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Finding Lines using y-y1=m(x-x1)

A useful way of finding the equation of a line is to use the formula y-y_1=m(x-x_1), where m is the gradient of the line and x_1 and y_1 are taken from the given point (x_1,y_y)

Finding the equation of the line given the gradient and a point

  1. Get the values of m, x_1, y_1
  2. Substitute into the formula
  3. Rearrange the formula to the desired form

Example

Find the line with a gradient of -2 that passes through the point (2,6)

So m=-2, x_1=2, y_1=6

\therefore y-6=-2(x-2) \Rightarrow y=-2x+10

Finding the equation of the line given a parallel line and a point

  1. Use the gradient of the parallel line for m
  2. Get the values of x_1 and y_1 from the given point
  3. Substitute into the formula
  4. Rearrange the formula to the desired form

Example

Find the equation of the line which  is parallel to y=3x+1 and passes through the point (2,1)

So m=3, x_1=2, y_1=1

\therefore y-1=3(x-2) \Rightarrow y=3x-5

Finding the equation of the line given a perpendicular line and a point

  1. Get the gradient of the perpendicular line and use the negative reciprocal for m
  2. Get the values of x_1 and y_1 from the given point
  3. Substitute into the formula
  4. Rearrange the formula to the desired form

Example

Find the equation of the line which  is perpendicular to y=-4x+7 and passes through the point (8,1)

The gradient of the given line is -4, so the perpendicular gradient is \frac{1}{4} \therefore m=\frac{1}{4}

From the point x_1 = 8 and y_1 = 1

\therefore y-1=\frac{1}{4}(x-8) \Rightarrow y=\frac{1}{4}x-1

Finding the equation of the line given two points

  1. Find the gradient from the points using \frac{rise}{run} or \frac{\delta y}{\delta x}
  2. Use either point in the formula, your choice
  3. Rearrange the formula to the desired form

Example

Find the line which passes through the points (-4, 7) and (3,10)

m = gradient = \frac{10-7}{3--4}=\frac{3}{7}

I will use the point (3,10), because both values are positive, but remember you would get the same result if you used (-4, 7)

\therefore x_1 = 3 and y_1 = 10

So substituting gives y-10=\frac{3}{7}(x-3) \Rightarrow y=\frac{3}{7}x+\frac{61}{7}

Finding the equation of the Perpendicular bisector between two points

  1. Find the gradient between the points and use the negative reciprocal for m
  2. Find the mid-point between the two points and use this for x_1 and y_1
  3. Substitute into the formula
  4. Rearrange the formula to the desired form

Example

Find the perpendicular bisector between the points (4, 6) and (2,10)

Gradient between the points is \frac{10-6}{2-4}=-2

\therefore m =\frac{1}{2}

The mid-point if \left(\frac{4+2}{2},\frac{6+10}{2}\right)=(3,8)

\therefore x_1 = 3 and y_1 = 8

So substituting gives y-8=\frac{1}{2}(x-3) \Rightarrow y=\frac{1}{2}x+\frac{13}{2}

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