Finding equations of lines How many lines have a gradient of 2? What do we know about the equation...

Finding equations of lines

How many lines have a gradient of 2?What do we know about the equation

of these lines?How many of these will pass thru (1,3)?

What we already know

• The equation of a line with a gradient of 2 and y intercept of -3 is

• y = 2x + 3

• The equation of a line with a gradient of m and a y-intercept of b is

• y = mx + b

What is the equation of the line with a gradient of 2 and passing thru (1,3) ?

y = 2x + 3

Here is an algebraic approach to this question

• What is the equation of the line with a gradient of 2 and passing thru (1,3) ?

• We know that the equation is of the form y = mx + b and we know m = 2

• y = 2x + b substitute (1,3) for x and y• 3 = 2(1) + b , now solve for b• b = 3 – 2 = 1• The equation of the line is y = 2x + 1

Using the algebraic method...

• What is the equation of the line with a gradient of 3 and passing thru (-2,2)?

• We know the equation is of the form y = mx + b and we know m = 3

• y = 3x + b substitute x=-2, y=2• 2 = 3(-2) + b• b = 2 + 6 = 8• The equation of the line is y = 3x + 8

What have we learnt so far..

• Given the gradient of a line and the y intercept you can find the equation by.....

• Given the gradient of a line and a point other than the y-intercept you can find the equation by....

• What could we do if we don’t know the gradient?

• Answer: Find the gradient!

How can we find the gradient of the line joining 2 points?

rise

run

Find the gradient of the line joining:

1. (1,2) and (4,5)2. (1,2) and (4,4)3. (1,2) and (1,5)4. (1,2) and (1,-1)5. (1,2) and (-1,-3)

Write a formula for finding the gradient of a line joining any 2 points, say (x1, y1) and (x2, y2)

The formula for finding the gradient is..

Now back to finding the equation of the line...

• Use your formula you have just developed to find the gradient of the line joining (1,4) and (3,5)

• And hence find the equation of the line passing thru (1,4) and (3,5)?

• y= ½ x + b substitute x = 1, y = 4• 4 = ½ (1) + b • b = 3 ½ • y = ½ x + 3 ½ or

In the previous question,

• Would it have mattered if we had substituted (3,5) instead of (1,4)?

• m = ½• y= ½ x + b substitute x = 3, y = 5 (instead of x=1 y= 4)• 5 = ½ (3) + b • b = 5 – 1 ½ = 3 ½ • y = ½ x + 3 ½ or No, it doesn’t matter which point

we substitute!!

• Find the gradient and hence equation of the line joining (4,7) and (-1,2)

• Gradient

• y = 1x + b substitute x = -1, y = 2• 2 = 1 (-1) + b• b = 3• y = x + 3

• Check graphically: Draw the line joining (4,7) and (-1,2) and find it’s equation

You should get y = x + 3 that we found algebraically

Explain a method for finding the equation of a line....

1. Given the gradient and y-intercept2. Given the gradient and a point other than the

y-intercept3. Given 2 points

2 special types of lines• What is the gradient of the line joining any 2

points that are horizontal ?• What can you conclude about the equation of

the line?

y = mx + b must become.......y = 0x + b y = b

Horizontal lines are of the form: y = c

2 special types of lines• What is the gradient of a line joining 2 points

that are vertical?• What can you conclude about the equation of

such a line?

y = mx + b ?Doesn’t apply as there is no y-interceptList 3 other points on your lineWhat do all these points have in common?

Vertical lines are of the form: x = c