Which lines are perpendicular to the line y – 1 = One-thirdx+2? Check all that apply.y + 2 = –3x – 4y − 5 = 3x + 11y = -3x – Five-thirdsy = One-thirdx – 23x + y = 7

Answer: First option. Third option. Fih option. The equation of the line in Slope-Intercept form is: \(y=mx+b\) Where "m" is the slope and "b" is the y-intercept. The equation of the line in Point-slope form is: \(y - y_1 = m(x-x_1)\) Where "m" is the slope and \((x_1,y_1)\) is a point on the line. By definition, the slopes of perpendicular line are negative reciprocals. Then, given the line: \(y- 1 = \frac{}1{}{}3{}(x+2)\) We know that a line perpendicular to it, must have this slope: \(m=-3\) Let's check each option: 1) \(y + 2 = -3(x -4)\) Since \(m=-3\), this line is perpendicular to the line \(y- 1 = \frac{}1{}{}3{}(x+2)\) 2) \(y - 5 = 3(x + 11)\) Since \(m\neq -3\), this line is not perpendicular to the line \(y- 1 = \frac{}1{}{}3{}(x+2)\) 3) \(y = -3x-\frac{}5{}{}3{}\) Since \(m=-3\), this line is perpendicular to the line \(y- 1 = \frac{}1{}{}3{}(x+2)\) 4) \(\frac{}1{}{}3{}x - 2\) Since \(m\neq -3\), this line is not perpendicular to the line \(y- 1 = \frac{}1{}{}3{}(x+2)\) 5) \(3x + y = 7\) Solving for "y": \(y =-3x+ 7\) Since \(m=-3\), this line is perpendicular to the line \(y- 1 = \frac{}1{}{}3{}(x+2)\)