santaclaus wrote:so, would I do, square root 3 = -1 + b? then b=square root 3 +1?

put it into the equation after finding slope like so? y=1/square root 3 + square root 3 +1?

I'm sorry, but I don't understand what you mean by the above...? (I'm assuming, though, that this is with respect to the first of the two exercises...?)

Assuming that the second point is meant to be (-1, sqrt[3]), you have two points; namely, (-1, sqrt[3]) and (0, 0). Plug these into

**the slope formula**.

Since the line passes through the origin, the y-value at which the line crosses the y-axis has been given to you; it is not any of the values you have listed above. (Think: If the point where y = 0 is on the line, then where does the line cross the y-axis?)

Plug the y-intercept and the slope value into the slope-intercept form of

**the line equation**.

santaclaus wrote:btw, I found slope to be 0- -1/0-square root 3 = -1/square root 3 and perpendicular slope is just square root 3

I think you mean [0 - (-1)] / (0 - sqrt[3]). (Division by zero, as written above, is not defined.)

However, the slope formula, as the lesson in the above link explains, is the subtraction of the y-values divided by the subtraction of the x-values. You appear to be subtracting the x-values in the numerator....