Question

Three lines are defined by the three equations: x + y = 0 x-y=0 2x+y=1 The three lines form a triangle with vertices at: A. (0,0),(33), 1,-1) B. (0,0), (17 2 ) (-1,-1) C. (1,1), (1, -1), (2, 1) D. (1,1),(3,-3), (-2,-1)

Answer 1

The triangle vertices are found by solving the system of equations: (0,0) from the first pair, (17/2, 1/2) from the second pair, and (-1,-1) from the third pair, confirming option B as the correct choice.

Step-by-step explanation:

The given system of equations represents three lines. To find the vertices of the triangle formed by these lines, we need to solve the system. Let's start by finding the intersection points of the pairs of lines.

The intersection of the first two lines, x + y = 0 and x - y = 0, yields x = 0 and y = 0, which corresponds to the point (0,0).

The second pair of lines, x - y = 0 and 2x + y = 1, intersect at x = 17/2 and y = 1/2.

The third pair, x + y = 0 and 2x + y = 1, intersect at x = -1 and y = -1.

Therefore, the vertices of the triangle formed by these lines are (0,0), (17/2, 1/2), and (-1,-1), which matches option B.

In summary, by solving the system of equations, we determined the intersection points of the lines, and these points correspond to the vertices of the triangle. The correct answer is B, as it accurately represents the vertices of the triangle formed by the given lines.

Answer 2

Comparing these points with the given options, we find that option B.
`$(0,0),\left((2)/(3), (2)/(3)\right),(-1,-1)$` is the correct answer.

To determine which vertices form a triangle using the given equations, we can solve the system of equations. Let's start by solving the first two equations:

Equation 1: x + y = 0

Equation 2: x - y = 0

To eliminate y, we can add the two equations together:

(x + y) + (x - y) = 0 + 0

2x = 0

Dividing both sides by 2, we get:

x = 0

Substituting x = 0 into Equation 1, we find:

0 + y = 0

y = 0

So the first two equations give us the point (0, 0).

Now let's solve the second and third equations:

Equation 2: x - y = 0

Equation 3: 2x + y = 1

To eliminate y, we can multiply Equation 2 by 2 and add it to Equation 3:

2(x - y) + (2x + y) = 2(0) + 1

2x - 2y + 2x + y = 1

4x - y = 1

Rearranging the equation, we have:

4x = 1 + y

4x = y + 1

Substituting x = 0, we find:

0 = y + 1

This implies that y = -1. So the second and third equations give us the point (0, -1).

Finally, let's solve the first and third equations:

Equation 1: x + y = 0

Equation 3: 2x + y = 1

To eliminate y, we can multiply Equation 1 by 2 and subtract it from Equation 3:

2(x + y) - (2x + y) = 2(0) - 1

2x + 2y - 2x - y = -1

y = -1

Substituting y = -1 into Equation 1, we find:

x + (-1) = 0

x - 1 = 0

x = 1

So the first and third equations give us the point (1, -1).

Therefore, the vertices of the triangle formed by the three lines are (0, 0), (0, -1), and (1, -1).