We first need to express the given line in standard form.

`y=6/5x+2`

`5y = 6x + 10`

`6x - 5y + 10 = 0`

Using the formula for the distance from a point to a line, we have:

`d=(|Am+Bn+C|)/(sqrt(A^2+B^2`

`=(|(6)(-3)+(-5)(7)+10|)/sqrt(36+25)`

`=|-5.506|`

`=5.506`

So the required distance is `5.506` units, correct to 3 decimal places.

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