Line Through A E5C633

1. **State the problem:** We are given the line $y = 16 - 8x$ passing through points $A$ and $B$. Point $N$ is the midpoint of segment $AB$. We need to find the equation of the line passing through point $A$ that has an $x$-intercept of $-18$ in the form $y = mx + c$. 2. **Find coordinates of points $A$ and $B$:** - Since $B$ lies on the positive $x$-axis, its $y$-coordinate is $0$. Set $y=0$ in the line equation: $$0 = 16 - 8x \implies 8x = 16 \implies x = 2$$ - So, $B = (2, 0)$. 3. **Find coordinates of midpoint $N$:** - $N$ is midpoint of $AB$, so its coordinates are: $$N = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)$$ 4. **Find coordinates of $A$ using $N$:** - $N$ lies on the line $y = 16 - 8x$, so $N$ satisfies this equation. - Let $A = (x_A, y_A)$. - Since $N$ is midpoint, $$x_N = \frac{x_A + 2}{2}, \quad y_N = \frac{y_A + 0}{2} = \frac{y_A}{2}$$ - Also, $y_N = 16 - 8x_N$. 5. **Express $y_A$ in terms of $x_A$:** - Since $A$ lies on the line $y = 16 - 8x$, $$y_A = 16 - 8x_A$$ 6. **Substitute $y_A$ and $x_N, y_N$ into midpoint equation:** $$y_N = \frac{y_A}{2} = \frac{16 - 8x_A}{2} = 8 - 4x_A$$ $$x_N = \frac{x_A + 2}{2}$$ - Since $N$ lies on the line, $$y_N = 16 - 8x_N = 16 - 8 \cdot \frac{x_A + 2}{2} = 16 - 4(x_A + 2) = 16 - 4x_A - 8 = 8 - 4x_A$$ 7. **Check consistency:** - Both expressions for $y_N$ are equal: $8 - 4x_A = 8 - 4x_A$, so $x_A$ can be any value on the line. 8. **Find coordinates of $A$:** - Since $A$ is not specified, we can use the fact that $A$ lies on the line and the $x$-intercept of the new line passing through $A$ is $-18$. 9. **Find slope $m$ of the new line passing through $A$ and $x$-intercept $-18$:** - The $x$-intercept is the point $(-18, 0)$. - Let $A = (x_A, y_A) = (x_A, 16 - 8x_A)$. - Slope $m$ is: $$m = \frac{y_A - 0}{x_A - (-18)} = \frac{16 - 8x_A}{x_A + 18}$$ 10. **Write equation of new line passing through $A$:** $$y - y_A = m(x - x_A)$$ Substitute $y_A$ and $m$: $$y - (16 - 8x_A) = \frac{16 - 8x_A}{x_A + 18}(x - x_A)$$ 11. **Simplify to slope-intercept form $y = mx + c$:** $$y = \frac{16 - 8x_A}{x_A + 18}x - \frac{16 - 8x_A}{x_A + 18}x_A + 16 - 8x_A$$ 12. **Simplify the constant term:** $$- \frac{16 - 8x_A}{x_A + 18}x_A + 16 - 8x_A = 16 - 8x_A - \frac{16x_A - 8x_A^2}{x_A + 18}$$ 13. **Choose $A$ to simplify:** - Since $A$ lies on the original line, pick $x_A = 0$ for simplicity. - Then $y_A = 16 - 8 \cdot 0 = 16$. - Slope: $$m = \frac{16 - 0}{0 + 18} = \frac{16}{18} = \frac{8}{9}$$ - Equation: $$y - 16 = \frac{8}{9}(x - 0) \implies y = \frac{8}{9}x + 16$$ 14. **Final answer:** $$\boxed{y = \frac{8}{9}x + 16}$$ This is the equation of the line passing through point $A$ with $x$-intercept $-18$.