Find the distance of the point (-1, 1) from t | vedclass

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(A) The given equation of the line is $12(x + 6) = 5(y - 2)$.$\Rightarrow 12x + 72 = 5y - 10$$\Rightarrow 12x - 5y + 82 = 0$ ... $(1)$On comparing equ

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$5$

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(A) The given equation of the line is $12(x + 6) = 5(y - 2)$.$\Rightarrow 12x + 72 = 5y - 10$$\Rightarrow 12x - 5y + 82 = 0$ ... $(1)$On comparing equation $(1)$ with the general equation of a line $Ax + By + C = 0$,we obtain $A = 12$,$B = -5$,and $C = 82$.The perpendicular distance $(d)$ of a line $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.The given point is $(x_1, y_1) = (-1, 1)$.Therefore,the distance of point $(-1, 1)$ from the given line is:$d = \frac{|12(-1) + (-5)(1) + 82|}{\sqrt{(12)^2 + (-5)^2}}$ units$d = \frac{|-12 - 5 + 82|}{\sqrt{144 + 25}}$ units$d = \frac{|65|}{\sqrt{169}}$ units$d = \frac{65}{13}$ units$d = 5$ units.

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

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