
How do you find the slope and intercept of $y = \dfrac{7}{3}x + 8$?
Answer
545.7k+ views
Hint: We can find the slope of the line by using the slope-intercept formula wherein we write the given equation in the form $y = mx + c$, where $m$ is the slope of the line and $c$ is the y-intercept. The intercepts are the points at which the line cuts the x-axis and y-axis which we can find by putting $y = 0$ and $x = 0$ respectively. Alternatively, to find the x-intercept and y-intercept of a linear graph we can write the equation in the form $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.
Complete step by step solution:
We have to find the slope and intercepts of the line given by the equation $y = \dfrac{7}{3}x + 8$.
We will use the slope-intercept formula to find the slope of the line.
The slope-intercept formula is given by $y = mx + c$.
We can see that the given equation in already in the form of $y = mx + c$
On comparing with the standard form of the slope-intercept formula, we see that
$m = \dfrac{7}{3}$ and $c = 8$
Thus, the slope of the given line is $\dfrac{7}{3}$ and the y-intercept is $8$.
Now we rewrite the given equation in the form $\dfrac{x}{a} + \dfrac{y}{b} = 1$. We can write,
$
y = \dfrac{7}{3}x + 8 \\
\Rightarrow \dfrac{7}{3}x - y = - 8 \\
\Rightarrow \dfrac{7}{{3 \times - 8}}x - \dfrac{y}{{ - 8}} = 1 \\
\Rightarrow \dfrac{7}{{ - 24}}x + \dfrac{y}{8} = 1 \\
\Rightarrow \dfrac{x}{{\left( {\dfrac{{ - 24}}{7}} \right)}} + \dfrac{y}{8} = 1 \\
$
Thus we get x-intercept $a = \dfrac{{ - 24}}{7}$ and y-intercept $b = 8$.
Hence, the slope of the line $y = \dfrac{7}{3}x + 8$ is $\dfrac{7}{3}$ and it cuts the x-axis at the point $(\dfrac{{ - 24}}{7},{\kern 1pt} {\kern 1pt} {\kern 1pt} 0)$ and the y-axis at the point $(0,{\kern 1pt} {\kern 1pt} {\kern 1pt} 8)$.
Note: For a line making acute angle with the x-axis, the slope is positive as the behavior of $y$ is same as that of $x$, i.e. the value of $y$ increases for increase in the value of $x$ and the value of $y$ decreases for decrease in the value of $x$. We can also find the x intercept of the line by putting $y = 0$ as when the line is cutting the x-axis the value of $y$ is $0$. Similarly, we can also find the y intercept of the line by putting $x = 0$ in the equation as when the line is cutting the y-axis the value of $x$ is $0$.
Complete step by step solution:
We have to find the slope and intercepts of the line given by the equation $y = \dfrac{7}{3}x + 8$.
We will use the slope-intercept formula to find the slope of the line.
The slope-intercept formula is given by $y = mx + c$.
We can see that the given equation in already in the form of $y = mx + c$
On comparing with the standard form of the slope-intercept formula, we see that
$m = \dfrac{7}{3}$ and $c = 8$
Thus, the slope of the given line is $\dfrac{7}{3}$ and the y-intercept is $8$.
Now we rewrite the given equation in the form $\dfrac{x}{a} + \dfrac{y}{b} = 1$. We can write,
$
y = \dfrac{7}{3}x + 8 \\
\Rightarrow \dfrac{7}{3}x - y = - 8 \\
\Rightarrow \dfrac{7}{{3 \times - 8}}x - \dfrac{y}{{ - 8}} = 1 \\
\Rightarrow \dfrac{7}{{ - 24}}x + \dfrac{y}{8} = 1 \\
\Rightarrow \dfrac{x}{{\left( {\dfrac{{ - 24}}{7}} \right)}} + \dfrac{y}{8} = 1 \\
$
Thus we get x-intercept $a = \dfrac{{ - 24}}{7}$ and y-intercept $b = 8$.
Hence, the slope of the line $y = \dfrac{7}{3}x + 8$ is $\dfrac{7}{3}$ and it cuts the x-axis at the point $(\dfrac{{ - 24}}{7},{\kern 1pt} {\kern 1pt} {\kern 1pt} 0)$ and the y-axis at the point $(0,{\kern 1pt} {\kern 1pt} {\kern 1pt} 8)$.
Note: For a line making acute angle with the x-axis, the slope is positive as the behavior of $y$ is same as that of $x$, i.e. the value of $y$ increases for increase in the value of $x$ and the value of $y$ decreases for decrease in the value of $x$. We can also find the x intercept of the line by putting $y = 0$ as when the line is cutting the x-axis the value of $y$ is $0$. Similarly, we can also find the y intercept of the line by putting $x = 0$ in the equation as when the line is cutting the y-axis the value of $x$ is $0$.
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