Question

find distance of (3,4)from 5x+5y+7=0

Answer 1

find the distance between of (3,4) from 5x+5y+7=0

Answer 2

There is one formula to solve this problem.

The distance of any point (x1,y1)" role="presentation" style="margin: 0px; padding: 0px; outline: 0px; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">(x1,y1)(x1,y1)from the line ax+by+c=0 is:

|ax1+by1+c|√(a2+b2)" role="presentation" style="margin: 0px; padding: 0px; outline: 0px; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">|ax1+by1+c|√(a2+b2)|ax1+by1+c|√(a2+b2)

In the given problem, x1=4,y1=1,a=3,b=−4,c=12" role="presentation" style="margin: 0px; padding: 0px; outline: 0px; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">x1=4,y1=1,a=3,b=−4,c=12x1=4,y1=1,a=3,b=−4,c=12

Substitute in the given values to get the answer.

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