Question

Write the slope-intercept form of an equation that passes through the given point and is perpendicular to the graph of the equation

(1,2), 2x+3y=-12​

Answer 1

First, let's rearrange this equation to solve for y in terms of x:

• 3y = -2x - 12
• y = (-2/3)x - 4

The slope of the given equation is -2/3. For a line perpendicular to this, the slope will be the negative reciprocal of -2/3, which is 3/2.

Now we can use the point-slope form to write the equation of the perpendicular line:

• y - y1 = m(x - x1)

Where:

• (x1, y1) is the given point (1, 2)
• m is the slope, which is 3/2
• Plug in the values:
• y - 2 = (3/2)(x - 1)

Simplify:

• y - 2 = (3/2)x - 3/2

Add 2 to both sides:

• y = (3/2)x - 3/2 + 2
• y = (3/2)x + 1/2

So, the equation of the line that passes through the point (1, 2) and is perpendicular to the graph of the equation 2x + 3y = -12 is:

• y = (3/2)x + 1/2