Coordinate axes - Ox and Oy. Let this perpendicular intersects AB at the point H. We will also draw a parallel line for Ox that is going through the point A. Let this line intersects CH at the point D. We also will take a point N(3;9). It will lie on the line AD (because the vector AN has coordinates {1; 0}, that means that it is collinear to the position vector that lines on Ox).
We will now find the angle α between AN and AB. For this we will find scalar product of the vectors AN and AB: vector AN has coordinates {1; 0}, and the vector AB has coordinates {6; -5}.
The scalar product of two vectors with coordinates {x1; y1} and {y1; y2} equals to x1 * x2 + y1 * y2. In this case, it equals to 6 * 1 + -5 * 0 = 6.
Also, it equals to the product of the lengthes of those vectors on the cos of angle between thise vectors. In this case, the length on AN equals to 1, the length of AB equals to √(6² + 5²) = √61.
So we can get that cosα * √61 = 6; cosα = 6/√61. Let β be the angle ADH. Because ADH is the right triangle, we get that cosα = sinβ, so sinβ = 6/√61; we know that β is acute, because it is the angle of the right triangle AHD, so cosβ > 0. We can find cosβ through the Pythagoren trigonometric identity. It tells us that cosβ = 5/√61, so tanβ = sinβ/cosβ = 6/5. But β is the interior alternate angle for the pair of parallel lines AD and Ox, so this is the angle between CD and Ox.
Reminder: for the line y = kx + b, k equals to the tan of the angle between this line and Ox.
So we have got that k = 6/5, and y = 6/5 * x + b. But we know that C lies on y, so we can substitute its coordinates in this equality:
-2 = 6/5 * -3 + b.
b = 18/5 - 2 = 8/5 = 1.6
k = 6/5 = 1.2
y = 1.2x + 1.6 - this is the answer.