Parallel and perpendicular lines ​

A. Slope = -³/2

B. y = -³/2x - 6

C. ³/2x + y = -6

Step-by-step explanation:

A. Slope of the line would be the negative reciprocal of the slope of 2x - 3y = 12, which is written in standard form. Rewrite in slope-intercept form, y = mx + b, where m = slope.

Thus:

2x - 3y = 12

-3y = -2x + 12

y = -2x/-3 + 12/-3

y = ⅔x - 4

The slope of 2x - 3y = 12 is therefore ⅔.

The slope of the line perpendicular to 2x - 3y = 12 would be the negative reciprocal of its slope, ⅔ which is:

-³/2

Therefore, the slope of the line perpendicular to 2x - 3y = 12 is -³/2

B. To find the equation of the line in slope-intercept form, first find the equation in point-slope form using the point given (-6, 3) and slope of the line, -³/2.

Substitute a = -6, b = 3, and m = -³/2 into y - b = m(x - a).

Thus:

y - 3 = -³/2(x - (-6))

y - 3 = -³/2(x + 6)

Rewrite in slope-intercept form, y = mx + b

Multiply both sides by 2

2(y - 3) = -3(x + 6)

2y - 6 = -3x - 18

2y = -3x - 18 + 6

2y = -3x - 12

y = -3x/2 - 12/2

y = -³/2x - 6

C. Rewrite y = -³/2x - 6 in standard form.

y = -³/2x - 6

³/2x + y = -6

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