Question

The distance of point (-1,3) from the line 3x + 4y -1=0 is

Answer 1

Answer:

1.3 equal 13 lines is 1.00.9

Answer 2

$\text{We know that the perpendicular distance of a point }(x_1,y_1) \text{ from a}\\\text{line }ax + by + c = 0 \text{ is given by:}$

$\implies \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^{2} +b^{2}}}$

$\text{In our case }(x_1,y_1) = (-1,3) \; ; \; a = 3 \; ; \; b = 4 \; ; \; c = -1$

$\implies \text{Distance} = \dfrac{|3(-1) + 4(3) - 1|}{\sqrt{3^{2} + 4^2}}$

$\implies \text{Distance} = \dfrac{|-3 + 12 -1|}{\sqrt{9 + 16}}$

$\implies \boxed{\text{Distance} = \dfrac{8}{5} \; units}$