Question

Give the slope-intercept form of the equation of the line that is perpendicular to -2x + 3y = 10 and contains P(10, -10).

a) y + 10 = -{3}/{2}(x - 10)

b) y + 10 = {3}/{2}(x - 10)

c) y = -{3}/{2}x - 5

d) y = -{3}/{2}x + 5

Answer 1

The slope-intercept form of the equation for the line perpendicular to -2x + 3y = 10 and containing P(10, -10) is a) y + 10 = -{3}/{2}(x - 10).

Step-by-step explanation:

To determine the slope-intercept form of the equation of the line that is perpendicular to -2x + 3y = 10 and contains the point P(10, -10), we first need to find the slope of the given line. We can convert the given equation to slope-intercept form (y = mx + b) to find the slope. Starting with the given equation:

-2x + 3y = 10 ⇒ 3y = 2x + 10 ⇒ y = ⅔x + ⅔3

The slope of the given line is ⅔. Since we are looking for a line perpendicular to the given line, we want the negative reciprocal of the given slope. The negative reciprocal of ⅔ is -{3}/{2}. Now we use the point-slope form (y - y1 = m(x - x1)) to write the equation of the desired line passing through point P(10, -10).

y - (-10) = -{3}/{2}(x - 10)
Simplifying the equation:

y + 10 = -{3}/{2}(x - 10)

Therefore, the correct slope-intercept form of the equation is choice a) y + 10 = -{3}/{2}(x - 10).