The closest point on the given line is the intersection of that line 6x+2y-5=0 (Let's say that line as A) and the line which is orthoganl to line A and at same time passes through (-1,-2).

First find the equation of the line which is orthoganal to A and which passes through (-1,-2) (Let's say it as B). For that we have to find the gradient of that line B.

Rearrange euqation of line A

`6x+2y-5 = 0`

`2y = -6x+5` 

`y = -3x+(5/3)`

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The gradient of line A is -3. Therefore the gradient of a line which is perpendicular to it is -1/(-3) (Because if the gradient is m, then the perpendicular gradient is -1/m).

So the gradient of B is +(1/3). Now we can find the equation of line which has gradient of 1/3 and passes through (-1,-2).

`(y - (-2))/(x - (-1)) = 1/3`

`3y+6=x+1`

`x-3y-5=0`

The equation of line B is x -3y-5=0 and the intersection point of these two lines can be found by solving these two as simultaneous equations.

6x+2y-5 = 0 -----A

x-3y-5=0     ------B

Multiply B ny 6 and subtract it from A, A - 6 *B

6x-2y-5 -(6x-18y-30) = 0

16y+25 = 0

y = -25/16

Substitute this y value in B, and solve for x,

x - 3*(-25/16) - 5 =0

x = 5 - 75/16

x = (80-75)/16

x = 5/16.

Therefore the point which is the closest to (-1,-2) on the given lien 6x+2y-5 =0 is (5/16,-25/16).

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