Length is easy

length = √[(x2-x1)² + (y2-y1)²]

So the length from (-3,1) to (-1,4) is √[(-1--3)² + (4-1)²] = √[4 + 9] = √13

And the length from (-3,1) to (0,-1) is √[(0--3)² + (-1-1)²] = √[9 + 4] = √13

To prove they are perpendicular to each other just find the gradients of each side and two of the lines will have gradient m whilst the other two sides will have gradient -1/m

The gradient can be found using (y2 - y1)/(x2 - x1)

So the gradient from (-3,1) to (-1,4) is => (4-1)/(-1-1) = -3/2

And the gradient from (-3,1) to (0,-1) is => (-1-1)/(0--3) = 2/3

So just do this for the other two sides, proving their lengths are root 13, and their gradients are m, -1/m, m and -1/m.

(b) is similar. Work out the equations of the diagonals using y = mx + c, filling in the values of y and x for each pair of diagnol coordinates. Then show that the gradient, m, for one diagonal is perpendicular to the other diagnol with gradient -1/m.

(c) is just working out the length of the diagonals using the formulae i gave you in part (a). Simple enough. The values of these lines should come out as √26 since this is what pythagorus theorem suggests.

Using pythagorus theorem => a² = b² + c²

a² = (√13)² + (√13)²

a² = 13 + 13 = 26

a = √26

Length = √[(2 - 1)² + (2 - -3)²] = √[1 + 25] = √26

]]>The dashed lines are the diagonals, and they do "bisect" (cut into two equal parts) each other, and they are equal in length

You can also do it mathematically, using:

length = √ (x² + y²)

for example, the diagonals are 1 unit in one one direction and 5 in the other apart:

length = √ (1² + 5²) = √ (1 + 25) = 5.1 (approx)

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A quadrilateral has vertices K(-1,4) L(2,2) M(0,-1) and N(-3,1). Verify that:

a) a quadrilateral is a square

b) each diagonal of the quadrilateral is the perpendicular bisector of the other diagonal

c) the diagonals of the quadrilateral are equal in length

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