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)`
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`
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
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).