3B - Finding the equation of a line
When determining the equation of a straight line we generally use one of two useful equations
- \(y=mx+c\)
- \(y-y_1=m(x-x_1)\)
Given the gradient and the \(y\)-intercept:
When you are given the gradient and the \(y\)-intercept, simply substitute the known information into the equation \(y=mx+c\). Remember that \(m\) is the gradient and \(c\) is the \(y\)-intercept.
When you are given the gradient and the \(y\)-intercept, simply substitute the known information into the equation \(y=mx+c\). Remember that \(m\) is the gradient and \(c\) is the \(y\)-intercept.
3B - Example 1: Finding the equation of a line given the gradient and y-intercept
Find the equation, in the form \(y=mx+c\), for each of the following
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3B - Example 1: Video solution
3B - Example 1: Practice
Question 1: Determine the equation of a straight line with a gradient of \(2\) and a \(y\)-intercept of \(-5\). Question 2: Determine the equation of a straight line with a gradient of \(\frac{-3}{4}\) and a \(y\)-intercept of \((0,2)\). 3B - Example 1: Solutions
Question 1: \[y=2x-5\] Question 2: \[y=-\frac{3x}{4}+2\] |
Given the gradient and a point:
When you are given the gradient, \(m\), and a point, \((x_1,y_1)\), that the line passes through, simply substitute the known information into the equation \(y-y_1=m(x-x_1)\).
When you are given the gradient, \(m\), and a point, \((x_1,y_1)\), that the line passes through, simply substitute the known information into the equation \(y-y_1=m(x-x_1)\).
3B - Example 2: Finding the equation of a line given the gradient and a point
Find the equation, in the form \(y=mx+c\), for each of the following
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3B - Example 2: Video solution
3B - Example 2: Practice
Question 1: Determine the equation of a straight line with a gradient of \(-3\) passing through the point \((5,6)\). Question 2: Determine the equation of a straight line with a gradient of \(\frac{1}{2}\) passing through the point \((-2,-3)\). 3B - Example 2: Solutions
Question 1: \[y-6=-3(x-5)\] \[\therefore y=-3x+21\] Question 2: \[y--3=\frac{1}{2}(x--2)\] \[\therefore y=\frac{x}{2}-2\] |
Given two points on the line:
When you are given two points, \((x_1,y_1)\) and \((x_2,y_2)\), that the line passes through:
When you are given two points, \((x_1,y_1)\) and \((x_2,y_2)\), that the line passes through:
- Determine the gradient of the line using teh formula \(m=\frac{y_2-y_1}{x_2-x_1}\)
- Select the simplest point that the lines goes through.
3B - Example 3: Finding the equation of a line given two points
Find the equation, in the form \(y=mx+c\), for a straight line that passes through the points \((-4,8)\) and \((1,3)\).
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3B - Example 3: Video solution
3B - Example 3: Practice
Question 1: Find the equation of a straight line that passes through the points \((-2,-7)\) and \((1,5)\). Question 2: Find the equation of a straight line that passes through the points \((-4,12)\) and \((2,3)\). 3B - Example 3: Solutions
Question 1: \[m=\frac{5--7}{1--2}=4\]\[y-5=4(x-1)\]\[\therefore y=4x+1\] Question 2: \[m=\frac{3-12}{2--4}=\frac{-3}{2}\]\[y-3=\frac{-3}{2}(x-2)\]\[\therefore y=-\frac{3x}{2}+6\] |
Horizontal and vertical lines: