Question

The points (1, 7), (4, 2), (-1, -1), (-4, 4) are the vertices of a _____

• A. Parallelogram
• B. Square
• C. Rhombus
• D. None of these

Answer 1

The correct option is B

Square

Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the give points.

Let's find the sides after joining these four points.

So, AB = $\sqrt{(1-4{)}^2+(7-2{)}^2}=\sqrt{9+25}=\sqrt{34}$

BC = $\sqrt{(4-1{)}^2+(2+1{)}^2}=\sqrt{25+9}=\sqrt{34}$

CD = $\sqrt{(-1+4{)}^2+(-1-4{)}^2}=\sqrt{9+25}=\sqrt{34}$

DA = $\sqrt{(1+4{)}^2+(7-4{)}^2}=\sqrt{25+9}=\sqrt{34}$

We observe that all the sides are equal.

If A, B, C and D are the vertices of the square, and then its diagonals should also be equal. Now check for

AC = $\sqrt{(1+1{)}^2+(7+1{)}^2}=\sqrt{4+64}=\sqrt{68}$

BD = $\sqrt{(4+4{)}^2+(2-4{)}^2}=\sqrt{64+4}=\sqrt{68}$

Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC & BD are also equal, therefore, ABCD is a square.