A point equidistant from the lines 4x + 3y + | vedclass

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(C) The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

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$(0, 0)$

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(C) The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.For the point $(0, 0)$:$1$. Distance from $4x + 3y + 10 = 0$ is $d_1 = \frac{|4(0) + 3(0) + 10|}{\sqrt{4^2 + 3^2}} = \frac{10}{\sqrt{16 + 9}} = \frac{10}{5} = 2$.$2$. Distance from $5x - 12y + 26 = 0$ is $d_2 = \frac{|5(0) - 12(0) + 26|}{\sqrt{5^2 + (-12)^2}} = \frac{26}{\sqrt{25 + 144}} = \frac{26}{13} = 2$.$3$. Distance from $7x + 24y - 50 = 0$ is $d_3 = \frac{|7(0) + 24(0) - 50|}{\sqrt{7^2 + 24^2}} = \frac{|-50|}{\sqrt{49 + 576}} = \frac{50}{25} = 2$.Since $d_1 = d_2 = d_3 = 2$,the point $(0, 0)$ is equidistant from all three lines.

$A$ point equidistant from the lines $4x + 3y + 10 = 0$,$5x - 12y + 26 = 0$ and $7x + 24y - 50 = 0$ is

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