Answer:
[tex] y = \dfrac{1}{25}x + \dfrac{209}{300} [/tex]
Step-by-step explanation:
To find the equation of the line passing through the points [tex]\left(\dfrac{4}{3}, \dfrac{3}{4}\right)[/tex] and [tex]\left(-\dfrac{3}{4}, \dfrac{2}{3}\right)[/tex], we can use the slope-intercept form of a linear equation, which is [tex]y = mx + b[/tex], where [tex]m[/tex] is the slope and [tex]b[/tex] is the y-intercept.
The slope ([tex]m[/tex]) can be found using the formula:
[tex] m = \dfrac{y_2 - y_1}{x_2 - x_1} [/tex]
Let's use the points [tex]\left(\dfrac{4}{3}, \dfrac{3}{4}\right)[/tex] and [tex]\left(-\dfrac{3}{4}, \dfrac{2}{3}\right)[/tex] to find the slope:
[tex] m = \dfrac{\dfrac{2}{3} - \dfrac{3}{4}}{-\dfrac{3}{4} - \dfrac{4}{3}} [/tex]
Now, simplify the expression:
[tex] m = \dfrac{\dfrac{8}{12} - \dfrac{9}{12}}{-\dfrac{9}{12} - \dfrac{16}{12}} [/tex]
[tex] m = \dfrac{-\dfrac{1}{12}}{-\dfrac{25}{12}} [/tex]
[tex] m = \dfrac{1}{25} [/tex]
Now that we have the slope ([tex]m[/tex]), we can use one of the points (let's use [tex]\left(\dfrac{4}{3}, \dfrac{3}{4}\right)[/tex] to find the y-intercept ([tex]b[/tex]):
[tex] \dfrac{3}{4} = \dfrac{1}{25} \cdot \dfrac{4}{3} + b [/tex]
Now, solve for [tex]b[/tex]:
[tex] \dfrac{3}{4} = \dfrac{4}{75} + b [/tex]
[tex] \dfrac{3}{4} - \dfrac{4}{75} = b [/tex]
[tex] \dfrac{225}{300} - \dfrac{16}{300} = b [/tex]
[tex] \dfrac{209}{300} = b [/tex]
Now that we have both the slope ([tex]m[/tex]) and the y-intercept ([tex]b[/tex]), we can write the equation in slope-intercept form:
[tex] y = \dfrac{1}{25}x + \dfrac{209}{300} [/tex]
So, the equation of the line passing through the points is in the slope-intercept form.
Answer:
[tex]y=\dfrac{1}{25}x +\dfrac{209}{300}[/tex]
Step-by-step explanation:
The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.
To find the slope (m), we can use the slope formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Slope formula}}\\\\\large\text{$m=\dfrac{y_2-y_1}{x_2-x_1}$}\\\\\textsf{where:}\\\phantom{w}\bullet\;\;m\; \textsf{is the slope.}\\\phantom{w}\bullet\;\;(x_1,y_1)\;\textsf{and}\;(x_2,y_2)\;\textsf{are two points on the line.}\end{array}}[/tex]
Given that points (4/3, 3/4) and (-3/4, 2/3) are on the line, we can plug their values into the slope formula:
[tex]m=\dfrac{\frac{2}{3}-\frac{3}{4}}{-\frac{3}{4}-\frac{4}{3}}[/tex]
To subtract fractions, they need to have the same denominator. The lowest common multiple of the two denominators, 3 and 4, is 12. Therefore, we need to multiply the numerator and denominator of the fractions with 3 as their denominator by 4, and those of the fractions with 4 as their denominator by 3. This will make the denominators of all fractions 12:
[tex]m=\dfrac{\frac{2\cdot 4}{3\cdot 4}-\frac{3\cdot 3}{4\cdot 3}}{-\frac{3\cdot 3}{4\cdot 3}-\frac{4\cdot 4}{3\cdot 4}}[/tex]
[tex]m=\dfrac{\frac{8}{12}-\frac{9}{12}}{-\frac{9}{12}-\frac{16}{12}}[/tex]
Subtract the fractions:
[tex]m=\dfrac{\frac{8-9}{12}}{\frac{-9-16}{12}}[/tex]
[tex]m=\dfrac{\frac{-1}{12}}{\frac{-25}{12}}[/tex]
[tex]\textsf{Apply the fraction rule:} \quad \dfrac{\frac{a}{c}}{\frac{b}{c}}=\dfrac{a}{b}[/tex]
[tex]m=\dfrac{-1}{-25}[/tex]
[tex]m=\dfrac{1}{25}[/tex]
Now, substitute the slope and one of the points into the point-slope formula. Let's use point (4/3, 3/4):
[tex]\begin{aligned}y-y_1&=m(x-x_1)\\\\y-\dfrac{3}{4}&=\dfrac{1}{25}\left(x-\dfrac{4}{3}\right)\end{aligned}[/tex]
Simplify:
[tex]\begin{aligned}y&=\dfrac{1}{25}x -\dfrac{1}{25}\cdot \dfrac{4}{3}+\dfrac{3}{4}\\\\y&=\dfrac{1}{25}x -\dfrac{4}{75}+\dfrac{3}{4}\\\\y&=\dfrac{1}{25}x -\dfrac{4\cdot 4}{75\cdot 4}+\dfrac{3\cdot 75}{4\cdot 75}\\\\y&=\dfrac{1}{25}x -\dfrac{16}{300}+\dfrac{225}{300}\\\\y&=\dfrac{1}{25}x +\dfrac{209}{300}\end{aligned}[/tex]
Therefore, the equation of the line that passes through the points (4/3, 3/4) and (-3/4, 2/3) is:
[tex]\Large\boxed{\boxed{y=\dfrac{1}{25}x +\dfrac{209}{300}}}[/tex]
