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#1 2005-10-05 10:31:05

Niruthik
Guest

Help me! Grade 10 algebra 1

Please, I really need this answer by today (before 12 hrs from now), for my take-home-quiz..
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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

#2 2005-10-05 15:30:11

MathsIsFun
Administrator

Offline

Re: Help me! Grade 10 algebra 1

Well, you could just sketch it on some graph paper !

http://www.mathsisfun.com/forum/images/square%20coords.gif

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 = √ (x + y)

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

length = √ (1 + 5) = √ (1 + 25) = 5.1 (approx)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#3 2005-10-06 03:31:23

rkman
Novice

Offline

Re: Help me! Grade 10 algebra 1

Mathematically for (a), you need to prove that each side of the quadrilateral is equal in magnitude and that opposing sides of the quadrilateral are parallel and perpendicular to the other sides.

Length is easy
length = √[(x2-x1) + (y2-y1)]
So the length from (-3,1) to (-1,4) is √[(-1--3) + (4-1)] = √[4 + 9] = √13
And the length from (-3,1) to (0,-1) is √[(0--3) + (-1-1)] = √[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 => a = b + c
                                            a = (√13) + (√13)
                                            a = 13 + 13 = 26
                                            a = √26

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

Last edited by rkman (2005-10-06 03:36:24)

#4 2005-10-06 07:09:22

niruthik
Guest

Re: Help me! Grade 10 algebra 1

thanks alot for the help rkman!

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