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-int. Essentially, we need to determine both the slope and y-int 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-int, we can use the following strategies:
- The y-int is given in the question.
- Solve for the slope first and then plug in a point \( (x,y) \) to solve for the y-int.
Let's practice a couple scenarios:
Given slope, y-intercept
We can write the equation directly using the slope-intercept form:
Find the equation of the line knowing its slope is -3 and y intercept is 5.
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We are given both the slope and y-int. Simply plug them into the equation of a line:
\(y = -3(x) + 5\)
Find the equation of the line knowing its slope is 5 and y intercept is -2.
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We are given both the slope and y-int. Simply plug them into the equation of a line:
\(y = 5(x) - 2\)
Slope and 1 point
Plug in the slope and point to solve for the y-int.
Find the equation of the line given its slope as 3 and pass through the point (2,4).
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\(y = 3(x) + b\)
\(4 = 3(2) + b\)
\(4 = 6 + b\)
\(4 - 6 = b\)
\(-2 = b\)
\(y = 3(x) - 2\)
Find the equation of the line given its slope as 5 and pass through the point (1,3).
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\(y = 5(x) + b\)
\(3 = 5(1) + b\)
\(3 = 5 + b\)
\(3 - 5 = b\)
\(-2 = b\)
\(y = 5(x) - 2\)
Parallel/Perpendicular Lines
Remember that when dealing with parallel and perpendicular lines, parallel lines have same slope while perpendicular lines have negative reciprocal slope.
Write the equation of the lines that are parallel and perpendicular to the equation \(y = 2(x) + 1\) and pass through the point (1, 5).
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For parallel lines, the slope is the same: \( m = 2\). Plug in the point and solve for \(b\):
\(5 = 2(1) + b\)
\(b = 3\)
\(y = 2(x) + 3\)
For perpendicular lines, the slope is the negative reciprocal: \( m = \cfrac{-1}{2}\). Plug in the point and solve for \(b\):
\(5 = \cfrac{-1}{2}(1) + b\)
\(b = (5) + \cfrac{1}{2}\)
\(y = \cfrac{-1}{2}(x) + (5)\cfrac{1}{2}\)
Write the equation of the lines that are parallel and perpendicular to the equation \(y = 5(x) - 3\) and pass through the point (2, -1).
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For parallel lines, the slope is the same: \( m = 5\). Plug in the point and solve for \(b\):
\(-1 = 5(2) + b\)
\(b = -11\)
\(y = 5(x) - 11\)
For perpendicular lines, the slope is the negative reciprocal: \( m = \cfrac{-1}{5}\). Plug in the point and solve for \(b\):
\(-1 = \cfrac{-1}{5}(2) + b\)
\(b = -1 + \cfrac{2}{5}\)
\(b = \cfrac{-3}{5}\)
\(y = \cfrac{-1}{5}(x) - \cfrac{3}{5}\)
Two points
When given two points, calculate the slope using the slope formula:
\( m = \cfrac{y_2 - y_1}{x_2 - x_1}\)
Find the equation of the line given the points (3, -5) and (6, 2).
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First find the slope using the two points:
\(m = \cfrac{2-(-5)}{6-3}\)
\(m = \cfrac{7}{3}\)
Now to find the y-intercept, we can just plug in one point.
\(-5 = \cfrac{7}{3}(3) + b\)
\(-5 = -7 + b\)
\(b = -12\)
Now we have the equation: \(y = \cfrac{7}{3}(x) - 12\)
Find the equation of the line given the points (1, 3) and (2, -2).
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First find the slope using the two points:
\(m = \cfrac{-2 - 3}{2 - 1}\)
\(m = \cfrac{-5}{1}\)
Now to find the y-intercept, we can just plug in one point.
\(3 = -5(1) + b\)
\(b = 8\)
Now we have the equation: \(y = -5(x) + 8\)