One side of the square has endpoints, (-3,1) and (2,5). Find the four equations of lines that create this square .

Question

You would like to create a square on the coordinate plane. 

Answer 1

We are thinking the same thing.

The line through the given points is y= 4/5x+17/5

 

The distance which we will need later is the square root of 41 or rad41 for short.

 

Two distinct lines are perpendicular to the given line at these endpoints.

 

The second line has slope -5/4 and passes through (-3,1). It's equation is then y=-5/4x +-11/4.

 

Likewise the 3rd line also has slope -5/4 but passes through (2,5).

It's equation is y=-5/4x+15/2

 

Now there are two such points (x,y) that satisfy 2 conditions:

 

(1) the distance from (x,y) to (-3,1) is rad41. Per the distance formula,

(X+3)^2 + (y-1)^2=41

 

(2) the point (x,y) must lie on the line y=-5/4x + -11/4

 

The algebra is left for you. You'll need the quadratic formula or some clever factoring.

 

The two points are (1,-4) and (-7,6)

 

(1,4) must lie on the line with slope 4/5. So it's equation is

y= 4/5x-24/5

 

It intersects the 3rd line at (6,0).

 

On the other hand, if its (-7,6), the slope is still 4/5. So it's equation is

Y=4/5x + 58/5

  Intersects at (-2,10)

 

Answer 2

As noted in my comment before there are 2 squares that

could be made.  One square would be plotting the new

corners to the right and down from the original corners;

The other would be plotting to the left and up.  I will work

this problem with the orientation of the new corners being

below the original line segment ... right and down.

 

We have a line segment from

(-3, 1) to (2, 5)

The slope of this line is m = (5-1)/(2-(-3)) = 4/5

 

Parallel slope for opposite line is m = 4/5

Perpendicular line for bisector lines is m_⊥ =  -5/4

 

We can find the corner point down and to the right from

(-3,1) from the perpendicular slope.  -5/4 means we go down

5 units in the y coordinate and right 4 in the x coordinate.

(-3, 1) becomes (-3+4, 1-5) = (1, -4)

 

We have a 2nd line segment from (-3, 1) to (1, -4)

 

with the slope of 4/5 between the original points

that means we went up 4 spaces and right 5 to

get from (-3,1) to (2,5)...

 

From our new corner point of (1, -4) we go up 4 in y,

and right 5 in x:  (1+5, -4 + 4) = (6, 0)

 

We have a 3rd line segment from (1, -4) to (6, 0)

 

Our 4th line segment is from (6, 0) back to (2, 5)

which is perpendicular to the original line segment

and has a slope of -5/4 as it should.

 

Line 1:  (-3, 1), (2,5), m = 4/5

            y = mx + b  for point slope form

            m = slope, b = y-intercept

            y = (4/5)x + b

            Using (-3, 1) as (x, y) solve for b

            1 = (4/5)(-3) + b

            1 = -12/5 + b

            b = 1 + 12/5 = 1 + 2 2/5 = 3 2/5 or 3.4

            y = (4/5)x + 3 2/5   ...   -3≤x≤2

           

 

Line 2:  (-3, 1) to (1, -4), m = -5/4

            y = (-5/4)x + b

            Again use one of the points to solve for b

            to get the equation of the line.

            Note, this is a line segment so it has the

            restriction that -3≤x≤1

 

Line 3:  (1, -4) to (6, 0), m = 4/5

             y = (4/5)x + b

             Follow same procedure as previous.

             1≤x≤6

 

Line 4:  (2, -5) to (6, 0), m = -5/4

            y = (-5/4)x + b

            Follow same procedure as previous.

            2≤x≤6