Question

The point (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a

• A. Square
• B. Rhombus
• C. Rectangle
• D. Quadrilateral

Answer 1

The correct option is A Square

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

$AB=\sqrt{(1-4{)}^2+(7-2{)}^2}=\sqrt{(-3{)}^2+(5{)}^2}$

$=\sqrt{9+25}$

$\therefore AB=\sqrt{34}$

$BC=\sqrt{(4-(-1){)}^2+(2-(-1){)}^2}=\sqrt{(5{)}^2+(3{)}^2}$

$=\sqrt{25+9}$

$\therefore BC=\sqrt{34}$

$CD=\sqrt{(-1-(-4){)}^2+(-1-4{)}^2}=\sqrt{(3{)}^2+(-5{)}^2}$

$=\sqrt{9+25}$

$CD=\sqrt{34}$

$AD=\sqrt{(1-(-4){)}^2+(7-4{)}^2}=\sqrt{(5{)}^2+(3{)}^2}$

$=\sqrt{25+9}$

$\therefore AD=\sqrt{34}$

$AC=\sqrt{(1-(-1){)}^2+(7-(-1){)}^2}=\sqrt{(2{)}^2+(8{)}^2}$

$=\sqrt{4+64}$

$AC=\sqrt{68}$

$BD=\sqrt{(4-(-4){)}^2+(2-4{)}^2}=\sqrt{(8{)}^2+(-2{)}^2}$

$=\sqrt{64+4}$

$\therefore BD=\sqrt{68}$

$\because AB=BC=CD=AD=\sqrt{34},$

$AC=BD=\sqrt{68}$

Thus, ABCD is a square.

So, option (a) is correct.