Question

[(2,0)((-1, 3){(1,0)[(-1, -3)(0, 2)[(-3,-1)]Line 1Line 2Point of Intersection(0, 2). (-5, -3)(0,5). (-5, -5)(-2, 1), (-3,-1)(0, 2), (2,0)(4,-2), (1,1)(4,2), (0, -2)(2,-2), (0, 2)(2, 1), (3, 2)ResetNext

Answer 1

First case

Line (0, 2), (-5, -3) ---> y = x + 2

Line (0, 5), (-5, -5) ---> y = 2x + 5

Point of intersection ---> we need to equate both equations:

`x+2=2x+5\Rightarrow-5+2=2x-x\Rightarrow-3=x`

If x = -3 ---> y = -3 + 2 ---> y = -1

The point of intersection in this case is (-3, -1) (the last one, from left to right).

Second case

Line (-2, 1), (-3, -1) ---> y = 2x + 5

Line (0, 2), (2, 0) ---> y = -x + 2

Equating both equations to find the intersection point:

`2x+5=-x+2\Rightarrow2x+x=2-5\Rightarrow3x=-3\Rightarrow x=-1`

If x = -1, then y = -(-1) + 2 ---> y = 1 + 2 ---> y = 3

The point of intersection is (-1, 3) (the second one, from left to right).

Third case

Line (4, -2), (1, 1) ---> y = -x +2

Line (4, 2), (0, -2) ---> y = x - 2

Then, we have ---> intersection point:

`-x+2=x-2\Rightarrow2+2=x+x\Rightarrow4=2x\Rightarrow x=2`

Using the second equation to find y ---> y = 2 -2 ---> y = 0

The point of intersection is (2, 0) (the first point, from left to right).

Fourth case

Line (2, -2) (0, 2) ---> y = -2x + 2

Line (2, 1), (3, 2) ---> y = x - 1

Then, we have:

`-2x+2=x-1\Rightarrow1+2=x+2x\Rightarrow3=3x\Rightarrow x=1`

Using the second equation to find y ---> y = 1 - 1 ---> y = 0

The point of intesection is (1, 0) ( the third point, from left to right).