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Equations of Lines in Another Format

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1 Equations of Lines in Another Format
Slideshow 31, Mathematics Mr Richard Sasaki, Room 307

2 Objectives Understand how to draw a line to an equation in the form π‘Žπ‘₯+𝑏𝑦+𝑐=0 by making 𝑦 the subject. Understand how to draw a line to an equation in the form π‘Žπ‘₯+𝑏𝑦+𝑐=0 by connecting two pairs of co-ordinates

3 Drawing Graphs We now know how to draw a graph based on two simple pieces of information: The line’s gradient The y-intercept it passes through This is great when the function is in the form 𝑦=π‘Žπ‘₯+𝑏 but not when it’s in a different form!

4 Changing The Subject Another common way to write down the name of a line is in the form π‘Žπ‘₯+𝑏𝑦+𝑐=0. But now it’s not so clear what the gradient or y-intercept is…so what should we do? That’s right, change the subject!

5 Changing The Subject Let’s try changing π‘Žπ‘₯+𝑏𝑦+𝑐=0 (the general form) so that 𝑦 becomes the subject. π‘Žπ‘₯+𝑏𝑦+𝑐=0⟹ π‘Žπ‘₯+𝑐=βˆ’π‘π‘¦ βŸΉπ‘¦=βˆ’ π‘Žπ‘₯ 𝑏 βˆ’ 𝑐 𝑏 So in this form, the gradient is βˆ’ π‘Ž 𝑏 and the constant is βˆ’ 𝑐 𝑏 . Sounds messy doesn’t it… Anyway, let’s try changing the subject with numbers!

6 Example Draw the graph 3π‘₯+2π‘¦βˆ’4=0. Let’s change it so 𝑦 is the subject.
3π‘₯+2π‘¦βˆ’4=0⟹ 3π‘₯βˆ’4=βˆ’2𝑦 βŸΉπ‘¦=βˆ’ 3π‘₯ βˆ΄π‘¦=βˆ’ 3π‘₯ 2 +2 The gradient is βˆ’ 3 2 and the y-intercept is 2. Let’s draw it!

7 Example 3π‘₯+2π‘¦βˆ’4=0 (𝑦=βˆ’ 3π‘₯ 2 +2) So it passes through y=2…
And it’s gradient is βˆ’ So we go right two and down three to meet our line again. Something like this…

8 Answers Shown are not ideal titles but the equation where y is the subject. 𝑦=π‘₯+2 𝑦= 3 2 π‘₯+3 𝑦=βˆ’ 1 2 π‘₯+ 1 2 𝑦=βˆ’6

9 Finding co-ordinates If we are given a line π‘Žπ‘₯+𝑏𝑦+𝑐=0, can you see an easy pair of co-ordinates that we could find? When written in this form, it’s easy to find the two points… Because when 𝑦=0, we get π‘Žπ‘₯+𝑐=0… And when π‘₯=0, we get 𝑏𝑦+ 𝑐=0. We can then easily find π‘₯ and 𝑦 at those points!

10 Example Draw the line 16π‘₯+4π‘¦βˆ’8=0 by finding two suitable pairs of co-ordinates. Let’s find co-ordinates π‘₯, 0 and (0, 𝑦) where π‘₯ and 𝑦 are numbers. When π‘₯=0… 4π‘¦βˆ’8=0 4𝑦=8 𝑦=2 So we get (0, 2). When 𝑦=0… 16π‘₯βˆ’8=0 16π‘₯=8 π‘₯= 1 2 So we get ( 1 2 , 0).

11 Example This time we are drawing a line! What do we do? 16π‘₯+4π‘¦βˆ’8=0
(Note: Last time when we had two pairs of co-ordinates we used π‘Ž= 𝑦 2 βˆ’ 𝑦 1 π‘₯ 2 βˆ’ π‘₯ 1 because we were calculating the line. This time we are drawing a line! What do we do? 16π‘₯+4π‘¦βˆ’8=0 Simply place our co- ordinates , 0 and (0, 2) onto the graph and join the dots! Note: You must use a ruler!!

12 Answers The square would have co-ordinates (5, 0), (0, 5), (-5, 0) and (0, -5) and would have 4 line segments 𝑦=π‘₯+5, 𝑦=π‘₯βˆ’5, 𝑦=βˆ’π‘₯+5, 𝑦=βˆ’π‘₯βˆ’5. 3, 0 π‘Žπ‘›π‘‘ (0, 3) 0, 6 π‘Žπ‘›π‘‘ (βˆ’2, 0) βˆ’2, 0 π‘Žπ‘›π‘‘ 0, 1 2 0, 9 π‘Žπ‘›π‘‘ (βˆ’4, 0)


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