# Let #P(x_1, y_1)# be a point and let #l# be the line with equation #ax+ by +c =0#. Show the distance #d# from #P->l# is given by: #d =(ax_1+ by_1 + c)/sqrt( a^2 +b^2)#? Find the distance #d# of the point P(6,7) from the line l with equation 3x +4y =11?

Now that

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To show the distance ( d ) from point ( P(x_1, y_1) ) to the line ( l ) with equation ( ax + by + c = 0 ) is given by ( d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} ), we follow these steps:

- Determine the perpendicular distance from point ( P ) to the line ( l ).
- Use the formula for the distance from a point to a line.

Given the point ( P(6, 7) ) and the line equation ( 3x + 4y = 11 ), we need to convert the line equation into the standard form ( ax + by + c = 0 ), where ( a ), ( b ), and ( c ) are coefficients.

Converting the line equation: [ 3x + 4y = 11 ] [ \Rightarrow 3x + 4y - 11 = 0 ]

Comparing with the standard form ( ax + by + c = 0 ), we have: [ a = 3, , b = 4, , c = -11 ]

Now, plug in the values into the distance formula: [ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} ] [ = \frac{|3(6) + 4(7) - 11|}{\sqrt{3^2 + 4^2}} ] [ = \frac{|18 + 28 - 11|}{\sqrt{9 + 16}} ] [ = \frac{|35 - 11|}{\sqrt{25}} ] [ = \frac{|24|}{5} ] [ = \frac{24}{5} ]

So, the distance ( d ) of the point ( P(6,7) ) from the line ( l ) with equation ( 3x + 4y = 11 ) is ( \frac{24}{5} ).

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