Question

The distance of the point (1,2) from the line $3x+4y–5=0$ is units.

• A. $\frac{6}{5}$
• B. $\frac{5}{6}$
• C. $\frac{16}{5}$

Answer 1

The correct option is A $\frac{6}{5}$

The distance of the point $(x_1,y_1)$ from the line $ax+by+c=0$ is given by
$\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

Thus , the distance of the point (1,2) from the line $3x+4y–5=0$ is

$d=\frac{|3\times 1+4\times 2-5|}{\sqrt{3^2+4^2}}$

$=\frac{|3+8-5|}{\sqrt{9+16}}$

$=\frac{|6|}{\sqrt{25}}=\frac{6}{5}$ units.