straight line equations?

Question

1)find an equation of the straight line which passes through the origin and is perpendicular to the line y = 1/2x + 3

2)find a cartesian equation for the perpendicular bisector of the line joining A(2,3) and B(0,6)

answer is 6y = 4x + 23 my friend told me i dont know the working though???

for question 2

Answer 1

1) We know a point on the line (0,0) and the slope of the line m= -2

Thus, using point slope form, (y-y1) = m(x-x1), we get
y - 0 = -2 (x - 0) or y = -2x (your linear equation in slope intercept form)

2) Again we can find a slope and a point. The slope of the line joing A and B can be found using the slope formula
m = (y2-y1)/(x2-x1) --> m = (6-3)/(0-2)= -3/2. The line we are looking for is perpendicular to this so, the slope of that line will be m = 2/3
Now, since the line we are looking for bisects the line joining A and B, we need to find the mid point. You find that by just averaging your x's and averaging your y's. The avg of your x's is 1 and the avg of your y's is 9/2. So your point is
(1, 9/2).
Now we have a slope m = 2/3 and a point (1, 9/2)
We again use the point slope formula to get
(y-9/2) = 2/3(x - 1)--> y - 9/2 = 2/3x - 2/3--->(adding 9/2 to both sides) y = 2/3x +23/6.

Answer 2

The slope of a line perpendicular to y=ax +b is -1/a

-1/(1/2) = 2

y=-2x is a line perpendicular to y = (1/2)x + 3. It passes through the origin because when x=0

y = -2x = -2*0 = 0

2) The slope of the line is (6 - 3)/(0 - 2) = 3/-2 = -1 1/2

Now, lets plug a=-1 1/2 and the coordinates of point B into the equation
y = ax + b

6 = a*0 +b

or b=6

y = (-1 1/2)x + 6

Answer 3

a perpendicular line has the opposite slope as that one
and a y intercept of 0

y=-2x

dont know about the second one though