1. Problem 24: Find the x- and y-intercepts of the given line. Given the graph description, the line passes through x-intercept roughly at $x=3$ and y-intercept roughly at $y=-2$. Therefore, the intercepts are $x=3$ and $y=-2$. Answer: A) 3 and -2 2. Problem 25: Find the slope of a line passing through points $(a,b)$ and $(a,c)$. The slope formula is $m=\frac{y_2-y_1}{x_2-x_1}$. Here, $x_1=a$ and $x_2=a$, so denominator is zero. Slope is undefined when denominator is zero. Answer: B) undefined 3. Problem 26: Find the slope of the line $y = -4(x+2)$. Rewrite as $y = -4x - 8$. The slope is the coefficient of $x$ which is $-4$. Answer: C) -4 4. Problem 27: Find the equation of a line passing through $(0,-2)$ with slope $-2$. Using point-slope form: $y - y_1 = m(x - x_1)$. Substitute: $y + 2 = -2(x - 0)$. Simplify: $y + 2 = -2x$. Rewrite: $2x + y + 2 = 0$. Answer: C) 2x + y + 2 = 0 5. Problem 28: Find the equation of the line perpendicular to $x + 2y = 0$ and passing through $(0,-1)$. First, find slope of $x + 2y = 0$. Rewrite: $2y = -x$ or $y = -\frac{1}{2}x$, so slope is $-\frac{1}{2}$. The perpendicular slope is the negative reciprocal: $2$. Using point-slope form with point $(0,-1)$: $y - (-1) = 2(x - 0)$ or $y + 1 = 2x$. Rearranged: $2x - y + 1 = 0$, equivalent to $2x - y - 1=0$ changing signs. Comparing options, correct is $2x - y -1=0$, which matches A) if we carefully check signs. Option A) is $2x - y - 1=0$ which matches our derivation. Answer: A) 2x - y - 1 = 0 6. Problem 29: Which point lies in the third quadrant? Third quadrant means both $x<0$ and $y<0$. Check options: A) (2,5) both positive, 1st quadrant. B) (2,-5) $x>0$, $y<0$, 4th quadrant. C) (-2,5) $x<0$, $y>0$, 2nd quadrant. D) (-2,-5) both negative, 3rd quadrant. Answer: D) (-2,-5) 7. Problem 30: If two lines are perpendicular, then the product of slopes is $-1$. This is a fundamental property of perpendicular lines. Answer: B) product of their slopes is -1