Question

4x-5y+16=0'.2x+y-6=0

Answer 1

Answer:

Step-by-step explanation:

We have system of equation:

A:4x−5y+16=0−−−(1)

B:2x+y−6=0−−−−(2)

For A

x0−41  

y16/504  

For B

x031

y604

From solving equation (1) & (2), we get only one point (1,4) as

−7y+28=0⇒y=4

4x−5y+16=04x+2y−12=0−−+

​  

,x=  

2

6−y

​  

=  

2

2−4

​  

=1

So, the vertices are (−4,0),(1,4) & (3,0)