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:** Set the two equations equal since at intersection $y$ values are the same: $$3x - 1 = -2x + 5$$ Add $2x$ to both sides: $$3x + 2x - 1 = 5$$ $$5x - 1 = 5$$ Add 1 to both sides: $$5x = 6$$ Divide both sides by 5: $$x = \frac{6}{5} = 1.2$$ Substitute $x=1.2$ into L₁ equation to find $y$: $$y = 3(1.2) - 1 = 3.6 - 1 = 2.6$$ 4. **Final answer:** - Equation of L₁: $y = 3x - 1$ - Intersection point: $\left(1.2, 2.6\right)$