Writing Linear Equations

We have learnt how to write a linear equation in the form \(y = mx + b\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. Essentially, we need to determine both the slope and \(y\)-intercept to write the equation.

To find the slope of the line we can use the following strategies:

  • The slope is given in the question.
  • We can calculate the slope using two points on the line using \( m = \cfrac{y_2 - y_1}{x_2 - x_1}\)
  • The slope is parallel to another line so the slopes are equal.
  • The slope is perpendicular to another line so we use the negative reciprocal: \( \cfrac{-1}{m} \)

To find the \(y\)-intercept, we can use the following strategies:

  • The \(y\)-intercept is given in the question.
  • Solve for the slope first and then plug in a point \((x,y)\) to solve for the \(y\)-intercept.

Let's practice a couple scenarios:

Given slope, y-intercept

Find the equation of the line knowing its slope is \(-3\) and \(y\)-intercept is \(5\).

We are given both the slope and \(y\)-intercept. Simply plug them into the equation of a line:

\(\boldsymbol{y = -3x + 5}\)


Find the equation of the line knowing its slope is \(5\) and \(y\)-intercept is \(-2\).

We are given both the slope and \(y\)-intercept. Simply plug them into the equation of a line:

\(\boldsymbol{y = 5x - 2}\)


Slope and 1 point

Find the equation of the line that has a slope of \(3\) and passes through the point \((2,4)\).

Since we are given the slope, we can write our initial equation as such:

\(y = 3x + b\)

In order to calculate the \(y\)-intercept, \(b\), we can substitute the point into the equation and solve:

\(4 = 3(2) + b\)

\(4 = 6 + b\)

\(b = 4 - 6\)

\(b = -2\)

Therefore, we can write the final equation as such:

\(\boldsymbol{y = 3x - 2}\)


Find the equation of the line that has a slope of \(5\) and passes through the point \((1,3)\).

Since we are given the slope, we can write our initial equation as such:

\(y = 5x + b\)

In order to calculate the \(y\)-intercept, \(b\), we can substitute the point into the equation and solve:

\(3 = 5(1) + b\)

\(3 = 5 + b\)

\(b = 3 - 5\)

\(b = -2\)

Therefore, we can write the final equation as such:

\(\boldsymbol{y = 5x - 2}\)


Parallel/Perpendicular Lines

Write the equation of lines that are parallel and perpendicular to the equation \(y = 2x + 1\) that pass through the point \((1, 5)\).

For parallel lines, the slope is the same: \( m = 2\). Plug in the point and solve for \(b\):

\(y = 2x + b\)

\(5 = 2(1) + b\)

\(b = 3\)

Therefore, we can write the parallel equation as such:

\(y = 2x + 3\)

For perpendicular lines, the slope is the negative reciprocal: \( m = -\cfrac{1}{2}\). Plug in the point and solve for \(b\):

\(y = -\cfrac{1}{2}x + b\)

\(5 = -\cfrac{1}{2}(1) + b\)

\(b = 5 + \cfrac{1}{2}\)

\(b = \cfrac{11}{2}\)

Therefore, we can write the perpendicular equation as such:

\(\boldsymbol{y = -\cfrac{1}{2}x + \cfrac{11}{2}}\)


Write the equation of lines that are parallel and perpendicular to the equation \(y = 5x - 3\) that pass through the point \((2, -1)\).

For parallel lines, the slope is the same, \( m = 5\). Plug in the point and solve for \(b\):

\(y = 5x + b\)

\(-1 = 5(2) + b\)

\(b = -11\)

Therefore, we can write the parallel equation as such:

\(y = 5x - 11\)

For perpendicular lines, the slope is the negative reciprocal, \(m = -\cfrac{1}{5}\). Plug in the point and solve for \(b\):

\(y = -\cfrac{1}{5}x + b\)

\(-1 = -\cfrac{1}{5}(2) + b\)

\(b = -1 + \cfrac{2}{5}\)

\(b = -\cfrac{3}{5}\)

Therefore, we can write the perpendicular equation as such:

\(\boldsymbol{y = -\cfrac{1}{5}x -\cfrac{3}{5}}\)


Two Points

Find the equation of the line given the points \((3, -5)\) and \((6, 2)\).

First, we can calculate the slope using the two points provided in the question:

\( m = \cfrac{y_2 - y_1}{x_2 - x_1}\)

\(m = \cfrac{2-(-5)}{6-3}\)

\(m = \cfrac{7}{3}\)

Next, to determine the \(y\)-intercept, we can just plug one of the points into the equation and solve:

\(y = \cfrac{7}{3}x + b\)

\(-5 = \cfrac{7}{3}(3) + b\)

\(-5 = -7 + b\)

\(b = -12\)

Then, we can write our final equation as such:

\(\boldsymbol{y = \cfrac{7}{3}x - 12}\)


Find the equation of the line given the points \((1, 3)\) and \((2, -2)\).

First, we can calculate the slope using the two points provided in the question:

\( m = \cfrac{y_2 - y_1}{x_2 - x_1}\)

\(m = \cfrac{-2 - 3}{2 - 1}\)

\(m = \cfrac{-5}{1}\)

Next, to determine the \(y\)-intercept, we can just plug one of the points into the equation and solve:

\(y = -5x + b\)

\(3 = -5(1) + b\)

\(b = 8\)

Then, we can write our final equation as such:

\(\boldsymbol{y = -5x + 8}\)