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What is the perpendicular distance from 2 5 that meets the line of 7x plus 13 -y equals 0 at right angles showing work?
Wiki User
∙ 9y ago
Equation: 7x+13-y = 0 => y = 7x+13
Slope: 7
Perpendicular slope: -1/7
Perpendicular equation: 7y = -x+37
Both equations intersect at: (-1.08, 5.44)
Distance: square root of [(-1.08-2)squared+(5.44-5)squared)] = 3.111 to 3 d.p.
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∙ 9y ago
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