M12-13

Official Solution:


What is the equation of the line \(k\) that is perpendicular to line \(y = 2x\) and passes through point \((a, b)\)?

Since line \(k\) is perpendicular to \(y=2x\), its slope must be the negative reciprocal of the slope of \(y=2x\), which is \(2\). Therefore, the slope of line \(k\) is \(-\frac{1}{2}\). Using the formula for slope, which is the "rise over run" or change in \(y\) divided by change in \(x\), we can find the equation of line \(k\) passing through the point \((a,b)\) with slope \(-\frac{1}{2}\). Thus, we have the equation \(-\frac{1}{2}=\frac{y-b}{x-a}\), which simplifies to \(y=-\frac{x}{2}+\frac{a}{2}+b\).

(1) \(a = -b\).

If \(a=b=0\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}\), but if \(a = 1\), and \(b= -1\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{1}{2}-1\). These are two different equations, so statement (1) alone is not sufficient.

(2) \(a - b = 1\).

If \(a=1\) and \(b=0\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{a}{2}\), but if \(a = 0\), and \(b= -1\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}-1\). Again, these are two different equations, so statement (2) alone is not sufficient.

(1)+(2) We have two distinct linear equations \(a = -b\) and \(a - b = 1\), which we can solve for \(a\) and \(b\), and get the exact equation of line \(k\). Solving gives \(a=\frac{1}{2}\) and \(b=-\frac{1}{2}\). Therefore, the equation of line \(k\) is \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{1}{4}-\frac{1}{2}=-\frac{x}{2}-\frac{1}{4}\). Sufficient.


Answer: C