1. **State the problem:** We have two lines: - Line L₁ passes through point $R(1, 2)$ with slope $3$. - Line L₂ is given by the equation $y = -2x + 5$. We need to: a) Find the equation of line L₁. b) Find the point where L₁ and L₂ intersect. 2. **Find the equation of L₁:** The point-slope form of a line is: $$y - y_1 = m(x - x_1)$$ where $m$ is the slope and $(x_1, y_1)$ is a point on the line. Given $m = 3$ and point $R(1, 2)$: $$y - 2 = 3(x - 1)$$ Simplify: $$y - 2 = 3x - 3$$ $$y = 3x - 3 + 2$$ $$y = 3x - 1$$ So, the equation of L₁ is: $$y = 3x - 1$$ 3. **Find the intersection point of L₁ and L₂:** At the intersection, both lines have the same $x$ and $y$ values. Set the equations equal: $$3x - 1 = -2x + 5$$ Solve for $x$: $$3x + 2x = 5 + 1$$ $$5x = 6$$ $$x = \frac{6}{5} = 1.2$$ Find $y$ by substituting $x=1.2$ into one of the line equations, e.g., L₁: $$y = 3(1.2) - 1 = 3.6 - 1 = 2.6$$ 4. **Final answer:** - Equation of L₁: $y = 3x - 1$ - Intersection point: $\left(\frac{6}{5}, \frac{13}{5}\right)$ or $(1.2, 2.6)$