MATH SOLVE

3 months ago

Q:
# Find the equation of a line perpendicular to y - 3x = – 8 that passes through the point (3, 2). (answer in slope-intercept form)A) y = -3x + 2 B) y = -3x + 3 C) y = - 1/3 x + 2D) y = - 1/3 x + 3

Accepted Solution

A:

We first rewrite the line y - 3x = – 8 in slope-intercept form as:

y=3x-8.

The form y=mx+n is called the "slope-intercept form" because m tells us the slope and n tells us the y-intercept of the line.

Thus, the slope of the line is 3. We know that if a is the slope of any line perpendicular to our line, then the product of these slopes, 3a, is -1.

This means that the slope a is equal to -1/3. We are also given that the perpendicular line contains (3, 2). Thus, we write the equation:

[tex]y-2= -\frac{1}{3}(x-3)\\\\y-2=-\frac{1}{3}x+1\\\\y=-\frac{1}{3}x+3[/tex]

Answer: [tex]y=-\frac{1}{3}x+3[/tex]

y=3x-8.

The form y=mx+n is called the "slope-intercept form" because m tells us the slope and n tells us the y-intercept of the line.

Thus, the slope of the line is 3. We know that if a is the slope of any line perpendicular to our line, then the product of these slopes, 3a, is -1.

This means that the slope a is equal to -1/3. We are also given that the perpendicular line contains (3, 2). Thus, we write the equation:

[tex]y-2= -\frac{1}{3}(x-3)\\\\y-2=-\frac{1}{3}x+1\\\\y=-\frac{1}{3}x+3[/tex]

Answer: [tex]y=-\frac{1}{3}x+3[/tex]