L1: 2x+4y-3=0 ..........(1) P: (2,0) The point on the line L1 closest to the given point P is at the intersection of L1 with L2, which is the perpendicular passing through P.
Slope of L1=-2/4=-1/2 Slope of L2=-1/(-1/2)=2 Since it passes throug P(2,0), we can use the point-slope formula: (y-0)=2(x-2) => L2: 2x-y-4=0.............(2)
Solve for x & y using (1) and (2) to get intersection point required: (1)-(2) 2x-2x + 4y-(-y) -3 -(-4) =0 5y=-1, y=-1/5 Substitute y=1/5 in equation (1) 2x+4(-1/5)-3=0 => 2x-19/5=0 x=19/10
=> the point on L1 closest to (2,0) is (19/10, -1/5)
