# Finding the equation of the straight line

• Oct 18th 2010, 06:12 AM
mastermin346
Finding the equation of the straight line
Diagram shows a kite $\displaystyle PQRS$.Points $\displaystyle P$ and $\displaystyle R$ lie on $\displaystyle Y-$axis and $\displaystyle X-$axis respectively.Point $\displaystyle S$ is

$\displaystyle (-1,5) Q(6,12)$

a)Given $\displaystyle PS = PQ$, Find

i)the equation of $\displaystyle PR$
• Oct 18th 2010, 06:20 AM
Unknown008
First, find the coordinates of the point P.

Let it be (x, y)

The distances PS and PQ are given by:

$\displaystyle |PS| = \sqrt{(12-y)^2 + (6-x)^2}$

$\displaystyle |PQ| = \sqrt{(5-y)^2 + (-1-x)^2}$

Since they are equal;

$\displaystyle \sqrt{(5-y)^2 + (-1-x)^2} = \sqrt{(12-y)^2 + (6-x)^2}$

$\displaystyle (5-y)^2 + (-1-x)^2 = (12-y)^2 + (6-x)^2$

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

$\displaystyle (5-y)^2 + (-1-0)^2 = (12-y)^2 + (6-0)^2$

$\displaystyle (5-y)^2 + 1 = (12-y)^2 + 36$

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.
• Oct 18th 2010, 06:49 AM
mastermin346
mid point of QS i get $\displaystyle \left(\frac{5}{2},\frac{17}{2}\right)$ normal i

the normal is -1

therefore,the equation is

$\displaystyle y- \frac{17}{2} = -1\left(x-\frac{5}{2}\right)$

$\displaystyle y= -x + 11$

right?
• Oct 18th 2010, 06:56 AM
Unknown008
Correct! (Smile)
• Oct 18th 2010, 06:59 AM
mastermin346
so,the point p is (0,11)?

and the point r is?
• Oct 18th 2010, 07:15 AM
Unknown008
You know that R has its y coordinates as 0. So, put y = 0 and x = ?? (Smile)
• Oct 18th 2010, 07:19 AM
mastermin346
owh..

point r =(11,0)

point p=(0,11)?
• Oct 18th 2010, 07:21 AM
Unknown008
Right! (Smile)