1. **Problem statement:** Given the lines $y = -x + 5$ and $y = x + 1$, which are perpendicular, identify all similar triangles formed by these lines and the coordinate axes, and justify your answer. 2. **Formula and rules:** Triangles are similar if their corresponding angles are equal and their sides are proportional. Since the lines are perpendicular, they form right angles with each other and with the axes. 3. **Step 1: Find intersection points of the lines with the axes and each other.** - Intersection of $y = x + 1$ and $y = -x + 5$: $$x + 1 = -x + 5 \implies 2x = 4 \implies x = 2$$ $$y = 2 + 1 = 3$$ So, $A = (2,3)$. - Intersection of $y = x + 1$ with $y$-axis ($x=0$): $$y = 0 + 1 = 1$$ So, $B = (0,1)$. - Intersection of $y = -x + 5$ with $y$-axis ($x=0$): $$y = -0 + 5 = 5$$ So, $C = (0,5)$. - Intersection of $y = x + 1$ with $x$-axis ($y=0$): $$0 = x + 1 \implies x = -1$$ So, $D = (-1,0)$. - Intersection of $y = -x + 5$ with $x$-axis ($y=0$): $$0 = -x + 5 \implies x = 5$$ So, $F = (5,0)$. - The origin is $E = (0,0)$. 4. **Step 2: Identify triangles formed:** - Triangle $ABC$ with vertices $A(2,3)$, $B(0,1)$, $C(0,5)$. - Triangle $BDE$ with vertices $B(0,1)$, $D(-1,0)$, $E(0,0)$. - Triangle $AEF$ with vertices $A(2,3)$, $E(0,0)$, $F(5,0)$. 5. **Step 3: Show similarity of triangles:** - Triangles $ABC$ and $AEF$ share angle at $A$. - Both have right angles (since lines are perpendicular). - By AA criterion, triangles $ABC \sim AEF$. - Triangles $BDE$ and $ABC$ also have right angles and share angle at $B$, so $BDE \sim ABC$. 6. **Step 4: Calculate lengths $BC$ and $DF$:** - $BC = |y_C - y_B| = |5 - 1| = 4$ - $DF$ is the segment between $D(-1,0)$ and $F(5,0)$ on the $x$-axis: $$DF = |x_F - x_D| = |5 - (-1)| = 6$$ 7. **Step 5: Calculate areas of triangles $CEF$ and $BDE$:** - Triangle $CEF$ vertices: $C(0,5)$, $E(0,0)$, $F(5,0)$. - Base $EF = 5$ (distance from $(0,0)$ to $(5,0)$). - Height $CE = 5$ (distance from $(0,0)$ to $(0,5)$). - Area: $$\text{Area}_{CEF} = \frac{1}{2} \times 5 \times 5 = \frac{25}{2} = 12.5$$ - Triangle $BDE$ vertices: $B(0,1)$, $D(-1,0)$, $E(0,0)$. - Base $DE = 1$ (distance from $(-1,0)$ to $(0,0)$). - Height $BE = 1$ (distance from $(0,0)$ to $(0,1)$). - Area: $$\text{Area}_{BDE} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} = 0.5$$ 8. **Step 6: Ratio of similarity and areas:** - Ratio of similarity (linear scale) between triangles $CEF$ and $BDE$ is the ratio of corresponding sides: $$\frac{EF}{DE} = \frac{5}{1} = 5$$ - Ratio of areas is the square of similarity ratio: $$\left(5\right)^2 = 25$$ - Check area ratio: $$\frac{12.5}{0.5} = 25$$ 9. **Step 7: Calculate area and perimeter of triangle $ABC$:** - Lengths: $$AB = \sqrt{(2-0)^2 + (3-1)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$ $$BC = 4$$ (from step 6) $$AC = \sqrt{(2-0)^2 + (3-5)^2} = \sqrt{4 + 4} = 2\sqrt{2}$$ - Area using base $BC$ and height from $A$: Height is horizontal distance from $A$ to $BC$ line (which is vertical at $x=0$), so height = $x$-coordinate of $A = 2$. $$\text{Area}_{ABC} = \frac{1}{2} \times BC \times \text{height} = \frac{1}{2} \times 4 \times 2 = 4$$ - Perimeter: $$P = AB + BC + AC = 2\sqrt{2} + 4 + 2\sqrt{2} = 4\sqrt{2} + 4$$ **Final answers:** - Similar triangles: $ABC \sim AEF \sim BDE$. - Lengths: $BC = 4$, $DF = 6$. - Areas: $\text{Area}_{CEF} = 12.5$, $\text{Area}_{BDE} = 0.5$. - Similarity ratio: $5$, area ratio: $25$. - Triangle $ABC$ area: $4$, perimeter: $4 + 4\sqrt{2}$.