// Numbas version: exam_results_page_options {"name": "Equation of line perpendicular to given line, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"tags": [], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "pickQuestions": 0, "name": ""}], "advice": "\n

The equation of the line is of the form $y=mx+c$.

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The gradient $m$ will be the $\\dfrac{-1}{n}$ where $n$ is the gradient of the line $\\displaystyle \\simplify{{(b-d)/n2}x+{(c-a)/n2}y={(b*c-a*d)/n2}}$, which is $\\displaystyle n= \\simplify{{b-d}/{a-c}}$. Having calculated $n$, calculate $\\displaystyle m=\\dfrac{-1}{n} = \\simplify{{a-c}/{d-b}}$. We can calculate the constant term $c$ by noting that $y=\\var{k}$ when $x=\\var{h}$.

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Using this we get:
\\[ \\begin{eqnarray} \\var{k}&=&\\simplify[std]{({a-c}/{d-b}){h}+c} \\Rightarrow\\\\ c&=&\\simplify[std]{{k}-({a-c}/{d-b}){h}={c*h-a*h+d*k-b*k}/{d-b}} \\end{eqnarray} \\]

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Hence the equation of the line is
\\[y = \\simplify[std]{({a-c}/{d-b})x+{c*h-a*h+d*k-b*k}/{d-b}}\\]

\n ", "name": "Equation of line perpendicular to given line, ", "preamble": {"js": "", "css": ""}, "statement": "\n

Find the equation of the straight line which:

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Input your answer in the form $mx+c$ for suitable values of $m$ and $c$.

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Input $m$ and $c$ as fractions or integers as appropriate and not as decimals.

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If you input $m$ as a fraction, put brackets ( ) around the fraction. For example, if your answer for $m$ is $\\dfrac{-2}{3}$ and your answer for $c$ is $\\dfrac{7}{5}$, you should write $(-2/3)x+7/5$.

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Click on Show steps if you need help, you will lose 1 mark if you do so.

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Find the equation of the straight line perpendicular to the given line that passes through the given point $(a,b)$.

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The equation of the line is of the form $y=mx+c$.

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The gradient $m$ will be the $\\dfrac{-1}{n}$ where $n$ is the gradient of the line $\\displaystyle \\simplify{{(b-d)/n2}x+{(c-a)/n2}y={(b*c-a*d)/n2}}$, so start by calculating the gradient of the second line. Having calculated $n$, calculate $m=\\dfrac{-1}{n}$ and finally calculate the constant term $c$ by noting that $y=\\var{k}$ when $x=\\var{h}$.

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$y=\\;\\phantom{{}}$[[0]]

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Input all numbers as fractions or integers as appropriate and not as decimals.

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