Find the distance of the point (3, -5) from t | vedclass

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(A) The given line is $3x - 4y - 26 = 0$. Comparing this with the general equation $Ax + By + C = 0$,we get $A = 3$,$B = -4$,and $C = -26$. The given

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$0.6$

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(A) The given line is $3x - 4y - 26 = 0$. Comparing this with the general equation $Ax + By + C = 0$,we get $A = 3$,$B = -4$,and $C = -26$. The given point is $(x_1, y_1) = (3, -5)$. The perpendicular distance $d$ of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$ is given by the formula: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ Substituting the values: $d = \frac{|3(3) + (-4)(-5) - 26|}{\sqrt{3^2 + (-4)^2}}$ $d = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}}$ $d = \frac{|3|}{\sqrt{25}} = \frac{3}{5} = 0.6$.

Find the distance of the point $(3, -5)$ from the line $3x - 4y - 26 = 0$.

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