Co-ordinate Geometry III

Equations of Lines II

By Mr Porter

0 2 4 6-2

X-axis

Y-axis

-2

2

4

Parallel and Perpendicular Lines.

m 1 = m 2

m1 x m

2 = -1

Assumed Knowledge:Gradient, m, using 2 points.

Standard form of a line: y = mx + b

General form of a line: Ax + By + C = 0

Ability to rearrange algebraic expression / equation: Change of Subject

ExamplesRearrange the general line Ax + By + C = 0, to standard form, y = mx + b. Where m is the gradient and b is the y-intercept, for each of the following.

a) 3x + 2y – 6 = 0

2y =-3x + 6

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y =−3 2

x + 3

b) 4x – 2y + 7 = 0

4x + 7 = 2y

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2x +72

= y

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i.e. y = 2x +72

y = mx + b

y = mx + b

Parallel LinesDefinition: Two lines y = m1x + b1 and y = m2 x + b2 are parallel, if their gradients are equal: m1 = m2.

Example 1Find the equation of the line parallel to y = 3x – 2, passing through the point (2,-5).To find the equation of a line, you need to have a gradient, m, and a point on the line. In this case, we have the point, need to find the gradient.

From, y =3x – 2, in standard form, y = mx + b.

The gradient of the given line is m = 3.

Gradient, mi, of ALL lines parallel to the given line are equal. m1 = m2

Therefore, m = 3.The point-gradient form of a lines is:

y - y1 = m(x - x1)

y – -5 = 3(x – 2)

y + 5 = 3x – 6

y = 3x – 1, is the required line.

The point-gradient form of a lines is:y - y1 = m(x - x1)

3y – 9 = -2x – 2

2x +3y – 7 = 0, is the required line.

Example 2Find the equation of the line parallel to , passing through the point (-1,3).

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y = −2 3 x − 5

Gradient, mi, of ALL lines parallel to the given line are equal. m1 = m2

From, , in standard form,

y = mx + b.

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y = −2 3 x − 5

The gradient of the given line is .

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m = −2 3

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m = −2 3Therefore,

y – 3 = (x – -1)

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−2 3

Parallel lines to a given general line: Ax + By + C = 0.

If the given line is in general form, it has to be rearrange to standard form to obtain the gradient, m.

Example 1Find the equation of the line parallel to 3x + 2y – 6 = 0, through the point (3,-5).Step 1: Rearrange the given equation to standard form: y = mx + b.

3x + 2y – 6 = 0

2y = -3x + 6

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y = −3 2 x + 3

i.e y = mx + b

€

m = −3 2

The point-gradient form of a lines is:y - y1 = m(x - x1)

€

y − −5 = −32 x − 3( )

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2y +10 = −3x + 9

3x + 2y + 1 = 0, is the required line.

Example 2Find the equation of the line parallel to 5x – 2y + 4 = 0, through the point (-2,8).Step 1: Rearrange the given equation to standard form: y = mx + b.

5x – 2y + 4 = 0

5x + 4 = 2y

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y = 52 x + 2

i.e y = mx + b

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m = 52

The point-gradient form of a lines is:y - y1 = m(x - x1)

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y − 8 = 52 x − −2( )

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2y −16 = 5x +10

5x – 2y + 26 = 0, is the required line.

Exercise 1: Find the equation of the line parallel to the given line passing through the given point.

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a) y = 3x −1, Point (2,6)

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b) y = 34 x + 5, Point (−2,1)

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c) 3x + 2y + 7 = 0, Point (1,2)

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d) 4x − y + 7 = 0, Point (−1,−5)

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e) y = −4 5 x − 4, Point (1,−2)

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f ) 3x − 5y −1= 0, Point (2,0)

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line : y = 3x

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line : 3x − 4y − 2 = 0

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line : 3x + 2y − 7 = 0

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line : y = 4x −1

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line : 4x + 5y + 6 = 0

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line : 3x − 5y − 6 = 0

Perpendicular Lines.Definition: Two lines y = m1x + b1 and y = m2 x + b2 are perpendicular, if the

 product of the gradients is equal to -1: m1 x m2 = -1.

Alternative: The gradients are negative reciprocals.

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m2 =−1m1

Example 1Find the equation of the line perpendicular to y = 3x – 2, passing through the point (2,-5).

From, y =3x – 2, in standard form, y = mx + b.

The gradient of the given line is m1 = 3.

The point-gradient form of a lines is:y - y1 = m2(x - x1)

Perpendicular gradient, m2, are such that 3 x m2 = -1.

Therefore,

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m2 = −1 3

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y − −5 = −1 3 x − 2( )

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3y +15 = −x + 2

x + 3y + 13 = 0, is the required line.

The gradient of the given line is .

The point-gradient form of a lines is:y - y1 = m2(x - x1)

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y − 3 = 32 x − −1( )

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2y − 6 = 3x + 3

3x – 2y + 9 = 0, is the required line.

From, , in standard form, y = mx + b.

€

y = −2 3 x − 5

Example 2Find the equation of the line perpendicular to , passing through the point (-1,3).

€

y = −2 3 x − 5

Perpendicular gradient, m2, are such that x m2 = -1.

Therefore,

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m2 = 32

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−2 3

€

m1 = −2 3

Negative reciprocal! Negative reciprocal!

Example 3Find the equation of the line perpendicular to 3x + 2y – 6 = 0, through the point (3,-5).Step 1: Rearrange the given equation to standard form: y = mx + b.

3x + 2y – 6 = 0 2y = -3x + 6

€

y = −3 2 x + 3

i.e y = mx + b

€

m = −3 2

The point-gradient form of a lines is:y - y1 = m(x - x1)

€

y − −5 = 23 x − 3( )

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3y +15 = 2x − 6

2x – 3y – 21 = 0, is the required line.

Perpendicular gradient, m2, are such that x m2 = -1.

Therefore,

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m2 = 23

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−3 2

Example 4Find the equation of the line perpendicular to 5x – 2y + 4 = 0, through the point (-2,8).Step 1: Rearrange the given equation to standard form: y = mx + b.

5x – 2y + 4 = 0 5x + 4 = 2y

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y = 52 x + 2

i.e y = mx + b

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m = 52

The point-gradient form of a lines is:y - y1 = m(x - x1)

€

y − 8 = −25 x − −2( )

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5y − 40 = −2x − 4

2x + 5y – 36 = 0, is the required line.

Perpendicular gradient, m2, are such that x m2 = -1.

Therefore,

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m2 = −2 5

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52

Exercise 2: Find the equation of the line perpendicular to the given line passing through the given point.

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a) y = 3x −1, Point (2,6)

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b) y = 34 x + 5, Point (−2,1)

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c) 3x + 2y + 7 = 0, Point (1,2)

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d) 4x − y + 7 = 0, Point (−1,−5)

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e) y = −4 5 x − 4, Point (1,−2)

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f ) 3x − 5y −1= 0, Point (2,0)

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line : x + 3y − 20 = 0

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line : 4x + 3y + 5 = 0

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line : 2x − 3y + 4 = 0

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line : x + 4y + 21= 0

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line : 5x − 4y −13 = 0

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line : 5x + 3y −10 = 0