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write the equation of the line parallel to x -3y = -12 that passes through [-3,4]
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from standard form of a line such as \[Ax+By+C=0\]Parallel lines have the same value as coefficients A and B of x and y, respectively. In case of the given line, \[x-3y+12=0\]therefore our A=1, B=-3. Therefore the line parallel to this has an equation \[x-3y+D=0\]And since this line passed through [-3,4], it satisfy the said equation. Just plug-in x and y to solve for the constant D...\[(-3)-3(4)+D=0\]\[-3-12+D=0\]\[D=15\]therefore the parallel line to \(x-3y=-12\) is \[x-3y+15=0\]or\[x-3y=-15\]
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