Question

find the equation of a line perpendicular to $3x + y = 3$ that passes through the point $(6,5)$. answer $\circ\\ x + 3y = 21$ $\circ\\ y = \frac{1}{3}x + 3$ $\circ\\ y = -\frac{1}{3}x + 3$ $\circ\\ 3x + y = 23$

Explanation:

Step1: Find slope of given line

Rewrite \(3x + y = 3\) as \(y=-3x + 3\). Slope (\(m_1\)) is \(-3\).

Step2: Find slope of perpendicular line

For perpendicular lines, \(m_1\times m_2=-1\). So \(-3\times m_2=-1\) gives \(m_2=\frac{1}{3}\).

Step3: Use point - slope form

Point - slope: \(y - y_1=m(x - x_1)\), with \((x_1,y_1)=(6,5)\) and \(m = \frac{1}{3}\).
\(y - 5=\frac{1}{3}(x - 6)\)
\(y-5=\frac{1}{3}x - 2\)
\(y=\frac{1}{3}x+3\)
Also, check by plugging \((6,5)\) into \(x + 3y=21\): \(6+3\times5=6 + 15 = 21\), which holds. But let's check slope of \(x + 3y=21\) (rewrite as \(y=-\frac{1}{3}x + 7\)), slope is \(-\frac{1}{3}\), not \(\frac{1}{3}\). Wait, no, earlier calculation for slope of perpendicular line: if \(m_1=-3\), then \(m_2=\frac{1}{3}\) (since \(-3\times\frac{1}{3}=-1\)). Now, let's check the equation \(y=\frac{1}{3}x + 3\): plug \(x = 6\), \(y=\frac{1}{3}\times6+3=2 + 3=5\), which matches the point \((6,5)\). And the slope is \(\frac{1}{3}\), which is perpendicular to \(-3\). Wait, but also \(x + 3y=21\) when \(x = 6\), \(y = 5\) (as \(6+15 = 21\)), but slope of \(x + 3y=21\) is \(-\frac{1}{3}\), which is not perpendicular to \(-3\) (since \(-3\times(-\frac{1}{3}) = 1
eq - 1\)). So the correct equation with slope \(\frac{1}{3}\) and passing through \((6,5)\) is \(y=\frac{1}{3}x + 3\). Also, let's check the other options:

• \(y=-\frac{1}{3}x + 3\): slope \(-\frac{1}{3}\), not perpendicular.
• \(3x + y=23\): slope \(-3\), same as original line, parallel.

So the correct equation is \(y=\frac{1}{3}x + 3\) (also, \(x + 3y=21\) was a miscalculation earlier, its slope is \(-\frac{1}{3}\), not \(\frac{1}{3}\)). Wait, no, let's re - do the slope of \(x + 3y=21\): \(3y=-x + 21\), \(y=-\frac{1}{3}x+7\), slope is \(-\frac{1}{3}\). The slope of the line perpendicular to \(3x + y = 3\) (slope \(-3\)) should be \(\frac{1}{3}\) (since \(-3\times\frac{1}{3}=-1\)). So the line with slope \(\frac{1}{3}\) passing through \((6,5)\) is \(y=\frac{1}{3}x + 3\) (when \(x = 6\), \(y = 2+3 = 5\)) and also, let's check \(x+3y = 21\) again: if we write it as \(y=-\frac{1}{3}x + 7\), slope is \(-\frac{1}{3}\), which is not perpendicular. So the correct equation is \(y=\frac{1}{3}x + 3\).

Answer:

\(y=\frac{1}{3}x + 3\) (the option with \(y=\frac{1}{3}x + 3\))