The lines are 3x + 4y - 2 = 0 and 2x - 3y + 7 = 0. They intersect at P(3, -1), and the angle between them is approximately -38.66 degrees.
To determine the two straight lines represented by the equation 6x² - xy - 12y² - 8x + 29y - 14 = 0, we can rewrite it as a quadratic form in terms of x and y. The given equation can be factored into (3x + 4y - 2)(2x - 3y + 7) = 0.
This results in two linear equations:
3x + 4y - 2 = 0
2x - 3y + 7 = 0
The points of intersection can be found by solving these two equations simultaneously. By solving, we get the point of intersection P(3, -1).
Now, to find the angle between the lines, we can use the formula tan(θ) = (m2 - m1) / (1 + m1 * m2), where m1 and m2 are the slopes of the two lines. The slopes are m1 = -3/4 and m2 = 3/2 for the first and second lines, respectively.
Substituting these values into the formula, we get tan(θ) = (3/2 + 3/4) / (1 - 3/4 * 3/2). Solving this yields tan(θ) = -5/7. Therefore, the angle between the lines is θ ≈ -38.66 degrees.
The question probable may be:
Determine the two straight lines represented 6x²-xy-12y² -8x +29y-14=0 Also the point intersection of the lines and Angle between.