Question

Show that (1,2) is equidistant from 4x-3y+7=0 and 5x+12y-16=0

Answer 1

Distance of point (a,b) from the line Ax +By+C=0 is given by $\frac{Aa+Bb+C}{\sqrt{a^{2} +b^{2}}}$

Distance of (1,2) from 4 x-3 y+7=0 is$\frac{4\times1-3\times2+7}{\sqrt{4^{2} +3^{2}} }=\frac{5}{5}=1$

Distance of (1,2) from 5 x+12 y-16=0 is $\frac{5\times 1+12 \times 2-16}{\sqrt{5^{2}+12^{2}}}=\frac{13}{13}=1$

Hence point (1,2) is equidistant from 4x-3y+7=0 and 5x+12y-16=0