Question

The distance of the point A(-2,3) from the line 12x-5y-13=0​

Answer 1

Hello !!

For you answer this question, you only need make use of this formula.

$\mathsf{d=\dfrac{|ax_{0}+by_{0}+c|}{\sqrt{(a^{2}+b^{2})}}}$

Now, you put the informations of the statement in the formula and find the result.

$\mathsf{d=\dfrac{|ax_{0}+by_{0}+c|}{\sqrt{(a^{2}+b^{2})}}} \\\\\\ \mathsf{d=\dfrac{|12(-2)-5(3)-13|}{\sqrt{(12^{2}+5^{2})}}} \\\\\\ \mathsf{d=\dfrac{|-24-15-13|}{\sqrt{(144+25)}}} \\\\\\ \mathsf{d=\dfrac{|-52|}{\sqrt{169}}} \\\\\\ \mathsf{d=\dfrac{|-52|}{13}} \\\\\\ \mathsf{d=\dfrac{52}{13}} \\\\\\ \boxed{\mathsf{\bf\red{d=4}}}$

Final result : 4 uc.

I hope I have collaborated !

Answer 2

The distance of the point A(-2,3) from the line $12x-5y-13=0$ is 4 unit.

Step-by-step explanation:

To find : The distance of the point A(-2,3) from the line $12x-5y-13=0$?

Solution :

The distance between a point (x_0,y_0) from the line ax+by+c=0 is given by,

$d=\dfrac{|ax_{0}+by_{0}+c|}{\sqrt{(a^{2}+b^{2})}}}$

On comparing, $x_0=-2,y_0=3,a=12,b=-5$ and $c=-13$

Substitute the value,

$d=\dfrac{|12(-2)-5(3)-13|}{\sqrt{(12)^{2}+(5)^{2}}}$

$d=\dfrac{|-24-15-13|}{\sqrt{144+25}}$

$d=\dfrac{|-52|}{\sqrt{169}}$

$d=\dfrac{|-52|}{13}$

$d=\dfrac{52}{13}$

$d=4$

Therefore, the distance of the point A(-2,3) from the line $12x-5y-13=0$ is 4 unit.

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