# Math Help - Finding the equation of the straight line

1. ## Finding the equation of the straight line

Diagram shows a kite $PQRS$.Points $P$ and $R$ lie on $Y-$axis and $X-$axis respectively.Point $S$ is

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

a)Given $PS = PQ$, Find

i)the equation of $PR$

2. First, find the coordinates of the point P.

Let it be (x, y)

The distances PS and PQ are given by:

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

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

Since they are equal;

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

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

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

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

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

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.

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.

3. mid point of QS i get $\left(\frac{5}{2},\frac{17}{2}\right)$ normal i

the normal is -1

therefore,the equation is

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

$y= -x + 11$

right?

4. Correct!

5. so,the point p is (0,11)?

and the point r is?

6. You know that R has its y coordinates as 0. So, put y = 0 and x = ??

7. owh..

point r =(11,0)

point p=(0,11)?

8. Right!