1. **State the problem:** Find the value of $b$ such that the line passing through points $(0,b)$ and $(2,0)$ is collinear with the line given by the equation $$x - 2y + 1 = 0.$$ 2. **Find the equation of the line through $(0,b)$ and $(2,0)$:** The slope is $$m = \frac{0 - b}{2 - 0} = \frac{-b}{2} = -\frac{b}{2}.$$ Using point-slope form with point $(0,b)$: $$y - b = -\frac{b}{2}(x - 0)$$ Simplifies to: $$y = -\frac{b}{2}x + b.$$ 3. **Rewrite the given line equation in slope-intercept form:** Given: $$x - 2y + 1 = 0,$$ Rearranged: $$-2y = -x - 1,$$ Divide both sides by $-2$: $$y = \frac{1}{2}x + \frac{1}{2}.$$ 4. **Set the slopes equal to ensure collinearity:** The slope from the given line is $\frac{1}{2}$. From step 2, slope is $-\frac{b}{2}$. Set equal: $$-\frac{b}{2} = \frac{1}{2}.$$ Multiply both sides by 2: $$-b = 1,$$ So: $$b = -1.$$ 5. **Check consistency:** Plug $b=-1$ back into the line through $(0,b)$ and $(2,0)$: Equation becomes: $$y = -\frac{-1}{2}x -1 = \frac{1}{2}x -1,$$ Rewrite as $$y = \frac{1}{2}x -1,$$ Given line: $$y = \frac{1}{2}x + \frac{1}{2}.$$ They differ in intercept, so the question must mean these lines are parallel (same slope), not the same line. If the task is to find $b$ such that the points lie on the line $x - 2y + 1=0$, check if $(0,b)$ lies on it: $$0 - 2b + 1=0 \implies -2b + 1=0 \implies b = \frac{1}{2}.$$ Considering the original question, the most consistent interpretation is that the line through $(0,b)$ and $(2,0)$ is the same as the line $x - 2y + 1=0$. To confirm, find $b$ such that both points satisfy the line: - For $(0,b)$: $$0 - 2b + 1=0 \implies b=\frac{1}{2}.$$ - For $(2,0)$: $$2 - 2(0) + 1=3 \neq 0,$$ so $(2,0)$ does not lie on the line. To find $b$ such that the line through $(0,b)$ and $(2,0)$ is parallel to $x - 2y + 1=0$ (slope $1/2$), then as found previously $b = -1$. **Final answer:** If the line through $(0,b)$ and $(2,0)$ is parallel to $x - 2y + 1=0$, then $$b = -1.$$