The slope of y=6x+3 is 6.
Therefore, the equation of the line we want to find is y=6x+a, with a to determine.
Because it passes through (0,-10), we have -10=6*0+a. So a=-10.
Thus the equatio of the line is
y=ax+bb) passes through the point (0,-1) and is perpendicular to the line with equation 3x-2y+5=0
It passes through the point 0,-1. Therefore, -1=a*0+b --> b=-1
A normal vector to it is .
A normal vector to 3x-2y+5=0 is .
Because the two lines are perpendicular, their normal vector are perpendicular too. In particular, the scalar product is 0.
You have the equation of the line now;..
i hope i didn't make any mistake...