Questions: Find an equation for the line with the given properties. Express your answer using either the general form or the slope-intercept form of the equation of a line. Perpendicular to the line (y=1/3 x+1); containing the point ((-1,3)) The equation is (square) (Type an equation. Simplify your answer.)

Find an equation for the line with the given properties. Express your answer using either the general form or the slope-intercept form of the equation of a line. Perpendicular to the line (y=1/3 x+1); containing the point ((-1,3)) The equation is (square) (Type an equation. Simplify your answer.)
Transcript text: Find an equation for the line with the given properties. Express your answer using either the general form or the slope-intercept form of the equation of a line. Perpendicular to the line $y=\frac{1}{3} x+1$; containing the point $(-1,3)$ The equation is $\square$ (Type an equation. Simplify your answer.)
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Solution

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Solution Steps

Step 1: Find the slope of the perpendicular line

To find a line perpendicular to the given line \(y = 0.3333333333333333x + 1\), we use the negative reciprocal of \(m\). The slope of the perpendicular line, \(m'\), is \(-\frac1{0.333}\) = -3.

Step 2: Use the point-slope form to find the equation of the perpendicular line

Using the point \((-1, 3)\) and the slope \(m' = -3\), we apply the point-slope form: \(y - 3 = -3(x + 1)\) Simplifying to the slope-intercept form \(y = mx + b\), we solve for \(y\): \(y = -3.0x + 0\)

Final Answer:

The equation of the line perpendicular to \(y = 0.3333333333333333x + 1\) passing through \((-1, 3)\) is \(y = -3.0x + 0\).

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