1. **Problem 1:** Find the equation of the line through $P = (2, 2)$ parallel to the line $y = 3x$. 2. The slope of the given line $y = 3x$ is $3$. 3. Lines that are parallel have the same slope, so the line through $P$ has slope $3$. 4. Use point-slope form: $$y - y_1 = m(x - x_1)$$ where $m=3$ and $(x_1, y_1) = (2, 2)$. 5. Substitute values: $$y - 2 = 3(x - 2)$$ 6. Simplify: $$y - 2 = 3x - 6$$ 7. Add $2$ to both sides: $$y = 3x - 4$$ --- 8. **Problem 2:** Given points $O = (0,0)$, $P = (x_1, 0)$ with $x_1 > 0$, and $Q = (x_2, y_2)$, show that $$|OQ| < |OP| + |PQ|$$. 9. This is the triangle inequality: the length of one side of a triangle is less than the sum of the other two sides. 10. Calculate lengths: $$|OP| = \sqrt{(x_1 - 0)^2 + (0 - 0)^2} = |x_1| = x_1$$ since $x_1 > 0. $$ $$|PQ| = \sqrt{(x_2 - x_1)^2 + (y_2 - 0)^2} = \sqrt{(x_2 - x_1)^2 + y_2^2}$$ $$|OQ| = \sqrt{(x_2 - 0)^2 + (y_2 - 0)^2} = \sqrt{x_2^2 + y_2^2}$$ 11. By the triangle inequality for vectors, $$|OQ| < |OP| + |PQ|$$ holds because $O$, $P$, and $Q$ form a triangle unless $Q$ lies on the segment $OP$. 12. This inequality is a fundamental property of Euclidean distances. **Final answers:** - Equation of the line parallel to $y=3x$ through $(2,2)$ is $$y = 3x - 4$$. - The inequality $$|OQ| < |OP| + |PQ|$$ holds by the triangle inequality.