3. Find the equation of the line that goes through the point (2, 3) perpendicular to the line whose equation is x + y = 3.

In slope intercept form, the equation of the line will be

y = mx + b

where m is the slope and b is the y-intercept.

In problem 2, we sqaw that the slope of the line whose equation is x + y = 3 was -1. The slope of a perpendicular line would be the negative reciprocal of that or

m = 1

As with problem 2, the point on the line is

(x1, y1) = (2, 3)

We put this into the slope intercept form of the equation

y = mx + b

and get

(3) = (1)(2) + b


3 = 2 + b

We subtract 2 from both sides to solve for b, and get

1 = b

so the equation of the perpendicular line is

y = x + 1

The foot of the point is where this perpendicular line through the point meets the given line. We set up a system of simultaneous linear equations.

This system is set up best for the substitution method. Substitute the formula which is equal to y from the second equation in for the y in the first equation.

x + (x + 1) = 3

Remove parentheses,

x + x + 1 = 3

combine like terms,

2x + 1 = 3

subtract 1 from both sides,

2x = 2

and divide both sides by 2.

x = 1

Now that we know that the x coordinate of the point is 1, we can substitute it in for the x in either one of the equations and get that

y = 2

So the coordinates for the foot are

(1, 2)

The graph looks like

To find the reflection of (2, 3) about the line whose equation is x + y = 3, note that the x-coordinate of the given point, 2, is 1 more than the x-coordinate of the foot, 1. As a result, the x- coordinate of the reflection will be 1 less than the x-coordinate of the foot, or 0. Since the reflection is on the line whose equation is

y = x +1,

the y coordinate will be

(0) + 1 = 1.

The reflection is then