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 – (– 4. simplifying, the line is y – 5 = (x + 6)

"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)

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