Question

Write the standard form of the equation of theline passing through the point (-5, 3) andperpendicular to the line - 2x – 3y = -6.

Answer 1

The first step we need to follow is to find the slope of the perpendicular line to -2x-3y = -6.

Then, we have:

$-2x-3y=-6\Rightarrow-3y=-6+2x\Rightarrow y=\frac{-6}{-3}+\frac{2}{-3}x\Rightarrow y=-\frac{2}{3}x+2$

Then, the equivalent equation for the line has a slope of m1 = - 2/3. The perpendicular line must have an inverse and reciprocal to this slope, that is m2 = 3/2. The product of these slopes must be -1, that is: m1 * m2 = -2/3 * 3/2 = -1.

The second step is to find the line equation for the perpendicular. We know that it has a slope of m = 3/2, and that passes through the point (-5, 3). Then, we can use the point-slope form of the line:

$y-y_1=m(x-x_1)$$y-3=\frac{3}{2}(x-(-5))\Rightarrow y-3=\frac{3}{2}(x+5)\Rightarrow y-3=\frac{3}{2}x+\frac{15}{2}$

Then

$y=\frac{3}{2}x+\frac{15}{2}+3\Rightarrow y=\frac{3}{2}x+\frac{21}{2}$

The Standard Form of the line is of the form:

$Ax+By=C$

Then, taking the previous equation:

$2y=3x+21\Rightarrow-3x+2y=21$

Therefore, the Standard Form of the equation is -3x + 2y = 21.