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$ a=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$. Finn stigningstallet til linja.
\n$a=$ [[0]]
\n", "variableReplacements": [], "scripts": {}, "showFeedbackIcon": true, "gaps": [{"showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "mustBeReduced": false, "variableReplacements": [], "scripts": {}, "showFeedbackIcon": true, "mustBeReducedPC": 0, "allowFractions": true, "type": "numberentry", "correctAnswerStyle": "plain", "marks": 1, "minValue": "m", "notationStyles": ["plain", "en", "si-en"], "correctAnswerFraction": true, "maxValue": "m"}]}, {"showCorrectAnswer": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "marks": 0, "prompt": "Ei rett linje som går gjennom punktet $(x_{1},y_{1})$ og har stigningstallet $a$, er gitt ved likninga
\n$ y-y_{1}=a(x-x_{1}) $.
\nFinn likninga for linja. Skriv svaret på formen $y=ax+b$
\n$\\displaystyle y=$ [[0]]
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"}, "answer": "{m}*x+{c}", "vsetrangepoints": 5, "vsetrange": [0, 1], "variableReplacements": [], "scripts": {}, "showFeedbackIcon": true, "checkingaccuracy": 0.001, "type": "jme", "expectedvariablenames": ["x"], "showpreview": true, "marks": 1, "answersimplification": "fractionNumbers", "checkingtype": "absdiff"}]}], "tags": [], "name": "Ida's copy of Find the equation of a line through two points - positive gradient", "variablesTest": {"condition": "\n", "maxRuns": 100}, "statement": "Hvilken likning beskriver den rette linjen gjennom de to punktene med koordinatene $A=(\\var{xa},\\var{ya})$ og $B=(\\var{xb},\\var{yb})$.
\n{plotPoints()}
\n", "rulesets": {}, "variable_groups": [], "variables": {"c": {"name": "c", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "ya-m*xa"}, "xa": {"name": "xa", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "random(-4..-1)"}, "yb": {"name": "yb", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "ya+random([2,4])"}, "m": {"name": "m", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "(ya-yb)/(xa-xb)"}, "ya": {"name": "ya", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "random(-4..2)"}, "xb": {"name": "xb", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "xa+random([2,4] except -xa)"}}, "metadata": {"description": "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"}, "ungrouped_variables": ["xa", "xb", "ya", "yb", "m", "c"], "type": "question", "contributors": [{"name": "Ida Landg\u00e4rds", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2336/"}]}]}], "contributors": [{"name": "Ida Landg\u00e4rds", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2336/"}]}