The points A(6,7), B(0,1) and C(9,4) are vertices of a triangle.Find the length AB and the area of the triangle ABC and hence obtain the perpendicular distance from C to the line AB.

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- Oct 19th 2010, 07:52 AMIlsaCoordinate geometry
The points A(6,7), B(0,1) and C(9,4) are vertices of a triangle.Find the length AB and the area of the triangle ABC and hence obtain the perpendicular distance from C to the line AB.

- Oct 19th 2010, 08:27 AMemakarov
Surely you can find the distance AB, can't you? The best way of figuring the area depends on what you covered about the topic. One way is to use cross product: the area is one half of the length of the cross product of the vectors AB and AC.

Please describe in more detail what you know and what your difficulty is. - Oct 19th 2010, 08:36 AMIlsa
how do we find the area using only the length AB?

how do we use the area to find the perpendicular distance?

Isn't the perpendicular distance the height of the triangle?

why is it 'hence' then?

Note: I haven't studied vectors as yet!

:)

Thanku 4 ur help! - Oct 19th 2010, 09:00 AMemakarovQuote:

how do we find the area using only the length AB?

The forum rules ask for the poster to show some effort in solving the problem, so finding AB seems to be a perfect way to do this.

Quote:

how do we use the area to find the perpendicular distance?

You still have to say what ways of finding the area you studied. The easiest way in this case is this formula, but you need to shift the triangle so that A is at the origin (0, 0). - Oct 19th 2010, 11:14 AMbigwaveuse distance forumula
If $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2,y_2)$ are any two points and if $\displaystyle d$ is the distance between them, then

$\displaystyle d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$

thus if $\displaystyle A(6,7)$ and $\displaystyle B(0,1)$ then

$\displaystyle \overline{AB} = \sqrt{(6-0)^2 + (7-1)^2}

\Rightarrow \sqrt{36 + 36}

\Rightarrow \sqrt{72}

\Rightarrow 6\sqrt{2}

$

also, the slope intercept form of $\displaystyle \overline{AB}$ is $\displaystyle y=x+1$ where $\displaystyle m=1$ thus the perpendicular to $\displaystyle \overline{AB}$ is the negative recipical of $\displaystyle 1$ which is $\displaystyle -1 $

Since $\displaystyle \overline{AC} $has $\displaystyle m= -1$ you can use that for $\displaystyle h$

thus $\displaystyle \overline{AC} = \sqrt{(9-6)^2+(4-7)^2} \Rightarrow 3\sqrt{2}$

then $\displaystyle \overline{AB}$ is $\displaystyle b$ then use

$\displaystyle A = \frac{1}{2}bh

\Rightarrow

\frac{1}{2} (3\sqrt{2}) (6\sqrt{2}) = 36$

draw this on graph paper then you can what is happening

try to take some steps then ask where you are stuck(Cool)