Question

Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x - 12y + 26 = 0 and 7x + 24y = 50.

Answer 1

$\text{Reduce}4x+3y+10=0\text{to perpendicular form}$

$4x+3y=-10$

$-4x-3y=10$

$\text{Dividing each term by}$

$\sqrt{(-4{)}^2+(-3{)}^2}=\sqrt{16+9}=\sqrt{25}=5$

$\frac{-4}{5}x-\frac{3}{5}y=\frac{10}{5}=2$

$\Rightarrow p_1=2...\left(1\right)$

$5x-12y+26=0$

$5x-12y=-26$

$\text{Dividing each term by}$

$\sqrt{(-5{)}^2+(12{)}^2}=\sqrt{25+144}=\sqrt{169}=13$

$\frac{-5}{13}x+\frac{12}{13}y=\frac{26}{13}=2$

$\Rightarrow p_2=2...\left(2\right)$

$7x+24y=50$

$\text{Dividing each term by}$

$\sqrt{(7{)}^2+(24{)}^2}=\sqrt{49+576}=\sqrt{625}=25$

$\frac{7x}{25}+\frac{24}{25}y=\frac{50}{25}=2$

$\Rightarrow p_3=2...\left(3\right)$

$\text{Hence, rom 1, 2 and 3 equation origin is equidistant from all three lines.}$