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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.

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The answer box for the third part plots the function which allows the student to check their answer against the graph before submitting.

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This particular example has a positive gradient.

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Calculate the gradient, $m$, of the straight line between these two points.

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$m=$ [[0]]

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Use this gradient and the coordinates of the points to calculate the $y$-intercept, $c$.

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$c=$ [[0]]

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Give the equation of the straight line through these points in the form $y=mx+c$.

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$\\displaystyle y=$ [[0]]

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Use the graph to plot your answer and check that it goes through these points.

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You must input your answer in the form y = mx +c where m and c are numbers.

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We find the equation of a straight line passing through two points by finding the gradient and the $y$-intercept of the line.

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#### a)

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We can find the gradient ($m$) using the points $A = (x_1,y_1)=(\\var{xa},\\var{ya})$ and $B = (x_2,y_2)=(\\var{xb},\\var{yb})$.

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As the definition of gradient is the ratio of vertical change ($y_2-y_1$) to horizontal change ($x_2-x_1$).

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\\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} \n #### b) \n 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{.} \

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We can then also use this equation with the other point's coordinates to check our answer.

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Let's use point $A$ first:

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\\\begin{align} c &= y_1-mx_1 \\\\ &= \\var{ya}-\\var[fractionnumbers]{m}\\times\\var{xa} \\\\ & = \\simplify[fractionnumbers]{{ya-m*xa}}\\text{.} \\end{align} \

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We then check this against point $B$:

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\\\begin{align} y_2 &= mx_2 + c \\\\[0.5em] &= \\simplify[fractionNumbers]{{m}{xb}+{c}} \\\\[0.5em] &= \\var[fractionnumbers]{m*xb+c}\\text{.} \\end{align} \

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#### c)

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We can now substitute these values for $m$ and $c$ into $y=mx+c$  to get:

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\$y=\\simplify[!noLeadingMinus,fractionNumbers,unitFactor]{{m} x+ {c}}\\text{.}\$

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The green line drawn on the graph represents the above line equation.

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{correctPoints()}

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$A = \\log_{10} [\\frac{I_o}{I}] = \\epsilon c l$

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In this question we will identify the equation of the straight line passing through points  $A=(\\var{xa},\\var{ya})$ and  $B=(\\var{xb},\\var{yb})$ $C=(\\var{xc},\\var{yc})$in the form $y = mx + c$.

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{plotPoints()}

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