QUESTION IMAGE
Question
write an equation of the line that passes through a pair of points: (-2, -2), (5, -5)
a. $y = \frac{3}{7}x - \frac{20}{7}$ c. $y = -\frac{3}{7}x + \frac{7}{20}$
b. $y = -\frac{3}{7}x + \frac{20}{7}$ d. $y = -\frac{3}{7}x - \frac{20}{7}$
please select the best answer from the choices provided
a
b
c
d
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 \((-2, -2)\) and \((5, -5)\), we have \( x_1=-2,y_1 = - 2,x_2 = 5,y_2=-5 \). So \( m=\frac{-5-(-2)}{5-(-2)}=\frac{-5 + 2}{5 + 2}=\frac{-3}{7}=-\frac{3}{7} \).
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 \((-2,-2)\) and \( m =-\frac{3}{7} \). Substitute into the formula: \( y-(-2)=-\frac{3}{7}(x - (-2)) \), which simplifies to \( y + 2=-\frac{3}{7}(x + 2) \).
Step3: Simplify the equation
Expand the right - hand side: \( y+2=-\frac{3}{7}x-\frac{6}{7} \). Then subtract 2 from both sides. Since \( 2=\frac{14}{7} \), we have \( y=-\frac{3}{7}x-\frac{6}{7}-\frac{14}{7}=-\frac{3}{7}x-\frac{20}{7} \)? Wait, no, wait. Wait, let's re - do the simplification.
Wait, \( y+2=-\frac{3}{7}(x + 2) \)
\( y+2=-\frac{3}{7}x-\frac{6}{7} \)
Subtract 2 (which is \( \frac{14}{7} \)) from both sides:
\( y=-\frac{3}{7}x-\frac{6}{7}-\frac{14}{7}=-\frac{3}{7}x-\frac{20}{7} \)? No, that's not right. Wait, wait, let's check the calculation of the slope again.
Wait, \( y_2 - y_1=-5-(-2)=-5 + 2=-3 \), \( x_2 - x_1=5-(-2)=5 + 2 = 7 \), so \( m =-\frac{3}{7} \) is correct.
Now, using the point \((5,-5)\) and \( m=-\frac{3}{7} \) in point - slope form \( y - y_1=m(x - x_1) \):
\( y-(-5)=-\frac{3}{7}(x - 5) \)
\( y + 5=-\frac{3}{7}x+\frac{15}{7} \)
Subtract 5 (which is \( \frac{35}{7} \)) from both sides:
\( y=-\frac{3}{7}x+\frac{15}{7}-\frac{35}{7}=-\frac{3}{7}x-\frac{20}{7} \)? No, that's not matching the options. Wait, maybe I made a mistake in the sign. Wait, let's check the options again. Wait, option b is \( y =-\frac{3}{7}x+\frac{20}{7} \), option d is \( y=-\frac{3}{7}x-\frac{20}{7} \). Wait, let's re - calculate the point - slope form with the first point \((-2,-2)\):
\( y - (-2)=-\frac{3}{7}(x - (-2))\)
\( y + 2=-\frac{3}{7}(x + 2) \)
\( y+2=-\frac{3}{7}x-\frac{6}{7} \)
\( y=-\frac{3}{7}x-\frac{6}{7}-2 \)
\( y=-\frac{3}{7}x-\frac{6}{7}-\frac{14}{7}=-\frac{3}{7}x-\frac{20}{7} \)? But that's not one of the options? Wait, no, wait, maybe I mixed up the points. Wait, let's check the options again. Wait, option b is \( y =-\frac{3}{7}x+\frac{20}{7} \), option d is \( y=-\frac{3}{7}x-\frac{20}{7} \). Wait, maybe I made a mistake in the slope calculation. Wait, \( y_2 - y_1=-5-(-2)=-3 \), \( x_2 - x_1=5-(-2)=7 \), so \( m =-\frac{3}{7} \) is correct.
Wait, let's plug the point \((5,-5)\) into option b: \( y =-\frac{3}{7}x+\frac{20}{7} \). When \( x = 5 \), \( y=-\frac{15}{7}+\frac{20}{7}=\frac{5}{7}
eq - 5 \). Plug into option d: \( y=-\frac{3}{7}x-\frac{20}{7} \), when \( x = 5 \), \( y=-\frac{15}{7}-\frac{20}{7}=-\frac{35}{7}=-5 \). When \( x=-2 \), \( y=-\frac{3}{7}\times(-2)-\frac{20}{7}=\frac{6}{7}-\frac{20}{7}=-\frac{14}{7}=-2 \). So the correct equation is \( y =-\frac{3}{7}x-\frac{20}{7} \), which is option d. Wait, but in the original options, option d is \( y =-\frac{3}{7}x-\frac{20}{7} \). Wait, maybe I made a mistake in the earlier simplification. Let's re - do the point - slope form:
Using \( m =-\frac{3}{7} \) and the point \((-2,-2)\):
\( y - y_1=m(x - x_1) \)
\( y-(-2)=-\frac{3}{7}(x - (-2)) \)
\( y + 2=-\frac{3}{7}(x + 2) \)
\( y=-\frac{3}{7}x-\frac{6}{7}-2 \)
\( 2=\frac{14}{7} \), so \( y=-\frac{3}{7}x-\frac{6 + 14}{7}=-\frac{3}{7}x-\frac{20}{7} \), which matches option d.
Wait, but let's check option b: \( y =-\frac{3}{7}x+\frac{20}{7} \). When \( x=-2 \), \( y=\frac{6}{7}+\frac{20}{7}=\…
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D. \( y =-\frac{3}{7}x-\frac{20}{7} \)