Find coordinates of a point of a line where function is perpendicular to this vector

Find point M of line x+5y=4 where vector OM would be perpendicular to vector 2a+3b, then a(2,-3) and b(-1,5). Provide coordinates of point M as an answer.

Line: x + 5y = 4 -----> x = 4 − 5y

Point M has coordinates (x, y) = (4−5y, y)
Assuming O is origin, then vector OM = (4−5y, y)

2a+3b = 2(2,−3) + 3(−1,5) = (4,−6) + (−3,15) = (1,9)

Two vectors are perpendicular if their dot product = 0
(4−5y, y) • (1,9) = 0
4 − 5y + 9y = 0
4y = −4
y = −1

x = 4 − 5(−1) = 9

M = (9, −1)

2a + 3b = 2(2,-3) + 3(-1,5) = (4-3,-6+15) = (1,9)

Perpendicular vectors are (-9,1), (9,-1)

I don't know where point O is, so I can't figure out the answer, but vector OM must be in a ratio:
[Mx-Ox] = -9[My-Oy]

Letting Mx = x, we can write My in terms of x.
[x-Ox] = -9[(4-x)/5-Oy]

Just plug in Ox and Oy, then solve for x.