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Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Q : 4     Find the distance of the point  (-1,1)  from the line   12(x+6)=5(y-2).
 

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Given the equation of the line is
12(x+6)=5(y-2)
we can rewrite it as
12x+72=5y-10
12x-5y+82=0
Now, we know that 
d= \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}       where A and B are the coefficients of x and y and  C is some constant  and  (x_1,y_1) is point from which we need to find the distance 
In this problem A = 12 , B = -5 , c = 82 and (x_1,y_1) = (-1 , 1)
Therefore,
d = \frac{|12.(-1)+(-5).1+82|}{\sqrt{12^2+(-5)^2}} = \frac{|-12-5+82|}{\sqrt{144+25}}=\frac{|65|}{\sqrt{169}}=\frac{65}{13}= 5
Therefore, the distance of the point  (-1,1)  from the line   12(x+6)=5(y-2) is 5 units

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