QUESTION IMAGE
Question
- points: (-1, -5) and (-3, -25)
equation: _____________
- points: (5, 2) and (-5, -6)
equation: _____________
- points: (6, 3) and (14, -1)
equation: _____________
- points: (-2, 1) and (4, 10)
equation: _____________
Step1: Recall the slope formula
The slope \( m \) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Then use the point - slope form \( y - y_1=m(x - x_1) \) to find the equation of the line.
For Problem 5: Points \((-1,-5)\) and \((-3,-25)\)
Step1: Calculate the slope
Let \((x_1,y_1)=(-1,-5)\) and \((x_2,y_2)=(-3,-25)\). Then \( m=\frac{-25-(-5)}{-3 - (-1)}=\frac{-25 + 5}{-3 + 1}=\frac{-20}{-2}=10 \)
Step2: Use point - slope form
Using the point \((-1,-5)\) and \( m = 10 \), the point - slope form is \( y-(-5)=10(x - (-1)) \), which simplifies to \( y + 5=10(x + 1) \). Expanding, we get \( y+5 = 10x+10 \), and then \( y=10x + 5 \)
For Problem 6: Points \((6,3)\) and \((14,-1)\)
Step1: Calculate the slope
Let \((x_1,y_1)=(6,3)\) and \((x_2,y_2)=(14,-1)\). Then \( m=\frac{-1 - 3}{14 - 6}=\frac{-4}{8}=-\frac{1}{2} \)
Step2: Use point - slope form
Using the point \((6,3)\) and \( m=-\frac{1}{2} \), the point - slope form is \( y - 3=-\frac{1}{2}(x - 6) \). Expanding, \( y-3=-\frac{1}{2}x + 3 \), and then \( y=-\frac{1}{2}x+6 \)
For Problem 7: Points \((5,2)\) and \((-5,-6)\)
Step1: Calculate the slope
Let \((x_1,y_1)=(5,2)\) and \((x_2,y_2)=(-5,-6)\). Then \( m=\frac{-6 - 2}{-5 - 5}=\frac{-8}{-10}=\frac{4}{5} \)
Step2: Use point - slope form
Using the point \((5,2)\) and \( m=\frac{4}{5} \), the point - slope form is \( y - 2=\frac{4}{5}(x - 5) \). Expanding, \( y-2=\frac{4}{5}x-4 \), and then \( y=\frac{4}{5}x-2 \)
For Problem 8: Points \((-2,1)\) and \((4,10)\)
Step1: Calculate the slope
Let \((x_1,y_1)=(-2,1)\) and \((x_2,y_2)=(4,10)\). Then \( m=\frac{10 - 1}{4-(-2)}=\frac{9}{6}=\frac{3}{2} \)
Step2: Use point - slope form
Using the point \((-2,1)\) and \( m=\frac{3}{2} \), the point - slope form is \( y - 1=\frac{3}{2}(x + 2) \). Expanding, \( y-1=\frac{3}{2}x+3 \), and then \( y=\frac{3}{2}x + 4 \)
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- \( y = 10x+5 \)
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