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:
To find the \(y\)-intercept, we can use the following strategies:
Let's practice a couple scenarios:
We are given both the slope and \(y\)-intercept. Simply plug them into the equation of a line:
\(\boldsymbol{y = -3x + 5}\)
We are given both the slope and \(y\)-intercept. Simply plug them into the equation of a line:
\(\boldsymbol{y = 5x - 2}\)
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:
\(\boldsymbol{y = 3x - 2}\)
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:
\(\boldsymbol{y = 5x - 2}\)
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:
\(\boldsymbol{y = -\cfrac{1}{2}x + \cfrac{11}{2}}\)
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:
\(\boldsymbol{y = -\cfrac{1}{5}x -\cfrac{3}{5}}\)
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:
\(\boldsymbol{y = \cfrac{7}{3}x - 12}\)
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:
\(\boldsymbol{y = -5x + 8}\)