Distance of $(4,2)$ from line $4 x+3 y-10=0$ is

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Professor PC Thomas Classes · Answer

Distance $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} = \frac{|4(4) + 3(2) - 10|}{\sqrt{4^2 + 3^2}} = \frac{|16 + 6 - 10|}{5} = \frac{12}{5} = 2.4$.