Parallel linesLines that are parallel have the same slope or gradient

x

ym1

m2

m1 = m2

Example 1What is the equation of the line parallel to 3x – 4y + 22 = 0, through the point (–3, 5)?

3x – 4y + 22 = 0 4y = 3x + 22 y = ¾x + 5½ m1 = ¾ m2 = ¾ y – y1 = m(x – x1)

y – 5 = ¾(x + 3) 4y – 20 = 3x + 93x – 4y + 29 = 0

Notice both equations start with 3x – 4y

Perpendicular linesLines that are perpendicular meet at 90°. The relationship we use is:

To show lines are perpendicular we use:m1m2 = –1

If we know the lines are perpendicular and need the second gradient we use:

Another way of thinking about this is the second gradient is the negative reciprocal of the first.

Example 2Show the lines 3x + 6y + 22 = 0 and y = 2x + 13 are perpendicular.

3x + 6y + 22 = 0 6y = –3x – 22 y = –½x – 32/3

m1 = –½

y = 2x + 13 m2 = 2 m1m2 = –½ × 2 = –1The lines are perpendicular

Example 3Find the equation of the line that is perpendicular to 3x – 5y + 15 = 0 passes through the point (3, 4).

3x – 5y + 15 = 0 5y = 3x + 15 y = 3/5x + 3 m1 = 3/5

y – y1 = m(x – x1)y – 4 = –5/3(x – 3)

3y – 12 = –5x + 155x + 3y – 27 = 0

Today’s work

Exercise 7.7Page 291→292Q1→4, 10 & 11

Yesterday’s work

Exercise 7.5Page 283Q1, 4, 7…

Exercise 7.6Page 287Q1 a, b, e & gQ2Q3 a & cQ5Q9