Answer:
a. The equation of the line is [tex]y = \frac{28}{15}x+\frac{41}{30}[/tex]
b. The equation of the line is [tex]y = \frac{-7}{5}x+\frac{1}{5}[/tex]
Explanation:
The general equation of the line in slope-intercept form is:
y = mx + c
where m is the slope and c is the y-intercept
1. To get the slope:
[tex]m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
where (x₁ , y₁) and (x₂ , y₂) are two points that belong to the line
2. To get the y-intercept:
We pick any point (x , y) that belong to the line and use this point to substitute in the general equation:
y = mx + c
m is calculated from part 1 and (x , y) is a known point on the line. So, we solve for c which is the only unknown.
Now, for the required lines:
Part a:
The two given points are (2, 5.1) and (-1, -0.5)
representing (x₁ , y₁) and (x₂ , y₂)
i. Getting the slope:
[tex]m = \frac{-0.5 - 5.1}{-1 -2}=\frac{28}{15}[/tex]
ii. Getting the y-intercept:
I will use the point (-1, -0.5)
y = mx + c
[tex]-0.5 = \frac{28}{15}(-1)+c\\\\-0.5=\frac{-28}{15}+c\\ \\ c = \frac{41}{30}[/tex]
iii. Based on the above calculations, the equation of the line is:
[tex]y = \frac{28}{15}x+\frac{41}{30}[/tex]
Part b:
The two given points are (-2, 3) and (3, -4)
representing (x₁ , y₁) and (x₂ , y₂)
i. Getting the slope:
[tex]m = \frac{-4-3}{3--2}=\frac{-7}{5}[/tex]
ii. Getting the y-intercept:
I will use the point (3, -4)
y = mx + c
[tex]-4 = \frac{-7}{5}(3) + c\\ \\-4 = \frac{-21}{5}+c\\ \\c=\frac{1}{5}[/tex]
iii. Based on the above calculations, the equation of the line is:
[tex]y = \frac{-7}{5}x+\frac{1}{5}[/tex]
Hope this helps :)
