Diagram shows a kite .Points and lie on axis and axis respectively.Point is

a)Given , Find

i)the equation of

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- October 18th 2010, 06:12 AMmastermin346Finding the equation of the straight line
Diagram shows a kite .Points and lie on axis and axis respectively.Point is

a)Given , Find

i)the equation of - October 18th 2010, 06:20 AMUnknown008
First, find the coordinates of the point P.

Let it be (x, y)

The distances PS and PQ are given by:

Since they are equal;

Now, you know that the x coordinate of P is zero... it becomes simpler!

Finish that to get the y coordinate of P.

Find the midpoint of QS. I assume you know how to find that.

The line through MP (where M is the midpoint of QS) also passes through R. You got two points (P and M). You can find the equation of PR. (Happy)

OR!

Find the midpoint of QS. (1)

Find the gradient of QS. (2)

Find the normal of the gradient of QS. (3)

You get a gradient (in (3)) and a point (in (1)). Find the line through PR!

This method is shorter and makes use the the property in kites that their diagonals are perpendicular. - October 18th 2010, 06:49 AMmastermin346
mid point of QS i get normal i

gradient QS = 1

the normal is -1

therefore,the equation is

right? - October 18th 2010, 06:56 AMUnknown008
Correct! (Smile)

- October 18th 2010, 06:59 AMmastermin346
so,the point p is (0,11)?

and the point r is? - October 18th 2010, 07:15 AMUnknown008
You know that R has its y coordinates as 0. So, put y = 0 and x = ?? (Smile)

- October 18th 2010, 07:19 AMmastermin346
owh..

point r =(11,0)

point p=(0,11)? - October 18th 2010, 07:21 AMUnknown008
Right! (Smile)