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There are a few ways to approach these questions:

Use the two point formula for a line $y-y_1=\\frac{y_2-y_1}{x_2-x_1}(x-x_1)$. Or,
Determine the gradient by $m=\\frac{\\text{rise}}{\\text{run}}$, or $m=\\frac{y_2-y_1}{x_2-x_1}$, and then use the one point gradient formula $y-y_1=m(x-x_1)$. Or,
Determine the gradient by $m=\\frac{\\text{rise}}{\\text{run}}$, or $m=\\frac{y_2-y_1}{x_2-x_1}$. Pick a point and substitute it and the value for $m$ into the gradient intercept form of a line $y=mx+b$ to determine $b$. Now you have $m$ and $b$ so you can write the equation of the line in the form $y=mx+b$.

\n\n

For example, suppose we had to find the equation of the line through $(2,3)$ and $(5,1)$. Let us call the first point $(x_1,y_1)$ and the second $(x_2,y_2)$. The following are examples of using the above approaches:

Substitute $(x_1,y_1)=(2,3)$ and $(x_2,y_2)=(5,1)$ into the equation $y-y_1=\\frac{y_2-y_1}{x_2-x_1}(x-x_1)$ to get $y-3=\\frac{1-3}{5-2}(x-2)$ (notice we leave $x$ and $y$ as variables). Simplifying the fraction we have $y-3=-\\frac{2}{3}(x-2)$. Expanding the brackets we have $y-3=-\\frac{2}{3}x+\\frac{4}{3}$. Making $y$ the subject gives $y=-\\frac{2}{3}x+\\frac{13}{3}$.
Find the gradient, $m=\\frac{y_2-y_1}{x_2-x_1}=\\frac{1-3}{5-2}=-\\frac{2}{3}$. Substitute this and $(x_1,y_1)=(2,3)$ into the equation $y-y_1=m(x-x_1)$. This gives $y-3=-\\frac{2}{3}(x-2)$. Expanding the brackets we have $y-3=-\\frac{2}{3}x+\\frac{4}{3}$. Making $y$ the subject gives $y=-\\frac{2}{3}x+\\frac{13}{3}$.
Find the gradient, $m=\\frac{y_2-y_1}{x_2-x_1}=\\frac{1-3}{5-2}=-\\frac{2}{3}$. Take a point, say $(2,3)$ as $(x,y)$ and substitute this and the gradient into the equation $y=mx+b$. This gives $3=-\\frac{2}{3}(2)+b$. Solving for $b$ gives $b=\\frac{13}{3}$. Therefore the equation of the line is $y=-\\frac{2}{3}x+\\frac{13}{3}$.

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The gradient intercept form of the line that passes through the points ({x0},{y0}) and ({x1},{y1}) is $y=$ [[0]]

{line_and_2points()}

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Identifying y=mx+b given two points

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "notes": ""}, "advice": "", "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Terry Young", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3130/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Terry Young", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3130/"}]}