It is known that the equation of the line L is 3x + 4y-12 = 0. The equation of the line L 'is solved so that the triangle area bounded by L' and l 'is 6

It is known that the equation of the line L is 3x + 4y-12 = 0. The equation of the line L 'is solved so that the triangle area bounded by L' and l 'is 6

50: Y = - 3x / 4 + 3 slope K1 = - 3 / 4 because l 'is perpendicular to L, the slope of the line L' is K2 = 4 / 3, so l ': 4x-3y = m (M is constant) deformation: l': 4x / M - 3Y / M = 1, so the triangle area enclosed by L 'and the coordinate axis is: S = (M / 4) * (M / 3) / 2 = (m ^ 2) / 24 = 6, so the square of the line L' is m = ± 12