Question

Find the distance of the point (2,1) from the line 12x-5y+9=0

Please answer only if u know it....

Answer 1

I hope this answer help you
Mark this answer as brainliest answer

Answer 2

Answer:

The distance of the point (2,1) from the line 12x-5y+9=0 is  $\frac{28}{13}$ units.

Step-by-step explanation:

The distance of the point $(x_1,y_1)$ from the line $ax+bx+c=0$ is

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

Using this formula, the distance of the point (2,1) from the line 12x-5y+9=0 is

$d=|\frac{12(2)-5(1)+9}{\sqrt{(12)^2+(-5)^2}}|$

$d=|\frac{24-5+9}{\sqrt{144+25}}|$

$d=|\frac{28}{\sqrt{169}}|$

$d=\frac{28}{13}$

Therefore the distance of the point (2,1) from the line 12x-5y+9=0 is $\frac{28}{13}$ units.