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User Mcchiz
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Answer:

Equation of line is given as y = mx + c, where m is the gradient and c is the y-intercept.

Qn A:

Parallel lines have the same gradient so gradient of line is 3.

Eqn of line: y = 3x + c

Since line passes through (1,4), subt coordinates into the equation to find c.

4 = 3(1) + c

c = 1

Hence, equation of the line is y = 3x + 1.

Qn B:

The product of the gradient of perpendicular lines is -1.

M₁ x M₂ = -1

To find the gradient of one perpendicular line,

M₂ = -1 ÷ M₁

Gradient of line = 3 ÷ (-1) = -3

Eqn of line: y = -3x + c

Since line passes through (-3,-1), subt coordinates into the equation to find c

-1 = -3(-3) + c

c = -10

Hence, equation of line is y = -3x - 10.

Qn C:

Rearrange the equation to y = mx + c form.

3y = x - 12

y = 1/3x - 4

Parallel lines have the same gradient so gradient of line is 1/3.

Eqn of line: y = 1/3x + c

Since line passes through (18,2), subt coordinates into the equation to find c

2 = 1/3(18) + c

c = -4

Hence, equation of line is y = 1/3x - 4.

Qn D:

Rearrange the equation to y = mx + c form.

2y + 6 = 3x

2y = 3x - 6

y = 3/2x - 3

Since line is perpendicular to y = 3/2x - 3, gradient of line = -1 ÷ 3/2 = -2/3

Eqn of line: y = -2/3x + c

Since line passes through (3,3), subt coordinates into the equation to find c

3 = -2/3x(3) + c

c = 5

Hence, equation of line is y = -2/3x + 5.

User Weeix
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