// Numbas version: exam_results_page_options {"name": "Linear graphs: find a line 1 - parallel", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Linear graphs: find a line 1 - parallel", "tags": ["Category: Linear Graphs"], "metadata": {"description": "

Find the equation of a line given a point and a parallel line.

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{apps}

Rearrange the given equation to find the gradient:

\n

\\[  \\simplify[all]{{a}x + {b}y + {c} = 0} \\]

\n

\\[ y = \\simplify[FractionNumbers]{{-a/b}x-{c/b}} \\]

\n

Therefore the gradient, $m_1$, is equal to $\\simplify{{m}}$.

Since we want the {pop} line, then the gradient of the new line, $m_2$, will {popgrad}, and therefore $m_2 = \\simplify[FractionNumbers]{{newm}}$.

{apps2}

\n

We can now take the equation $y = mx + c$, and subsitute in $m_2$ along with $(x,y) = (\\var{x},\\var{y})$, to give:

\\[ \\var{y} = \\simplify[FractionNumbers]{{newm}} \\times \\simplify[FractionNumbers]{{x}} + c \\]

and rearrange to calculate 

\n

\\[ c = \\simplify[FractionNumbers]{{y} - {newm}*{x}} \\]
\\[ c = \\simplify[FractionNumbers]{{y - newm*x}} \\].

Therefore

\\[y = \\simplify[fractionNumbers]{{newm}*x +{y - newm*x}}\\].

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Find the equation of the line {pop} to

$\\simplify[all]{{a}x + {b}y + {c} = 0}$,

passing through the point $(\\var{x},\\var{y})$. 

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