Coordinate geo question once again ! (1 Viewer)

Aysce · Jun 26, 2011

Question:

The coordinates of the vertices of a triangle ABC are (2,5) , (6,1) and (4,-3) respectively. Find:

a. The equation of the line through C perpendicular to AB.

Originally I received the answer of y = 2x - 11, but when i checked the answers it was y=x-7.

If anyone could explain and solve this question I will reward you with a cookie

SpiralFlex · Jun 26, 2011

Aysce said:
Question:

The coordinates of the vertices of a triangle ABC are (2,5) , (6,1) and (4,-3) respectively. Find:

a. The equation of the line through C perpendicular to AB.

Originally I received the answer of y = 2x - 11, but when i checked the answers it was y=x-7.

If anyone could explain and solve this question I will reward you with a cookie

I want that cookie!

So let's see what it means by C being perpendicular to AB.

Firstly, find the gradient of AB. We will called this $m1$

Remember our gradient formula?

$m1=\frac{y2-y2}{x2-x1}$

In this case we want to use the points A (2, 5) and B (6, 1)

$m1=\frac{1-5}{6-2}$

$\therefore m1 = -1$

So the gradient of AB = -1.

So then what is the gradient of C? [We will call this $m2$ ]

Remember $m1*m2 = -1$ if they are perpendicular.

So, $-1 * m2 = -1$

$m2 = 1$

Now we must use this formula:

$y-y1=m(x-x1)$ with point C (4, -3)

$y - (-3) = 1(x-4)$

$y + 3 = x- 4$

$y = x - 7$

I prefer my cookies boiled, thanks!

funnytomato · Jun 26, 2011

@spiralflex , nice solution
@aysce, out of curiosity, how did u get gradient 2 for that line?

Aysce · Jun 26, 2011

funnytomato said:
@spiralflex , nice solution
@aysce, out of curiosity, how did u get gradient 2 for that line?

Well I didn't really understand the question, so I got gradient 2 by using the gradient formula with point B and point C. *Facepalm* @SpiralFlex Cheers once again ! Double cookies for you

Aysce · Jun 27, 2011

Guys, can you verify my answer, I've seriously checked it thousands of times and yet it's different to the given answer

anyways:

Find the equation of the line:

b. Passing through (-3,1) and perpendicular to 4x - 2y + 3 = 0

Here's my working:
$4x-2y+3=0$

$-2y=-4x-3$

$2y=4x+3$

$y=2x+\frac{3}{2}$

$\therefore$ $m_{1}=2$

$m_{1} * m_{2}=-1$

$m_{2} * 2 =-1$

$m_{2}=-\frac{1}{2}$

$y-y_{1}=m(x-_x{1})$

$y-1=-\frac{1}{2}(x+3)$

$2(y-1)=-1(x+3)$

$2y-2=-x-3$

$x+2y-2+3=0$

$x+2y+1=0$

The answer given at the back was: x+2y-5=0

K4M1N3 · Jun 27, 2011

I can't see anything wrong in your working, have a look back at the question, including the point given and the line. Elsewise the answer might just be wrong in the book.

Aysce · Jun 28, 2011

K4M1N3 said:
I can't see anything wrong in your working, have a look back at the question, including the point given and the line. Elsewise the answer might just be wrong in the book.

Yeah, I assume it's correct but I'll get it checked. Cheers for the approval ^^

jnney · Jun 28, 2011

up in this hour. lol.

Aysce · Jun 28, 2011

jnney said:
up in this hour. lol.

I could say the same for you

Coordinate geo question once again ! (1 Viewer)

Aysce

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