Coordinate geometry with given parameters.

[Aguy1]

Hello,

Please forgive me if it's in the wrong sub-forum because don't know where to place it.

I need help solving this problem it's chapter 1 and in our class we are already in chapter 5 so I might sound like a fool asking the teacher about it, I was revising and decided to do some questions then I stumbled on this question and got stuck.

Q.The coordinates of the points A and B are (2,3) and (4,-3) respectively,coordinates of P is (2+2t,3-t)

1.Find the values of t such that the length of AP is 5 units.

2. Find the value of t such that OP is perpendicular to l (where O is the origin). Hence, find the length of the perpendicular from O to l.

Just in case here are the answers 1. sqrt5 or -sqrt5 2. t= (-1/5,8/5 sqrt5)

I know in the first I should use the distance formula but I'm not getting the right answer.

I'm sorry if I sound too demanding.

 

[Petrus]

Re: Coordinate geometry.

Hello aguy1,
lets say we got \(\displaystyle A=(x,y)\) and \(\displaystyle B=(p,g)\) Then \(\displaystyle AB=(p-x,g-y)\) and the length of A is \(\displaystyle |A|=\sqrt{x^2+y^2}\) does this help you?

Regards,
\(\displaystyle |\pi\rangle\)

 

[Aguy1]

Re: Coordinate geometry.

Hello aguy1,
lets say we got \(\displaystyle A=(x,y)\) and \(\displaystyle B=(p,g)\) Then \(\displaystyle AB=(p-x,g-y)\) and the length of A is \(\displaystyle |A|=\sqrt{x^2+y^2}\) does this help you?

Regards,
\(\displaystyle |\pi\rangle\)

Thanks a lot, I got the answer for the first one :).

what about second question? I'm guessing O coordinates is (0,0) since it's the origin and I know the gradient should be perpendicular m1*m1=-1, so which equation should I use now?

Again thank you.

 

[Deveno]

Re: Coordinate geometry.

You haven't told us what "l" is...

if l (I assume this is a lower case L) is the line traced out by the point P as t varies:

First, we need to find the slope of this line. we can write:

$l = (2,3) + t(2,-1)$

which makes it clear the slope of l is -1/2. Thus the perpendicular line has slope 2. Now solve:

$2 + 2t = 2(3 - t)$

(that is: find the point (x,2x) (the point on the line through the origin y = 2x) that lies on l).

 

[Aguy1]

Re: Coordinate geometry.

You haven't told us what "l" is...

Oh I'm very sorry, no coordinates of l just the line equation which is x+2y=8.

I hope you can forgive my foolishness.

It's a four part question but I finished both and the other two (which are the above, one which I just solved)

just for more info here are the two questions which I already solved.

(a) Find the equation of the line l through the point A(2,3) with gradient -1/2 (that's how I got l equation x+2y=8)

(b) Show that the point P with coordinates (2+2t, 3-t) will always lie on l whatever the value of t.

 

[Deveno]

Re: Coordinate geometry.

So my guess was correct, then, as:

$(2,3) + t(2,-1)$ passes through the points (2,3) (t = 0), and (4,2) (t = 1), so the slope is:

$\frac{2 - 3}{4 -2} = -\frac{1}{2}$.

Using the point-slope formula:

$y - y_1 = m(x - x_1)$ and the point (4,2)

We get:

$y - 2 = -\frac{1}{2}(x - 4)$
$y - 2 = \frac{-x}{2} + 2$
$y - 4 = \frac{-x}{2}$
$2y - 8 = -x$
$x + 2y = 8$

so it is the same line.