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Home  >>  CBSE XI  >>  Math  >>  Straight Lines

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

$\begin{array}{1 1}(A)\;5\: units \\(B)\; 10\: units \\(C)\; \large\frac{65}{\sqrt{13}}\: units \\(D)\;\large\frac{\sqrt {65}}{13}\: units \end{array} $

1 Answer

Toolbox:
  • Distance of a point $(x_1,y_1)$ from the given line is $ d=\large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
The given equation of the line is
$12(x+6)=5(y-2)$
This can be written as
$12x+72=5y-10$
$ \Rightarrow 12x-5y+82=0$
Let us now compare this with the general equation $Ax+By+C=0$
Here $A = 12, B=-5, C=82$
$d = \large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
Since the line passes through the point (-1,1)
$ d = \large\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^2+(-5)^2}}$
$ = \large\frac{|-12-5+82|}{\sqrt{161}}$$ = \large\frac{|65|}{13}$$=5$ units.
Hence the distance is 5 units.
answered May 9, 2014 by thanvigandhi_1
 

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