A triangle has vertices L(-5,6), M(-2,-3), and N(4,5). Write an equation for the line perpendicular to side LM that contains the point N.

Answer 1

The equation of a line takes the form y = mx + b

Where m is the slope of the line and b is the y-intercept. The values of y and x are the y and x values of any given point on that line.

Slope is the ratio of the change of our y values to the change of our x values.

If you want the slope of a line which is perpendicular to another line, you can find the other line's slope first and then recognize that the slope of a line perpendicular to that is the negative reciprocal.

Line LM goes from point (-5, 6) to point (-2, -3), so it's change in y is (-3) - (6) = -9

and it's change in x is (-2) - (-5) = +3

The slope of line LM is then -9/3 or -3.

The slope of a line perpendicular to LM would be -(1/(-3)) or simply 1/3.

To find the y-intercept, we will use the y = mx + b formula, and plug in a point that we know is on the line, (4, 5), because the problem asked us to be sure that the line passes through this point.

5 = (1/3)*(4) + b, when solved for b gives us b = 11/3

The equation of the line that is perpendicular to line LM and passes through point (4, 5) is then:

y = (1/3)x + 11/3

3y = x + 11 is also acceptable.

Answer 2

We can use the "slope-intercept" form: y = mx + b, where m is the slope and b is the y-intercept.

First find the slope of the line LM: slope = delta y / delta x = (6 + 3)/(-5 + 2) = -3. The perpendicular line is the negative inverse, which is +1/3 (m in the standard form)

Plug in m and the give point to the standard format:

5 = 1/3 * 4 + b

Solve for b: b = 5 - 4/3 = 11/3

So: y = x/3 + 11/3

Answer 3

slope of LM is 9/-3 = -3

slope of the perpendicular is 1/3

which eats N(4,5)

B = y - mx

= 5 - (1/3)(4)

= 5 - 4/3

= (15 - 4)/3

= 11/3

y = (1/3)x + 11/3