Question

Simon wanted to find the equation of a line that passes through (–6, 5) and is perpendicular to the graph of 3x + 2y = –11. His work is shown below.
1. 3x + 2y = –11 written in slope-intercept form is y = x , so the slope is .
2. The slope of the perpendicular line is .
3. Substitute the point and the new slope into point-slope form to get y – 5 = (x – (–6)).
4. Simplifying, the line is y – 5 = (x + 6)

Answer 1

Answer:

The answer is C:

No, the slope of the line perpendicular to the line 3x + 2y = -11 should be 2/3.

Step-by-step explanation:

I did the question.

Answer 2

"3x + 2y = –11 written in slope-intercept form is y = x." Not so.

To find the slope of this line, solve for y:

2y = -3x - 11 => y = (-3/2)x - 11/2. The slope of the given line is -3/2. The

graph of the new (perpendicular) line will then be the negative reciprocal of -3/2, or +2/3.

Now use the point-slope formula for the equation of a straight line to determine the equation of this new (perpendicular) line thru (-6,5):

y-5 = (2/3)(x+6) (answer)