For this case we have that by definition, the equation of a line in the slope-intersection form is given by:
[tex]y = mx + b[/tex]
Where:
m: It's the slope
b: It is the cutoff point with the y axis
Equation 1:
Go through the points:
[tex](x1, y1) :( 1,1)\\(x2, y2) :( 2,4)[/tex]
[tex]m = \frac {y2-y1} {x2-x1} = \frac {4-1} {2-1} = \frac {3} {1} = 3[/tex]
So, the line is:
[tex]y = 3x + b[/tex]
We substitute a point:
[tex]1 = 3 (1) + b\\1 = 3 + b\\b = 1-3\\b = -2[/tex]
Finally the line is:
[tex]y = 3x-2[/tex]
Equation 2:
[tex](x1, y1) :( 2, -2)\\(x2, y2): (- 1, -5)[/tex]
[tex]m = \frac {y2-y1} {x2-x1} = \frac {-5 - (- 2)} {- 1-2} = \frac {-5 + 2} {- 3} = \frac {-3 } {- 3} = 1[/tex]
The equation is:
[tex]y = x + b[/tex]
Substituting a point:
[tex]-2 = 2 + b\\-2-2 = b\\b = -4[/tex]
Finally the equation is:
[tex]y = x-4[/tex]
Equaling both equations:
[tex]3x-2 = x-4\\3x-x = -4 + 2\\2x = -2\\x = \frac {-2} {2}\\x = -1[/tex]
We look for the value of y:
[tex]y = x-4\\y = -1-4\\y = -5[/tex]
The point (-1, -5) is system solution
ANswer:
[tex](x, y): (- 1, -5)[/tex]
Answer:
your answer would be -1, -5 for Plato users
Step-by-step explanation:
