Use two points on a line graph to calculate the gradient and $y$-intercept and hence the equation of the straight line running through both points.
\nThe answer box for the third part plots the function which allows the student to check their answer against the graph before submitting.
\nThis particular example has a positive gradient.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "In this question we will identify the equation of the straight line passing through points $A=(\\var{xa},\\var{ya})$ and $B=(\\var{xb},\\var{yb})$ in the form $y = mx + c$.
\n{plotPoints()}
", "advice": "We find the equation of a straight line passing through two points by finding the gradient and the $y$-intercept of the line.
\nWe can find the gradient ($m$) using the points $A = (x_1,y_1)=(\\var{xa},\\var{ya})$ and $B = (x_2,y_2)=(\\var{xb},\\var{yb})$.
\nAs the definition of gradient is the ratio of vertical change ($y_2-y_1$) to horizontal change ($x_2-x_1$).
The equation for gradient is,
\\begin{align}
m &= \\frac{y_2-y_1}{x_2-x_1} \\\\[0.5em]
&= \\frac{\\simplify[!collectNumbers]{{yb}-{ya}}}{\\simplify[!collectNumbers]{{xb}-{xa}}} \\\\[0.5em]
&= \\frac{\\simplify[]{{yb}-{ya}}}{\\simplify{{xb}-{xa}}} \\\\[0.5em]
&= \\simplify[simplifyFractions,unitDenominator]{({yb-ya})/({xb-xa})}\\text{.}
\\end{align}
Rearranging the equation $y=mx+c$ and substituting either of the points gives
\n\\[c = y_1-mx_1 \\quad \\mathrm{or} \\quad c = y_2-mx_2 \\,\\text{.} \\]
\nWe can then also use this equation with the other point's coordinates to check our answer.
\nLet's use point $A$ first:
\n\\[
\\begin{align}
c &= y_1-mx_1 \\\\
&= \\var{ya}-\\var[fractionnumbers]{m}\\times\\var{xa} \\\\
& = \\simplify[fractionnumbers]{{ya-m*xa}}\\text{.}
\\end{align}
\\]
We then check this against point $B$:
\n\\[
\\begin{align}
y_2 &= mx_2 + c \\\\[0.5em]
&= \\simplify[fractionNumbers]{{m}{xb}+{c}} \\\\[0.5em]
&= \\var[fractionnumbers]{m*xb+c}\\text{.}
\\end{align}
\\]
We can now substitute these values for $m$ and $c$ into $y=mx+c$ to get:
\n\\[y=\\simplify[!noLeadingMinus,fractionNumbers,unitFactor]{{m} x+ {c}}\\text{.}\\]
\nThe green line drawn on the graph represents the above line equation.
\n{correctPoints()}
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\n$m=$ [[0]]
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\n$c=$ [[0]]
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\n$\\displaystyle y=$ [[0]]
\nUse the graph to plot your answer and check that it goes through these points.
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