Question

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the line.​

Answer 1

`\huge\boxed{y=(2)/(3)x-(4√(13))/(3)\ \vee\ y=(2)/(3)x+(4√(13))/(3)}`

Explanation:

The equation of a line:

`y=mx+b`

We have

`m=(2)/(3)`

substitute:

`y=(2)/(3)x+b`

The formula of a distance between a point and a line:

General form of a line:

`Ax+By+C=0`

Point:

`(x_0,\ y_0)`

Distance:

`d=(|Ax_0+By_0+C|)/(√(A^2+b^2))`

Convert the equation:

`y=(2)/(3)x+b` |subtract
`y` from both sides

`(2)/(3)x-y+b=0` |multiply both sides by 3

`2x-3y+3b=0\to A=2,\ B=-3,\ C=3b`

Coordinates of the point:

`(0,\ 0)\to x_0=0,\ y_0=0`

substitute:

`d=4`

`4=(|2\cdot0+(-3)\cdot0+3b|)/(√(2^2+(-3)^2))\\\\4=(|3b|)/(√(4+9))`

`4=(|3b|)/(√(13))\qquad|` |multiply both sides by
`√(13)`

`4√(13)=|3b|\iff3b=-4√(13)\ \vee\ 3b=4√(13)` |divide both sides by 3

`b=-(4√(13))/(3)\ \vee\ b=(4√(13))/(3)`

Finally:

`y=(2)/(3)x-(4√(13))/(3)\ \vee\ y=(4√(13))/(3)`