If the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square then what is the length of the diagonal of square?

$\sqrt{68}$ units

The correct answer is Option (2) → $\sqrt{68}$ units

1. Identify the Vertices

Let the vertices be:

• $A = (1, 7)$
• $B = (4, 2)$
• $C = (-1, -1)$
• $D = (-4, 4)$

In a square, the diagonals connect opposite vertices. Here, the diagonals are $AC$ and $BD$.

2. Use the Distance Formula

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let's calculate the length of diagonal $AC$:

• $x_1 = 1, y_1 = 7$
• $x_2 = -1, y_2 = -1$

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

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

$AC = \sqrt{4 + 64}$

$AC = \sqrt{68} \text{ units}$

3. Verification (Diagonal $BD$)

• $x_1 = 4, y_1 = 2$
• $x_2 = -4, y_2 = 4$

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

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

$BD = \sqrt{64 + 4}$

$BD = \sqrt{68} \text{ units}$

Since $AC = BD$, the length of the diagonal is consistent.

Conclusion

The length of the diagonal of the square is $\sqrt{68}$ units.