Line intersection

**Glitch** · July 9th 2010, 06:58 PM

This is the question:

4x + 3y = 24, x^2 + y^2 = 25

Find where the line intersects the circle.

I tried solving for y, the substituting it into the circle equation, but I don't get the correct answer. It ends up becomes a messy quadratic equation.

Can someone help me out? Thanks.

**mr fantastic** · July 9th 2010, 07:00 PM

Originally Posted by Glitch

This is the question:

4x + 3y = 24, x^2 + y^2 = 25

Find where the line intersects the circle.

I tried solving for y, the substituting it into the circle equation, but I don't get the correct answer. It ends up becomes a messy quadratic equation.

Can someone help me out? Thanks.

Please show all the work you have done.

**Glitch** · July 9th 2010, 07:21 PM

3y = -4x + 24
y = (-4/3)x + 8

Sub this into the circle eqn:

$x^2 + \frac{-4}{3}x^2 + \frac{64}{3}x + 64 = 25$

$-\frac{1}{3}x^2 + \frac{64}{3}x = -39$

$x^2 - 64x = 117$

$x^2 - 64x - 117 = 0$

I don't know where to go from here. The quadratic formula seems to make a mess.

**mr fantastic** · July 9th 2010, 07:29 PM

Originally Posted by Glitch

3y = -4x + 24
y = (-4/3)x + 8

Sub this into the circle eqn:

$x^2 + \frac{-4}{3}x^2 + \frac{64}{3}x + 64 = 25$ Mr F says: This is wrong. And who knows why since you have not shown all your work (as I requested).

[snip]

Actually I do know why, but I want to see every step that lead you to this. (And when you start typing those steps you will probably find the mistake yourself, which is another reason why I asked you to show all your work).

BY the way: http://www.wolframalpha.com/input/?i...B+y%5E2+%3D+25

**Glitch** · July 9th 2010, 07:34 PM

Sorry, I this is the bit I'm missing:

x^2 + y^2 = 25 (circle eqn)

x^2 + ((-4/3)^2 + 8)^2 = 25 (substituting into circle eqn)

Am I using the wrong expansion method?

**mr fantastic** · July 9th 2010, 07:39 PM

Originally Posted by Glitch

Sorry, I this is the bit I'm missing:

x^2 + y^2 = 25 (circle eqn)

x^2 + ((-4/3)^2 + 8)^2 = 25 (substituting into circle eqn)

Am I using the wrong expansion method?

So you went straight from x^2 + ((-4/3)^2 + 8)^2 = 25 to $x^2 + \frac{-4}{3} x^2 + \frac{64}{3} + 64 = 25$ , did you?

For the last time, please show ALL your work:

Originally Posted by Mr F, who said (bold face added)

[snip]but I want to see every step that lead you to this.

**Glitch** · July 9th 2010, 08:00 PM

Yes I did, using this equation:

$(a + b)^2 = a^2 + 2ab + b^2$

**mr fantastic** · July 9th 2010, 08:06 PM

Originally Posted by Glitch

Yes I did, using this equation:

$(a + b)^2 = a^2 + 2ab + b^2$

OK, for your question what is a and what is b?

Now, please show me exactly how you calculated a^2 + 2ab (b^2 = 64 was correct). Show every detail. Pretend you are explaining it to a 14 year old student.

**Glitch** · July 9th 2010, 08:35 PM

I squared $-\frac{4}{3}x$ to get $\frac{16}{9}x^2$ , which is a^2.

For 2ab, I did $2*(\frac{-4}{3}* 8) = \frac{64}{3}x$

Hmm, looks like I squared it wrong before. I think I also expanded it using the wrong equation, since (-4/3 + 8)^2 is equivallent to (8 - 4/3)^2, which expands using the (a - b)^2 equation.

I think I've got the correct answers now.

**mr fantastic** · July 9th 2010, 08:54 PM

Originally Posted by Glitch

I squared $-\frac{4}{3}x$ to get $\frac{16}{9}x^2$ , which is a^2.

For 2ab, I did $2*(\frac{-4}{3}* 8) = \frac{64}{3}x$ Mr F says: This should be negative.

Hmm, looks like I squared it wrong before. I think I also expanded it using the wrong equation, since (-4/3 + 8)^2 is equivallent to (8 - 4/3)^2, which expands using the (a - b)^2 equation.

I think I've got the correct answers now.

..

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