Problem

Find the equation of the line perpendicular to the line \(2x - 3y = 6\) and passing through the point \((1, 2)\).

Answer

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Answer

Step 4: Substitute \(m = -\frac{3}{2}\) and \(b = \frac{7}{2}\) into the equation \(y = mx + b\) to get the equation of the line.

Steps

Step 1 :Step 1: Find the slope of the given line by rearranging \(2x - 3y = 6\) into slope-intercept form \(y = mx + b\). This gives \(y = \frac{2}{3}x - 2\). The slope of this line is \(\frac{2}{3}\).

Step 2 :Step 2: The slope of the line perpendicular to this line is the negative reciprocal of \(\frac{2}{3}\), which is \(-\frac{3}{2}\).

Step 3 :Step 3: Use the slope-intercept form \(y = mx + b\) to find the equation of the line. Substitute \(m = -\frac{3}{2}\), \(x = 1\), and \(y = 2\) into the equation to solve for \(b\). This gives \(2 = -\frac{3}{2} * 1 + b\), or \(b = \frac{7}{2}\).

Step 4 :Step 4: Substitute \(m = -\frac{3}{2}\) and \(b = \frac{7}{2}\) into the equation \(y = mx + b\) to get the equation of the line.

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