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perpendicular distance from the point (x1,y1) to the line ax+by+c=0

is d=(ax1+by1+c)/sqrt(a^2+b^2)

given point (2,4) line equation 2x-y+10=0

d= (2(2)-4+10)/sqrt(2^2+1^2)

d= 10/sqrt(5)

d= 4.472

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chinthaginjala venka...

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1y ago
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11y ago

Straight line equation: y = 2x+10

Perpendicular equation: y = -1/2x+5

Point of intersection: (-2, 6)

Distance: the square root of [(2--2)2+(4-6)2] = 4.472 to 3 decimal places

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Q: What is the perpendicular distance from the point of 2 and 4 to the line of y equals 2x plus 10?
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