Linear Value C15602

1. **State the problem:** We are given two points from a table: $(-12, -45)$ and $(6, 45)$, and we know the graph of the linear equation passes through $\left(\frac{1}{4}, a\right)$. We need to find the value of $a$. 2. **Find the slope $m$ of the line:** The slope formula is $$m = \frac{y_2 - y_1}{x_2 - x_1}.$$ Using the points $(-12, -45)$ and $(6, 45)$: $$m = \frac{45 - (-45)}{6 - (-12)} = \frac{45 + 45}{6 + 12} = \frac{90}{18} = 5.$$ 3. **Write the equation of the line:** Using the slope-intercept form $y = mx + b$, substitute $m=5$ and use one point to find $b$. Using point $(-12, -45)$: $$-45 = 5(-12) + b$$ $$-45 = -60 + b$$ $$b = -45 + 60 = 15.$$ So the equation is: $$y = 5x + 15.$$ 4. **Find $a$ when $x = \frac{1}{4}$:** Substitute $x = \frac{1}{4}$ into the equation: $$a = 5 \times \frac{1}{4} + 15 = \frac{5}{4} + 15 = \frac{5}{4} + \frac{60}{4} = \frac{65}{4}.$$ **Final answer:** $$a = \frac{65}{4}.$$