// Numbas version: finer_feedback_settings {"name": "Linear graphs: Find the equation of a line 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Linear graphs: Find the equation of a line 2", "tags": [], "metadata": {"description": "

Find the equation of the line that passes through given points

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You can calculate the gradient first, and then substitute this into the equation $y=mx+c$, and rearrange to find $c$.

It is more efficient, however, to calculate the whole thing in one go, using the equation:

$\\frac{y - y_1}{x - x_1} = \\frac{y_2 - y_1}{x_2 - x_1}$

and then rearrange this into any suitable form.

In this question we have:
$(x_1, y_1) = (\\var{ax}, \\var{ay})$ and $(x_2, y_2) = (\\var{bx}, \\var{by})$

Therefore

\\[ \\begin{split} \\frac{y - y_1}{x - x_1} &= \\frac{y_2 - y_1}{x_2 - x_1} \\\\ \\\\
\\frac{y - \\var{ay}}{x - \\var{ax}} &= \\frac{\\var{by} - \\var{ay}}{\\var{bx} - \\var{ax}} \\\\ \\\\
\\frac{y - \\var{ay}}{x - \\var{ax}} &= \\frac{\\var{by-ay}}{\\var{bx-ax}} \\\\ \\\\
\\var{bx-ax}(y - \\var{ay}) &= \\var{by-ay}(x - \\var{ax})\\\\ \\\\
\\simplify{{bx-ax}(y - {ay})} &= \\simplify{{by-ay}(x - {ax})} \\\\ \\\\
\\var{(bx-ax)}y - \\var{(bx-ax)*ay} &= \\var{(by-ay)}x + \\var{(ay-by)*ax} \\\\ \\\\
\\var{(bx-ax)}y &= \\var{(by-ay)}x + \\var{(ay-by)*ax+(bx-ax)*ay} \\\\ \\\\
y &= \\simplify{({by-ay}/{bx-ax})x + {(ay-by)*ax/(bx-ax)+(bx-ax)*ay/(bx-ax)}}
\\end{split} \\]

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Find the equation of the line that passes through the points:

$(\\var{ax},\\var{ay})$ and $(\\var{bx}, \\var{by})$.

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