Question

The length of perpendicular from the origin to the line 12x+5y+7 =0 Is?

Answer 1

Hiiii. ....friend

The answer is here,

The length of the perpendicular from the origin to the line 12x+5y+7 =0.

$= > \: \frac{ |c| }{ \sqrt{ {a}^{2} + {b}^{2} } }$

Here , c = constant.

a= x- coefficient.

b=y- coefficient.


$= > \: \frac{ |7| }{ \sqrt{ {5}^{2} + {12}^{2} } }$

$= > \: \frac{7}{ \sqrt{ {13}^{2} } }$

$= > \: \frac{7}{13}$

So, The perpendicular distance from origin to the point 12x+5y+7 =0. is 7/13 units.

:-)Hope it help u.

Answer 2

The length of perpendicular from the origin to the line 12x + 5y + 7 = 0 is $\frac{7}{13}$ units.

Step-by-step explanation:

The perpendicular distance to a straight line ax + by + c = 0 from a given point $(x_{1},y_{1})$ is given by the formula

$\frac{|ax_{1} + by_{1} + c|}{\sqrt{a^{2} + b^{2}}}$.

Now, the point is here (0,0) i.e. origin and the straight line is 12x + 5y + 7 = 0.

Therefore, the distance will be, d = $\frac{|12 \times 0 + 5 \times 0 + 7|}{\sqrt{12^{2} + 5^{2}}}$ = $\frac{7}{13}$ units.

Hence, the length of perpendicular to the line 12x + 5y + 7 = 0 from the origin is $\frac{7}{13}$ units. (Answer)