Equation from a gradient and a point
Recover the intercept by substitution
\[y=mx+c\]
- Use both coordinates Substitute the point into the rule with the stated gradient \(\displaystyle y_1=mx_1+c\)
- Isolate the constant Remove the gradient contribution from the output \(\displaystyle c=y_1-mx_1\)
- Recognise the direct cases A point on the vertical axis gives the intercept directly. On a horizontal line every point has that same vertical coordinate \(\displaystyle mx_1=0\;\Longrightarrow\;c=y_1\)
Worked example
Find the equation with the stated gradient through the point
\[m=-\frac{3}{2},\quad P(4,-1)\]
- \(\displaystyle y=-\frac{3}{2}x+c\)Put the given gradient into the general form
- \(\displaystyle -1=-\frac{3}{2}(4)+c\)Substitute both coordinates of the point
- \(\displaystyle -1=-6+c\)Simplify the product and keep the rest of the equation
- \(\displaystyle -1+6=c\)Add the opposite of the gradient contribution to both sides
- \(\displaystyle 5=c\)Evaluate the remaining sum
- \(\displaystyle y=-\frac{3}{2}x+5\)Replace the constant with the recovered intercept
- \(\displaystyle \begin{aligned}-\frac{3}{2}(4)+5&=-6+5\\&=-1\end{aligned}\)The point's vertical coordinate is reproduced
Common mistakes
Taking the point's height as the intercept
✕\(\displaystyle y=-\frac{3}{2}x-1\)✓\(\displaystyle y=-\frac{3}{2}x+5\)
This point is off the vertical axis and the line is not horizontal, so its height is not the interceptRemoving a negative product
✕\(\displaystyle \begin{aligned}c&=-1-6\\c&=-7\end{aligned}\)✓\(\displaystyle \begin{aligned}c&=-1+6\\c&=5\end{aligned}\)
Subtracting a negative contribution is the same as adding its positive oppositeDifficulty levels
introRecover an integer intercept from a signed integer gradient and point
standardRecover the intercept using signed coordinates, simple fractional gradients or a horizontal line
challengeRecover the exact intercept when one coordinate is fractional
examRecognise the equation matching both the stated gradient and point
Questions about this topic
Can I use the point's height as the intercept?+
Yes if the point is on the vertical axis, or if the line is horizontal. Otherwise account for the horizontal position
Why do I need the gradient as well as the point?+
Many lines pass through one point. The gradient determines which direction the line takes
Must the intercept be an integer?+
No. Keep an exact fraction when the calculation gives one
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 9 Mathematics, chapter 4: Linear relations
In this chapter
Construct a table of values from a formulaRead coordinates and points from a straight-line graph4A Points, tables and linear relations: Decide whether a point lies on a lineFind the axis intercepts of a line4C Horizontal, vertical and through-origin lines: Recognise horizontal, vertical and through-origin graphsClassify and compare gradients from graphsFind the gradient of a line through two points4E Direct proportion: Use a direct-proportion model4E Direct proportion: Read and interpret direct-proportion graphsRead the vertical-axis intercept and gradient from an equation4F Gradient-intercept form: Recognise a line or find points for its graphConstruct a line on a coordinate gridFind the equation of a line from its graph or from its intercept and slopeRecover an intercept or recognise a line from a gradient and pointFind the equation of a line through two pointsMidpoint, endpoint and distance of a coordinate segmentForm an equation of a parallel or perpendicular lineDetermine a perpendicular bisector (optional extension)4J Linear modelling: Construct a linear model from a descriptionInterpret the graph of a linear model4K Graphical simultaneous equations: Check a candidate point against two lines (optional extension)Read the point of intersection of two lines (optional extension)