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

Answer:-3x + 2y = 21.

Explanation:

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

Then, we have:

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:

Then

The Standard Form of the line is of the form:

Then, taking the previous equation:

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

Answer 2

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=(-6)/(-3)+(2)/(-3)x\Rightarrow y=-(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=(3)/(2)(x-(-5))\Rightarrow y-3=(3)/(2)(x+5)\Rightarrow y-3=(3)/(2)x+(15)/(2)`

Then

`y=(3)/(2)x+(15)/(2)+3\Rightarrow y=(3)/(2)x+(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.