Question

Find the distance of 7x-y-5=0 from the origin.

Answer 1

GIven:

The equation is given as 7x-y-5 = 0.

The objective is to find the distance between the line and the origin.

Step-by-step explanation:

Consider the coordinate of the origin as,

`(x_1,y_1)=(0,0)`

The general form of straight line is,

`ax_1+by_1+c=0`

By comparing the coefficients of given equation with the general equation,

`\begin{gathered} a=7 \\ b=-1 \\ c=-5 \end{gathered}`

To find the distance:

The formula to find the distance between a point and the equation of a line is,

`d=\frac{\lvert ax_1+by_1+c\rvert}{\sqrt[]{a^2+b^2}}\text{ . . . . ..(1)}`

Substitute the obtained values in equation (1).

`\begin{gathered} d=\frac{\lvert7(0)-1(0)-5\rvert}{\sqrt[]{(7)^2+(-1)^2}} \\ =\frac{\lvert-5\rvert}{\sqrt[]{49+1}} \\ =\frac{5}{\sqrt[]{50}} \\ =\frac{5}{5\sqrt[]{2}} \\ =\frac{1}{\sqrt[]{2}} \end{gathered}`

Hence, the distance between the line and the origin is (1/√2).