Question

The perpendicular distance from 0,0 to the line 7x+24y+15=0 is

Answer 1

Step-by-step explanation:

Given:The perpendicular distance from 0,0 to the line 7x+24y+15=0 is

To find: Perpendicular distance

Solution:

We know that perpendicular distance of a line from origin is given by the

$\bold{\red{d = \bigg| \frac{c}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg |}}$

here equation of line is ax+by+c=0

Here,line is 7x+24y+15=0

a= 7

b= 24

c= 15

Put these values in the formula

$d = \bigg| \frac{15}{ \sqrt{ {(7)}^{2} + {(24)}^{2} } } \bigg | \\ \\ d = \bigg| \frac{15}{ \sqrt{ 49 + 576 } } \bigg | \\ \\ d = \bigg| \frac{15}{ \sqrt{ 625 } } \bigg | \\ \\ d = \frac{15}{25} \\ \\ d = \frac{3}{5} \\ \\ \bold{d = 0.6 \: units}$

Distance between (0,0) and line 7x+24y+15=0 is 0.6 units.

Hope it helps you.

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

https://brainly.in/question/3038830

2)the perpendicular distance of the point (3,-4) from the line 2x-5y+2=0 is

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Answer 2

Step-by-step explanation:

this is the required answer