Question

find the distance of the point (0,9) from the line x + 2y - 3 = 0​

Answer 1

Answer:

The answer is $3\sqrt 5$

Step-by-step explanation:

Concept: Distance of the point P$(x_1,y_1)$ from the line$ax+by+c=0$ is

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

Therefore distance of the point $(0,9)$ from the line $x+2y-3=0$ is

                                          $=\dfrac{|0+2(9)-3|}{\sqrt{1^2+2^2}}=\dfrac{15}{\sqrt5}=\dfrac{3\times 5}{\sqrt5}=3\sqrt 5$