Question

The vertices of square PQRS are P(−4, 7), Q(5, 4), R(2,−5) and S(−7,−2). Which of the following shows that its diagonals are congruent perpendicular bisectors of each other?

Answer 1

Answer: The first option is correct.

Step-by-step explanation: Given that the vertices of square PQRS are P(−4, 7), Q(5, 4), R(2,−5) and S(−7,−2). So, PR and QS are its two diagonals.

First, we will measure PR and QS as follows.

`PR=√((2+4)^2+(-5-7)^2)=√(36+144)=√(180)=3√(20),\\\\QS=√((-7-5)^2+(-2-4)^2)=√(144+36)=√(180)=3√(20).`

Therefore, PR = QS = 3√20.

Now, slope of PR is

`m_1=(-5-7)/(2+4)=-2`

and slope of QS is

`m_2=(-2-4)/(-7-5)=(1)/(2).`

Hence, PR ⊥ QS.

Again, mid-point of PR is

`\left((-4+2)/(2),(7-5)/(2)\right)=(-1,1),`

and mid-point of QS is

`\left((5-7)/(2),(4-2)/(2)\right)=(-1,1).`

Thus, the correct option is

`(a)~~\textup{PR}=\textup{QS}=3√(20).\\\\\textup{Slope of PR} =-2,~\textup{and slope of QS}=(1)/(2).\\\\so~PR\perpQS,~(-1,1)~\textup{is the mid-point of PS and QS},\\\\so,~\textup{PR and QS bisect each other}.`

Answer 2

Option (A)

Explanation:

The vertices of square PQRS are P(−4, 7), Q(5, 4), R(2,−5) and S(−7,−2).

Now, Join the diagonals PR and QS,

now, PR=
`\sqrt{(2+4)^(2)+(-5-7)^(2)}`

=
`√(36+144)`

=
`3√(20)`

Also, QS=
`\sqrt{(5+7)^(2)+(4+2)^(2)}`

=
`√(144+36)`

=
`3√(20)`

Therefore, PR is congruent to QS that is PR≅QS.

Slope of PR=
`(y_(2)-y_(1))/(x_(2)-x_(1))`

=
`(-5-7)/(2+4)=(-12)/(6)=-2`

Slope of QS=
`(-2-4)/(-7-5)=(-6)/(-12)=(1)/(2)`

Thus, PR⊥QS.

Now, Mid point of PR=
`((x_(1)+x_(2))/(2), (y_(1)+y_(2))/(2))`

=
`((-4+2)/(2), (7-5)/(2))`

=
`(-1,1)`

Also, mid point of QS=
`((5-7)/(2), (-2+4)/(2))`

=
`(-1,1)`

Therefore, (-1,1) is the mid point of both PR and QS, so PR and QS bisect each other.

Hence, option (A) is correct.