Homework Help

The graph has picture of square with vertices at A(-3,1) B(1,4) C(4,0) D(0,-3). Is it...

user profile pic

sheriff97 | Student, Grade 9 | Honors

Posted October 27, 2012 at 5:06 AM via web

dislike 2 like

The graph has picture of square with vertices at A(-3,1) B(1,4) C(4,0) D(0,-3). Is it really a square? Justify.

geometry..do I need to do midpoint formula or m= y-y x-x

2 Answers | Add Yours

Top Answer

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 27, 2012 at 10:37 AM (Answer #1)

dislike 1 like

You need to prove that two adjacent sides of the square are perpendicular and they have equal lengths.

You may prove that the sides AB and BC are perpendicular if the product of the slopes `m_(AB)`  and `m_(BC)`  is equal to -1 such that:

`m_(AB) = (y_B - y_A)/(x_B - x_A)`

`m_(AB) = (4-1)/(1+3) => m_(AB) = 3/4`

`m_(BC) = (y_C - y_B)/(x_C - x_B)`

`m_(BC) = (0-4)/(4-1) = -4/3`

`m_(AB)*m_(BC)= (3/4)*(-4/3) = -1`

Notice that evaluating the product of the slopes yields -1, hence the lines AB and BC are perpendicular to each other.

You need to evaluate the lengths of AB and BC such that:

`AB = sqrt((x_B - x_A)^2 + (y_B - y_A)^2)`

`AB = sqrt((1+3)^2 + (4-1)^2) => AB = sqrt(16+9) => AB = sqrt25 => AB = 5`

`BC = sqrt((x_C - x_B)^2 + (y_C - y_B)^2)`

`BC = sqrt((4-1)^2 + (0-4)^2) => BC = sqrt(9+16) => BC = sqrt25 => BC = 5`

You also need to prove that the length of CD is of 5 such that:

`CD = sqrt((0-4)^2+(-3-0)^2) => CD = sqrt(25) => CD = 5`

Hence, the lengths of AB,BC,CD are equal, thus, since AB is also perpendicular to BC, you may state that ABCD is a square.

Top Answer

user profile pic

Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted October 27, 2012 at 10:47 AM (Answer #2)

dislike 1 like

To prove, we may use these four properties of a square which are:

(1) The fours sides are equal in length.

(2) The two diagonals are equal.

(3) Adjacent sides are perpendicular to each other.

(4) And, opposite sides are parallel.

To do so, plot the points and connect them starting from the left moving clockwise or counterclockwise. So the graph is:

To show that the length of the sides are equal, use the distance formula between two points which is

`d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)`

So the length of the red side with endpoints (-3,1) and (1,4) is:

`d_r = sqrt((1-(-3))^2+(4-1)^2) = sqrt(4^2+3^2)=sqrt25=5`

For the blue side with endpoints (1,4) and (4,0), its length is:

`d_b=sqrt((4-1)^2 + (0-4)^2) = sqrt(3^2+(-4)^2) =sqrt25=5`

For the green side with endpoints (4,0) and (0,-3), its length is:

`d_g=sqrt((0-4)^2+(-3-0)^2) = sqrt((-4)^2+ (-3)^2) = sqrt25=5`

And length of the  purple side with endpoints (0,-3) and (-3,1) is:

`d_p=sqrt((-3-0)^2+(1-(-3))^2)=sqrt((-3)^2+4^2) = sqrt25=5`

Since all four sides have a length of 5 units, this satisfy the first property of the square mentioned above.

Also, use the distance formula between two points to determine the length of the two diagonals.

The length of the diagonal with endpoints (-3,1) and (4,0) is:

`d_1 = sqrt((4-(-3))^2 + (0-1)^2)=sqrt(7^2+(-1)^2 )= sqrt50`

And the length of the diagonal with endpoints (1,4) and (0,-3) is:

`d_2=sqrt((0-1)^2+(-3-4)^2)=sqrt((-1)^2+(-7)^2)=sqrt50`

Since the two diagonals have the same length, this satisfies the second property listed above.

Next, let's determine the slope of each side. Use the formula of the slope between two points which is:

`m= (y_2-y_1)/(x_2-x_1)`

So the slope of the red side is:

`m_r = (4-1)/(1-(-3))= 3/4`

Slope of the blue side is:

`m_b=(0-4)/(4-1) = -4/3`

Slope of the green side is:

`m_g=(-3-0)/(0-4)=3/4`

And, slope of the purple side is:

`m_p=(1-(-3))/(-3-0) = -4/3`

Base on this values of slope, it shows that as we go from one side to the next, the slope of the next side is negative reciprocal of the slope of the side before it. When the slopes of the two lines are negative reciprocal of each other, the two lines are perpendicular. Hence, this satisfies the third property listed above.

Base on the above graph, the side opposite to the red line is the green line. And the slope of these two lines are both 3/4. Since the slopes of the two lines are the same, hence the red line is parallel to its opposite side which is green line.

Likewise, the blue line is parallel to the side opposite to it which is the purple line since both have the same slope.Thus it satisfies the fourth property mentioned above.

---------------------------------------------------------------------------

Therefore points (-3,1), (1,4) , (4,0) and (0,-3) are vertices of a square.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes