Using the formula for distance from a point to a line, d = |Ax1 + By1 + C| / sqrt(A^2 + B^2), we find d = |2(3) + 3(4) - 12| / sqrt(2^2 + 3^2) = 3.

Since the line is parallel, it has the same slope. Using point-slope form: y - 5 = 3(x - 4) gives y = 3x - 7.

Since the line is parallel, it has the same slope (3). Using point-slope form: y - 1 = 3(x - 4) gives y = 3x - 11.

Using the slope-intercept form y = mx + b, with m = 5 and b = 0, we get y = 5x.

Using the slope-intercept form, the equation is y = -3x.

The slope of the given line is 3, so the slope of the perpendicular line is -1/3. Using point-slope form, we get y + 1 = -1/3(x - 4), which simplifies to y = -1/3x + 11/3.

The slope m = (7 - 3) / (4 - 2) = 2. Using point-slope form: y - 3 = 2(x - 2) gives y = 2x + 1.

Setting x = 0 in the equation gives y = 8. Therefore, the y-intercept is 8.

Rearranging gives y = (3/4)x + 3. The slope of the line is 3/4, so the slope of the parallel line is -3/4.

Parallel lines have the same slope. The slope of the given line is 3/4, which is the same as the slope of 6x - 8y + 24 = 0.

The slopes are m1 = 2 and m2 = -1/2. The angle θ = tan^(-1) |(m1 - m2) / (1 + m1*m2)| = tan^(-1)(5/4) which is approximately 60 degrees.

The slope m = (6 - 2) / (3 - 1) = 2. Using point-slope form, y - 2 = 2(x - 1) gives y = 2x.

Solving the equations simultaneously, we find x = 2 and y = 3. Thus, the point of intersection is (2, 3).

Setting 3x + 1 = -x + 5 gives 4x = 4, thus x = 1. Substituting x back gives y = 4. The intersection point is (1, 4).

The slope of the line is -5/2. The slope of the perpendicular line is the negative reciprocal, which is 2/5.

The slope of the line is 7/2. The slope of the perpendicular line is the negative reciprocal, which is -2/7.

The slope of the line is 7/2. The slope of the perpendicular line is the negative reciprocal, which is -2/7.

Setting y = 0 in the equation gives 5x + 10 = 0, thus x = -2.

Substituting (0, -4) into the equation gives 4(0) - 5(-4) + 20 = 0, which is true.