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

Answers (1)
G Gautam harsolia

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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