The slope-intercept form of the equation of a line that passes through point (–2, –13) is y = 5x – 3. what is the point-slope form of the equation for this line?

y – 13 = 5(x – 2)
y + 13 = 5(x + 2)
y – 2 = 5(x – 13)
y + 2 = 5(x + 13)

Y - y1 = m(x - x1)
slope(m) = 5
(-2,-13)...x1 = -2 and y1 = -13
now sub and pay attention to ur signs
y - (-13) = 5(x - (-2) =
y + 13 = 5(x + 2) <

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