Using diagonals from a common vertex, how many triangles could be formed from a hexagon?
1 answer:
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Answer:
A system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2
Step-by-step explanation:
The equation of the given line is 8·x - 4·y = 12
Which gives;
8·x- 12= 4·y
y = 2·x - 3
Given that the line passes through the points (0, -3) and (1, -1), we have;
When (x₁, y₁) = (0. -3) and (x₂, y₂) = (1, -1), we have;
y - (-3) = 2×(x - 0)
y = 2·x - 3 which is the equation of the given line
For the lines 8·x - 4·y = 12, which is the sane as y = 2·x - 3 and the line y = m·x - 6 to have no solution, the slope of the two lines should be equal that is m = 2
Given that the line passes through the point (1.5, 0), we have;
y - 0 = 2×(x - 1.5)
y = 2·x - 3...................(1)
For the equation, y = m·x - 6, when m = 2, we have;
y = 2·x - 6..................(2)
Solving equations (1) and (2) gives;
2·x - 3 = 2·x - 6, which gives;
2·x - 2·x= - 3 - 6
0 = 9
Therefore, a system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2.