Question

Find the distance between 2x-3y=4 and 6y=4x+3?

Answer 1

Perpendicular distance between lines is `15/(2sqrt(61))`

Explanation:

Converting given equations into slope intercept form:
`2x-3y=4`
`rarrcolor(white)("XXXX")y = 2/3x-4/3`

`6y=4x+3`
`rarrcolor(white)("XXXX")y=2/3x+1/2`

Note the equations represent parallel lines with a slope of `2/3`

By solving for the `x` and `y` intercepts for both equations,
we note that the vertical distance between the lines of the equations is
`color(white)("XX")abs(1/2-(-4/3)) = 3/2`
and the horizontal distance between the lines is
`color(white)("XX")abs((-3/4) - 1/2) = 5/4`

`abs(AC) = sqrt((5/4)^2+(3/2)^2)`
`color(white)("XXX")= sqrt(61)/4`

`triangle ABC` and `triangle ADC` are similar so

`(abs(BC))/(abs(AC)) = (abs(DB))/(abs(AB))`

`(3/2)/(sqrt(61)/4) = abs(DB)/(5/4)`

`rarr abs(DB) = 5/4*6/sqrt(61)`

`color(white)("XXX")=15/(2sqrt(61))`