Question

determine another point on the line given two points on the line: (3, 0), (12, 3)
a) (0, -1)
b) (1, -6)
c) (6, -1)
d) (8, 2)

Explanation:

Step1: Find the slope of the line

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). For the points \((3,0)\) and \((12,3)\), we have \( x_1 = 3,y_1 = 0,x_2=12,y_2 = 3 \). So \( m=\frac{3 - 0}{12 - 3}=\frac{3}{9}=\frac{1}{3} \).

Step2: Find the equation of the line

Using the point - slope form \( y - y_1=m(x - x_1) \), with \((x_1,y_1)=(3,0)\) and \( m=\frac{1}{3} \), we get \( y-0=\frac{1}{3}(x - 3) \), which simplifies to \( y=\frac{1}{3}x - 1 \).

Step3: Check each option

• Option A: Substitute \( x = 0 \) into \( y=\frac{1}{3}x-1 \), we get \( y=\frac{1}{3}(0)-1=-1 \). But we need to check if the slope between \((3,0)\) and \((0, - 1)\) is \(\frac{1}{3}\). The slope between \((3,0)\) and \((0,-1)\) is \( \frac{-1 - 0}{0 - 3}=\frac{-1}{-3}=\frac{1}{3} \)? Wait, \( \frac{-1-0}{0 - 3}=\frac{-1}{-3}=\frac{1}{3} \), but let's check the equation. Wait, when \( x = 0 \), from \( y=\frac{1}{3}x-1 \), \( y=-1 \), but let's check the other points.
• Option B: Substitute \( x = 1 \) into \( y=\frac{1}{3}x-1 \), we get \( y=\frac{1}{3}(1)-1=\frac{1 - 3}{3}=-\frac{2}{3}

eq - 6 \).

• Option C: Substitute \( x = 6 \) into \( y=\frac{1}{3}x-1 \), we get \( y=\frac{1}{3}(6)-1=2 - 1 = 1

eq - 1 \).

• Option D: Substitute \( x = 8 \) into \( y=\frac{1}{3}x-1 \), we get \( y=\frac{1}{3}(8)-1=\frac{8 - 3}{3}=\frac{5}{3}

eq2 \). Wait, there is a mistake above. Let's recalculate the slope. The two points are \((3,0)\) and \((12,3)\). The slope \( m=\frac{3 - 0}{12 - 3}=\frac{3}{9}=\frac{1}{3} \). The equation of the line using point - slope form: \( y - 0=\frac{1}{3}(x - 3)\), so \( y=\frac{1}{3}x - 1 \). Now check option A: when \( x = 0 \), \( y=\frac{1}{3}(0)-1=-1 \). Now check the slope between \((3,0)\) and \((0,-1)\): \( m=\frac{-1 - 0}{0 - 3}=\frac{-1}{-3}=\frac{1}{3} \), which is the same as the original slope. Wait, but let's check the other options again. Wait, maybe I made a mistake in option D. Wait, \( y=\frac{1}{3}x-1 \). For \( x = 8 \), \( y=\frac{8}{3}-1=\frac{5}{3}\approx1.666
eq2 \). For option A, when \( x = 0 \), \( y=-1 \), and the slope between \((3,0)\) and \((0,-1)\) is \( \frac{-1-0}{0 - 3}=\frac{1}{3} \), which matches. But wait, let's check the difference in x and y between the two given points. From \((3,0)\) to \((12,3)\), \( \Delta x=12 - 3 = 9 \), \( \Delta y=3 - 0 = 3 \), so the ratio of \( \Delta y/\Delta x=\frac{3}{9}=\frac{1}{3} \). Now, from \((3,0)\) to \((0,-1)\), \( \Delta x=0 - 3=-3 \), \( \Delta y=-1 - 0=-1 \), \( \Delta y/\Delta x=\frac{-1}{-3}=\frac{1}{3} \), which is correct. Wait, but the options: Let's re - examine the problem. Maybe I made a mistake in the equation. Wait, the two points are \((3,0)\) and \((12,3)\). The slope is \( \frac{3 - 0}{12 - 3}=\frac{1}{3} \). The equation is \( y=\frac{1}{3}(x - 3)\), so \( y=\frac{1}{3}x-1 \). Now check option A: \((0,-1)\) satisfies \( y=\frac{1}{3}x - 1 \) (when \( x = 0 \), \( y=-1 \)). The slope between \((3,0)\) and \((0,-1)\) is \( \frac{-1-0}{0 - 3}=\frac{1}{3} \), which is the same as the original slope. So option A is correct.

Answer:

A) (0, -1)