Question

write an equation of the line that passes through a pair of points: (-5, -2), (3, -1)
a. $y = \frac{1}{8}x + \frac{11}{8}$
b. $y = \frac{1}{8}x - \frac{11}{8}$
c. $y = -\frac{1}{8}x - \frac{11}{8}$
d. $y = \frac{1}{8}x + \frac{8}{11}$
please select the best answer from the choices provided
a
b
c
d

Explanation:

Step1: Calculate the slope

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 \((-5, -2)\) and \((3, -1)\), we have \( x_1=-5,y_1 = - 2,x_2=3,y_2=-1 \). So \( m=\frac{-1-(-2)}{3-(-5)}=\frac{-1 + 2}{3 + 5}=\frac{1}{8} \).

Step2: Use point - slope form

The point - slope form of a line is \( y - y_1=m(x - x_1) \). Let's use the point \((3,-1)\) (we could also use \((-5,-2)\)). Substitute \( m = \frac{1}{8},x_1 = 3,y_1=-1 \) into the formula: \( y-(-1)=\frac{1}{8}(x - 3) \), which simplifies to \( y + 1=\frac{1}{8}x-\frac{3}{8} \). Then \( y=\frac{1}{8}x-\frac{3}{8}-1=\frac{1}{8}x-\frac{3 + 8}{8}=\frac{1}{8}x-\frac{11}{8} \)? Wait, no, wait. Wait, if we use the point \((-5,-2)\): \( y-(-2)=\frac{1}{8}(x - (-5)) \), so \( y + 2=\frac{1}{8}(x + 5) \), \( y=\frac{1}{8}x+\frac{5}{8}-2=\frac{1}{8}x+\frac{5-16}{8}=\frac{1}{8}x-\frac{11}{8} \)? Wait, no, I made a mistake. Wait, let's recalculate the slope again. \( y_2 - y_1=-1-(-2)=-1 + 2 = 1 \), \( x_2 - x_1=3-(-5)=3 + 5 = 8 \), so slope \( m=\frac{1}{8} \). Now use point \((3,-1)\): \( y - (-1)=\frac{1}{8}(x - 3)\Rightarrow y + 1=\frac{1}{8}x-\frac{3}{8}\Rightarrow y=\frac{1}{8}x-\frac{3}{8}-1=\frac{1}{8}x-\frac{3 + 8}{8}=\frac{1}{8}x-\frac{11}{8} \)? Wait, but when we plug \( x=-5 \) into option a: \( y=\frac{1}{8}\times(-5)+\frac{11}{8}=\frac{-5 + 11}{8}=\frac{6}{8}=\frac{3}{4} \), but the \( y \)-value at \( x = - 5 \) should be - 2. Wait, no, wait my calculation is wrong. Wait, let's use point \((-5,-2)\) in option a: \( y=\frac{1}{8}\times(-5)+\frac{11}{8}=\frac{-5 + 11}{8}=\frac{6}{8}=\frac{3}{4} eq - 2 \). Wait, no, I messed up the point - slope. Wait, let's do it again. The two points are \((x_1,y_1)=(-5,-2)\) and \((x_2,y_2)=(3,-1)\). Slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-1-(-2)}{3-(-5)}=\frac{1}{8} \). Now, the slope - intercept form is \( y=mx + b \). Let's plug in one of the points, say \((3,-1)\) into \( y=\frac{1}{8}x + b \). So \( -1=\frac{1}{8}\times3 + b\Rightarrow b=-1-\frac{3}{8}=-\frac{8 + 3}{8}=-\frac{11}{8} \)? No, that's not right. Wait, plug in \((-5,-2)\): \( -2=\frac{1}{8}\times(-5)+b\Rightarrow -2=-\frac{5}{8}+b\Rightarrow b=-2+\frac{5}{8}=-\frac{16}{8}+\frac{5}{8}=-\frac{11}{8} \). Wait, but option a is \( y=\frac{1}{8}x+\frac{11}{8} \). Let's check if \((-5,-2)\) satisfies option a: \( y=\frac{1}{8}\times(-5)+\frac{11}{8}=\frac{-5 + 11}{8}=\frac{6}{8}=\frac{3}{4} eq - 2 \). Wait, no, I think I made a mistake in the sign when calculating \( b \). Wait, let's use the two - point formula correctly. The equation of a line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \( y-y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \). So \( y-(-2)=\frac{-1-(-2)}{3-(-5)}(x - (-5)) \), \( y + 2=\frac{1}{8}(x + 5) \), \( y=\frac{1}{8}x+\frac{5}{8}-2=\frac{1}{8}x+\frac{5-16}{8}=\frac{1}{8}x-\frac{11}{8} \). But let's check option a with the point \((3,-1)\): \( y=\frac{1}{8}\times3+\frac{11}{8}=\frac{3 + 11}{8}=\frac{14}{8}=\frac{7}{4} eq - 1 \). Wait, no, I must have messed up the points. Wait, the first point is \((-5,-2)\) and the second is \((3,-1)\). Let's recalculate the slope: \( m=\frac{-1-(-2)}{3-(-5)}=\frac{1}{8} \). Now, let's use the slope - intercept form \( y = mx + b \). Let's take \( x = 3,y=-1 \): \( -1=\frac{1}{8}\times3 + b\Rightarrow b=-1-\frac{3}{8}=-\frac{11}{8} \). Take \( x=-5,y=-2 \): \( -2=\frac{1}{8}\times(-5)+b\Rightarrow b=-2+\frac{5}{8}=-\frac{11}{8} \). Wait, but option a is \( y=\frac{1}{8}x+\frac{11}{8} \). Let's check \( x=-5 \) in option a: \( y=\frac{-5 + 11}{8}=\frac{6}{8}=\frac{3}{4} eq - 2 \). Option b: \( y=\frac{1}{8}x-\frac{11}{8} \). Check \( x = 3 \): \( y=\frac{3-11}{8}=\frac{-8}{8}=-1 \), which matches. Check \( x=-5 \): \( y=\frac{-5-11}{8}=\frac{-16}{8}=-2 \), which matches. Wait, but earlier when I calculated using \((3,-1)\), I got \( y=\frac{1}{8}x-\frac{11}{8} \), which is option b. Wait, but the initial calculation when I used \((-5,-2)\) I think I made a sign error. So the correct equation is \( y=\frac{1}{8}x-\frac{11}{8} \), which is option b? Wait, no, wait, let's check option a again. Wait, maybe I mixed up the points. Wait, the first point is \((-5,-2)\) and the second is \((3,-1)\). Let's use the formula \( y = mx + b \). \( m=\frac{1}{8} \). Let's plug in \((-5,-2)\): \( -2=\frac{1}{8}\times(-5)+b\Rightarrow b=-2+\frac{5}{8}=-\frac{11}{8} \). So the equation is \( y=\frac{1}{8}x-\frac{11}{8} \), which is option b. Wait, but when I first calculated with \((3,-1)\), I also got \( y=\frac{1}{8}x-\frac{11}{8} \). So the correct answer is option b? Wait, no, wait, let's check option a: \( y=\frac{1}{8}x+\frac{11}{8} \). If \( x = 3 \), \( y=\frac{3 + 11}{8}=\frac{14}{8}=\frac{7}{4} eq - 1 \). Option b: \( x = 3 \), \( y=\frac{3-11}{8}=-1 \); \( x=-5 \), \( y=\frac{-5-11}{8}=-2 \). So it's correct.

Answer:

B. \( y=\frac{1}{8}x-\frac{11}{8} \)