a slope of 3. Find the equation of K. Answer: y – y 1 = m(x – x 1 ) y – 4 = 3(x – 2) y – 4 = 3x – 6 y = 3x – 6 + 4 y = 3x – 2 y – y 1 = m(x – x 1 )<- Formula (x 1, y 1 ) m Question 2. The line K contains the point (4,5). K has a slope of 2. Find the equation of K. Answer: y – y 1 = m(x – x 1 ) y – 5 = 2(x – 4) y – 5 = 2x – 8 y = 2x – 8 + 5 y = 2x – 3 (x 1, y 1 ) m Question 3. The line K contains the point (1,3). K has a slope of 5. Find the equation of K. Answer: y – y 1 = m(x – x 1 ) y – 3 = 5(x – 1) y – 3 = 5x – 5 y = 5x – 5 + 3 y = 5x – 2 (x 1, y 1 ) m
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Equation of the Line Question 4. The line K contains the point (3,1). K has a slope of 3. Find the equation of K. y – y 1 = m(x – x 1 )<- Formula Question 5. The line K contains the point (4,2). K has a slope of 2. Find the equation of K. Question 6. The line K contains the point (-3,2). K has a slope of 4. Find the equation of K. Question 7. The line K contains the point (1,5). K has a slope of 2. Find the equation of K. Question 8. The line K contains the point (3,-2). K has a slope of 5. Find the equation of K. Question 9. The line K contains the point (4,3). K has a slope of 1. Find the equation of K.
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Equation of the Line Question 4. The line K contains the point (3,1). K has a slope of 3. Find the equation of K. y – y 1 = m(x – x 1 )<- Formula Question 5. The line K contains the point (4,2). K has a slope of 2. Find the equation of K. Question 6. The line K contains the point (-3,2). K has a slope of 4. Find the equation of K.
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Equation of the Line y – y 1 = m(x – x 1 )<- Formula Question 7. The line K contains the point (1,5). K has a slope of 2. Find the equation of K. Question 8. The line K contains the point (3,-2). K has a slope of 5. Find the equation of K. Question 9. The line K contains the point (4,3). K has a slope of 1. Find the equation of K.
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Line Intersecting x and y axis Question 1. l is the line 2x + y = 4 L cuts the y-axis at the point T By Letting x = 0 find the co- ordinates of the point T. Answer: 2x + y = 4 0 + y = 4 y = 4 (The line l cuts the y axis at the point (0,4) )
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Find Co-Ordinates of the points where these lines cut the x and y axis
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Revision (x 1, y 1 )(x 2, y 2 ) (x 1, y 1 )(x 2, y 2 ) Steps Formula Labels Replace Simplify
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Revision (x 1, y 1 )(x 2, y 2 ) Steps Formula Labels Replace Simplify Equation of a Line Point (1,3) Slope = 5 y – y 1 = m(x – x 1 ) y – 3 = 5(x – 1) y – 3 = 5x – 5 y = 5x – 5 + 3 y = 5x – 2 (x 1, y 1 ) m
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