Write an equation of the line that passes through (-3, 3) and is perpendicular to the line 2y = 8x - 6.

My method is not the most common, but is correct:

Realize that if 2 lines are perpendicular, then the slope of one (call it "m") is the negative reciprocal of the slope of the other.

Solving 2y = 8x - 6 for y:  y = 4x - 3.  This has a slope of "m"=4, and so the slope of any line perpendicular to y = 4x - 3  will be -1/4.

Now we have the slope of the new line and know that this line passes through (-3,3).  Use the point-slope formula here:

y-3 = (-1/4)(x-[-3]), or     y-3 = (-1/4)(x+3)

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12