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How would you find the perpendicular distance from the point 7 5 to the line 3x plus 4y -16 equals 0 on the Cartesian plane?
Wiki User
∙ 6y ago
Equation of line: 3x+4y-16 = 0 or y = -3/4x+4
Slope of line: -3/4
Slope of perpendicular line: 4/3
Perpendicular equation: y-5 = 4/3(x-7) or 3y = 4x-13
Both equations intersect at: (4, 1)
Perpendicular distance from (7, 5) to (4, 1) = 5 units using distance formula
Wiki User
∙ 6y ago
Wiki User
∙ 6y agoThe perpendicular distance from the point (p, q) to the line ax + by + c = 0 is
|ap + bq + c|/sqrt(a2 + b2) = |3*7 + 4*5 - 16|/sqrt(32 + 42) = 25/5 = 5 units.
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