1. **Problem statement:** Given two lines AB and CD that are perpendicular and intersect at point (1, 4). Line AB crosses the x-axis at (3, 0). We need to find the coordinates of points P and Q, where P lies on AB at the y-axis and Q lies on CD at the x-axis. 2. **Find the equation of line AB:** We know AB passes through (1, 4) and (3, 0). Slope of AB is $$m_{AB} = \frac{0 - 4}{3 - 1} = \frac{-4}{2} = -2$$ Using point-slope form with point (1, 4): $$y - 4 = -2(x - 1)$$ $$y = -2x + 2 + 4 = -2x + 6$$ 3. **Find coordinates of point P on AB at the y-axis:** At the y-axis, $x=0$. Substitute into AB equation: $$y = -2(0) + 6 = 6$$ So, $$P = (0, 6)$$ 4. **Find the equation of line CD:** Since CD is perpendicular to AB, its slope is the negative reciprocal of $m_{AB}$: $$m_{CD} = \frac{1}{2}$$ Line CD passes through (1, 4), so: $$y - 4 = \frac{1}{2}(x - 1)$$ $$y = \frac{1}{2}x - \frac{1}{2} + 4 = \frac{1}{2}x + \frac{7}{2}$$ 5. **Find coordinates of point Q on CD at the x-axis:** At the x-axis, $y=0$. Set $y=0$ in CD equation: $$0 = \frac{1}{2}x + \frac{7}{2}$$ Multiply both sides by 2: $$0 = x + 7$$ $$x = -7$$ So, $$Q = (-7, 0)$$ **Final answers:** $$P = (0, 6)$$ $$Q = (-7, 0)$$