Prove that: If the diagonals of a quadrilateral bisect each other at right angles, then the quadrilateral is a rhombus

Proved below

Step-by-step explanation:

I have attached an image showing the diagram of a quadrilateral with its diagonals bisecting each other at right angles.

From the image attached, the quadrilateral is ABCD and the diagonals AC and BD are seen to bisect each other at right angles at the point O.

In the 2 triabgles ΔAOB and ΔAOD, we see that;

AO = CO (Common parts of a line)

Also; OB = OD

(since point O is the mid-point of BD)

Angles; ∠AOB = ∠AOD

(Since Each are both at right angles of 90°)

Thus;

ΔAOB ≈ ΔAOD

(from SAS criteria for triangle similarity)

This means that their corresponding parts will be equal and we have;

AB = AD

AB = BC

BC = CD

CD = AD

From the 4 equal corresponding parts deduced, we can say that;

AB = BC and CD = DA

Therefore, the quadrilateral ABCD is a rhombus since all the sides are equal.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)