Please help! I will mark Brainliest. Write an equation to a linear line in Standard Form that is perpendicular to -2x-3y=5 and through the point (1/3, -2).

You are not allowed to you point slope formula to solve this.

y = (3/2)x - 5/2

Step-by-step explanation:     5 + 2x

-2x-3y=5, solved for y, is y = , with a slope of -2/3.  A line

                                                     -3

perpendicular to this one has a slope which is the negative reciprocal of -2/3, that is, +3/2.  Using the slope-intercept formula, we calculate the y-intercept of the line that is perpendicular to the given one and passes through (1/3, -2):  y = mx + b => -2 = (3/2)(1/3) + b, or

-2 = 1/2 + b.  Thus, b must be -2 1/2, or -5/2.

The desired equation is y = (3/2)x - 5/2.

y - 3/2x = - 5/2

Step-by-step explanation:

we calculate the slope of -2x - 3y = 5, by rewriting in slope-intercept form:

- 3y = 2x + 5,

y = -2/3x - 5/3

slope = -2/3

perpendicular slope (negative reciprocal) = 3/2

So the perpendicular equation is in the form y = 3/2x + b. Provided this passes through the point (1/3, -2), let's substitute and solve for b:

-2 = 3/2(1/3) + b,

b = -5/2

In slope-intercept form we have the equation as y = 3/2x - 5/2. But it asks for it in standard form. Therefore it's y - 3/2x = - 5/2.