Question

Find the perpendicular distance from the origin to the line 6x + 8y - 50 = 0.​

Answer 1

Answer:

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Answer 2

The perpendicular distance from the origin to the line 6x + 8y - 50 = 0 is 5 unit

Given :

The equation of the line 6x + 8y - 50 = 0

To find :

The perpendicular distance from the origin to the line

Solution :

Step 1 of 2 :

Write down the given equation of the line

The given equation of the line is

6x + 8y - 50 = 0

Step 2 of 2 :

Find the perpendicular distance from the origin to the line

The required perpendicular distance from the origin to the line

$\displaystyle \sf{ = \bigg| \: \frac{(6 \times 0) + (8 \times 0) - 50}{ \sqrt{ {6}^{2} + {8}^{2} } } \: \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg| \: \frac{0 + 0 - 50}{ \sqrt{ 36 + 64 } } \: \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg| \: \frac{ - 50}{ \sqrt{ 100 } } \: \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg| \: - \frac{ 50}{ 10} \: \bigg| } \: \: unit$

$\displaystyle \sf{ =5 } \: \: unit$

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