# The sides of a triangle are given by the lines........find the area of this triangle

• Jul 16th 2008, 06:47 AM
sweetG
The sides of a triangle are given by the lines........find the area of this triangle
The sides of a triangle are given by the lines:

2x - y + 3 = 0

x - 2y - 3 = 0

and 2x + y - 11 = 0

Find the area of this triangle.

Now something that I'm sure of is that this triangle is NOT a right angled triangle.

What I'm not sure of is how I did the question. First I found the x and y intercepts of all equations by saying x = 0, y = ? and when y = 0, x = ?
But a lot of people have told me to solve the lines together yet my tutor told me to do it the way I did so I'm not sure if I'm doing it right or not. I hope you guys understand me coz i guess I'm not making much sense right now anyway can someone help me out and I know I'm supposed to use some perpindicular distance forumla but I forgot how to use it. I would appreciate it if someone could help me out thanks :)
• Jul 16th 2008, 07:01 AM
masters
Quote:

Originally Posted by sweetG
The sides of a triangle are given by the lines:

2x - y + 3 = 0

x - 2y - 3 = 0

and 2x + y - 11 = 0

Find the area of this triangle.

Now something that I'm sure of is that this triangle is NOT a right angled triangle.

The intersection between x-2y-3=0 and 2x+y-11=0 is indeed a right angle. Their slopes are negative reciporcals of each other.

First, solve for the intersections of your 3 lines by solving simultaneous equations.

Then use the distance formula to determine the lengths of the two legs of your right triangle.

Then use A=1/2 bh, where b and h are your two legs.
• Jul 18th 2008, 07:51 AM
sweetG
Well I solved them simutaneously and got the points:
A(-3,-3)
B(-5,1)
C(2,7)

I'm not sure if these are right so could someone please check and tell me thanks :)

And I think I'm supposed to use the distance from a point to a line' formula except I've forgotten the formula and how to use it so could someone please tell me it if they know it.

And my tutor told me to do it that way otherwise I would've just used the normal area of a triangle formula after finding the legnths of the lines.
• Jul 18th 2008, 07:58 AM
masters
Quote:

Originally Posted by sweetG
Well I solved them simutaneously and got the points:
A(-3,-3)
B(-5,1)
C(2,7)

I'm not sure if these are right so could someone please check and tell me thanks :)

And I think I'm supposed to use the distance from a point to a line' formula except I've forgotten the formula and how to use it so could someone please tell me it if they know it.

And my tutor told me to do it that way otherwise I would've just used the normal area of a triangle formula after finding the legnths of the lines.

Check the intersection at B. I believe you'll find it to be (5,1).
• Jul 18th 2008, 07:59 AM
TKHunny
Try B again. s/b (5,1)
• Jul 18th 2008, 08:12 AM
masters
Quote:

Originally Posted by masters
Check the intersection at B. I believe you'll find it to be (5,1).

Now apply the distance formula to find BC and AB. Those two segments meet at right angles because the slopes of the two lines are negative reciprocals of each other.

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

$AB=\sqrt{(5- -3)^2+(1- -3)^2}=\sqrt{80}=\sqrt{16\cdot5}=4\sqrt{5}$

$BC=\sqrt{(5-2)^2+(1-7)^2}=\sqrt{45}=\sqrt{9\cdot5}=3\sqrt{5}$

Find area of triangle using $A=\frac{1}{2} \cdot AB \cdot BC$

What do you get?
• Jul 19th 2008, 06:08 AM
sweetG
I got 30.

Is that right ?
• Jul 19th 2008, 07:53 AM
masters
Quote:

Originally Posted by sweetG
I got 30.

Is that right ?

That's what I get. Congrats!!(Clapping)