Question

find the distance of the origin from the line 7 x + 24 Y - 50 = 0​

Answer 1

Answer:

2 units

Step-by-step explanation:

7x +24y -50 =0

so the distance of the origin from the line is =║(7×0 +24×0 - 50)÷(7²+24²)║

                                                                             = 50÷25 units = 2 units

Answer 2

The distance between original and given line is 2 units.

Step-by-step explanation:

We are given the following in the question:

$7 x + 24y - 50 = 0$

We have to find the distance of the origin from the line.

Coordinates of origin:

(0,0)

Formula:

$\text{Distance}(ax + by + c = 0,(x_0,y_0)) = \dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$

Comparing, we get

a = 7

b = 24

c = -50

Putting the values, we get:

$D = (7x + 24y -50 = 0,(0,0)) = \dfrac{|7(0) + 24(0)+-50|}{\sqrt{(7)^2 + (24)^2}}\\\\D = \dfrac{50}{25}\\\\D = 2\text{ units}$

Thus, the distance between original and given line is 2 units.

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