A line has equation \(3x - 4y = 12\). (a) Find the gradient of this line. [2] (b) Find the equation of the line parallel to it that passes through the point...
Question 1 Report
A line has equation \(3x - 4y = 12\).
(a) Find the gradient of this line. [2]
(b) Find the equation of the line parallel to it that passes through the point \((8,\ -1)\). Give your answer in the form \(y = mx + c\). [1]
Answer Details
An equation given in the form \(ax+by=c\) does not show its gradient directly. Rearranging it into \(y=mx+c\) makes the gradient visible as the coefficient of \(x\). Parallel lines then share that gradient and differ only in their intercept.
(a) Starting from \(3x-4y=12\), add \(4y\) and subtract 12 from both sides:
\[4y=3x-12\quad\text{oe}\quad y=\frac{3x-12}{4}\][M1]
Dividing every term by 4 gives \(y=\frac{3}{4}x-3\), so the gradient is
\[\frac{3}{4}\quad\text{oe}\quad 0.75\][A1]
Note the sign: dividing \(-4y\) across correctly leaves a positive \(\frac{3}{4}\), and answering \(-\frac{3}{4}\) is the usual slip.
(b) A parallel line has the same gradient \(\frac{3}{4}\), so its equation is \(y=\frac{3}{4}x+c\). Substituting the point \((8,\ -1)\):
\[-1=\frac{3}{4}\times 8+c=6+c\quad\Longrightarrow\quad c=-7\]The line is
\[y=\frac{3}{4}x-7\]oe [B1]
A quick verification: at \(x=8\), \(\frac{3}{4}\times 8-7=6-7=-1\), which is the given \(y\)-coordinate, so the point does lie on the line.
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