The slope of the line perpendicular to the line whose equation is 6x + 4y = - 56 is 2/3.
Two lines with slopes m₁ and m₂ are parallel when m₁ = m₂ and perpendicular when m₁*m₂ = -1.
In the question, we are asked for the slope of the line perpendicular to the line whose equation is 6x + 4y = - 56.
Representing the given equation is the slope-intercept form y = mx + b, we get:
6x + 4y = - 56,
or, 4y = - 6x - 56,
or, y = (-6/4)x + (-56/4),
or, y = (-3/2)x + (-14),
we get the slope, m is (-3/2), and the y-intercept, b is (-14).
Assuming the slope of the required line to be m₁, we know that m₁*m = -1, as the two lines are perpendicular to each other.
Thus, m₁*(-3/2) = -1,
or, m₁ = 2/3.
Thus, the slope of the line perpendicular to the line whose equation is 6x + 4y = - 56 is 2/3.
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