Question

determine algebraically the vertices of the triangle formed by the lines.
5x-y=5 , x+2y=1 and 6x+y=17

Answer 1

We will deal with two lines at a time and solve them simultaneously:

L1: 5x-y=5

L2: x+2y=1

L3: 6x+y=17


L1 and L2
5x-y=5

(x+2y=1) x5

5x+10y=5
5x-y=5
11y=0
y=0

x=1
point (1, 0)


L1 and L3
5x-y=5
6x+y=17
Add
11x = 2
X =2
Y= 5

Point (2, 5 )



L2 and L3
(x+2y=1) x 6
6x+y=17

6x+12y=6
6x+y=17

11y= - 11
y= - 1
x = 3

Point (3, -1)

The vertices of the triangle formed are:

(1, 0), (2, 5 ), (3, -1)

Answer 2

1) 5x-5y=5  

2) x+2y=1  

3)6x+y=17  

Now to find the vertices of the triangle formed by these lines we have to solve two equation at a time.  

Solve equation 1 and 2  

5x-5y=5  

(x+2y=1) * -5  

----------------  

0-15y=0 => y=0 Thus substitute the value of y in equation 2 we get x=1-2*0=1  

There fore one of the vertices is (1,0)  

Similarly solve equation 1 and 3 and get (x,y)=(18/7,11/7) and from equation 2 and 3 get (x,y)=(3,-1)  

Thus the coordinates of triangle are:  

(1,0),(18/7,11/7) & (3,-1).