Question

Find the distance of the point p(2,5) from the line 3x+4y+14=0

Answer 1

distance from point P(2, 5) from the line 3x + 4y + 14 is 8 unit.

the distance of the point P(x1, y1) from the line ax

the distance of the point P(x1, y1) from the line ax+ by + c = 0 is

= |ax1 + by1 + c|/√(a² + b²)

using above application,

distance of he point P(2, 5) from the line 3x + 4y + 14 = 0 is |3(2) +4(5) + 14|/√(3² + 4²)

= |6 + 20 + 14|/5

= 40/5 = 8 unit.

hence, distance from point P(2, 5) from the line 3x + 4y + 14 is 8 unit.

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Answer 2

Answer:

Find the distance of the point p(2,5) from the line 3x+4y+14=0