Home > Graphing > Finding Lines using y-y1=m(x-x1)

## 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 $y-8=\frac{1}{2}(x-3) \Rightarrow y=\frac{1}{2}x+\frac{13}{2}$