GR11 Math: finding endpoints of chord within a circle.

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May 6th 2009, 06:43 PM
nathan02079

GR11 Math: finding endpoints of chord within a circle.

Given...

1. Point P is on the line $3x+y=26$ and is 10 units from the origin. Determine the coordinates of P.

I've gotten this far...

So:
$3x+y=26$
$x^2+y^2=\sqrt{10}$ .

Then $y=26-3x$ .

Plugging in y...

$x^2+(26-3x)^2=\sqrt{10}$

$10x^2-156x+676=\sqrt{10}$

This is where I'm stuck since I cannot factor it. When the quadratic formula is applied, the number inside the brackets is a negative number.

2. Given...

From a lighthouse, the range of visibility on a clear day is 40km. On a coordinate system, where O(0,0) represents the lighthouse, a ship is travelling on a course represented by $y=2x+80$ . Between which two points on the course can the ship be seen from the lighthouse?

I got this far...

$y=2x+80$
$x^2+y^2=\sqrt{40}$

Plugging it in...

$x^2+(2x+80)^2=\sqrt{40}$

$5x^2+320x+6400=\sqrt{40}$

Again, I am stuck with the same problem. I get a negative number when the quadratic formula is applied.

What am I doing wrong here? Thank you so much!

N.
May 6th 2009, 07:45 PM
pickslides

Quote:

Originally Posted by nathan02079

Given...

1. Point P is on the line $3x-y=26$ and is 10 units from the origin. Determine the coordinates of P.

I've gotten this far...

So:
$3x+y=26$

(Itwasntme)

Are these signs supposed to be different?
May 6th 2009, 07:46 PM
Soroban

Hello, nathan02079!

In your "distance formula", you're square-rooting when you should square.

Quote:

2. From a lighthouse, the range of visibility on a clear day is 40 km.
On a coordinate system, where O(0,0) represents the lighthouse,
a ship is travelling on a course represented by: $y\:=\:2x+80$ .
Between which two points on the course can the ship be seen from the lighthouse?

Code:

| | * | * Q * | o | * * | P * * | o * | * * * | * *| - - - - - - - - - - - * - - - - |O

$P$ and $Q$ have coordinates $(x,y)$ such that: . $OP = OQ = 10.$

We want: . $x^2 + y^2 \:=\:10^2$

So we have: . $x^2 + (2x+80)^2 \:=\:10^2 \quad\Rightarrow\quad 5x^2 + 320x + 4800 \:=\:0$

Factor: . $5(x+40)(x+24) \:=\:0$

Then: . $x \:=\:\text{-}40,\:\text{-}24 \quad\Rightarrow\quad y \:=\:0,\:32$

The ship can be seen from $(\text{-}40,\:0)$ to $(\text{-}24,\:32)$
May 6th 2009, 08:44 PM
nathan02079

Quote:

Originally Posted by pickslides

(Itwasntme)

Are these signs supposed to be different?

Whoops, typo there. :/

Quote:

Originally Posted by Soroban

Hello, nathan02079!

In your "distance formula", you're square-rooting when you should square.

Code:

| | * | * Q * | o | * * | P * * | o * | * * * | * *| - - - - - - - - - - - * - - - - |O

$P$ and $Q$ have coordinates $(x,y)$ such that: . $OP = OQ = 10.$

We want: . $x^2 + y^2 \:=\:10^2$

So we have: . $x^2 + (2x+80)^2 \:=\:10^2 \quad\Rightarrow\quad 5x^2 + 320x + 4800 \:=\:0$

Factor: . $5(x+40)(x+24) \:=\:0$

Then: . $x \:=\:\text{-}40,\:\text{-}24 \quad\Rightarrow\quad y \:=\:0,\:32$

The ship can be seen from $(\text{-}40,\:0)$ to $(\text{-}24,\:32)$

Wow, duh! Thanks for stating the obvious! Darn, I never pay attention.

Thank you!

N.

SOLUTION #1:

$10x^2-156x+576=0$
$x=9.6, 6$

Points of intersection: (9.6, -2.8), (6,8)

SOLUTION #2:

$5x^2+320x+4800=0$
$x=-24, -40$

Points of intersection: (-24, 32), (-40,0)

Thanks again!