Question

How do you find the slope of the line that passes through $\left( 2.5,3 \right),\left( 1,-9 \right)$?

Answer 1

Hint: We will understand the definition of slope and use the concept of line equation to find the slope of the line passing through the point $\left( 2.5,3 \right),\left( 1,-9 \right)$. The slope of the line passing through the point $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ is given as: $\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$. So, we will put the value of ${{x}_{1}},{{x}_{2}},{{y}_{1}},\text{and }{{y}_{2}}$ to get the slope of line.

Complete step by step answer:
We know that slope is the tangent of the angle made by the line with x-axis and we generally denote slope with ‘m’. Let us say that the line passes through two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$. Then, the slope
of the line passing through points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ is given as $\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$.
Since, we have to find the slope of the line passing through the points $\left( 2.5,3 \right),\left( 1,-9 \right)$.
Let us say that the slope of the line is ‘m’.
We will compare both the points with general points and find the value of ${{x}_{1}},{{x}_{2}},{{y}_{1}},\text{and }{{y}_{2}}$.
So, after comparing the points $\left( 2.5,3 \right),\left( 1,-9 \right)$ with $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$, we will get:
${{x}_{1}}=2.5,{{x}_{2}}=1,{{y}_{1}}=3,{{y}_{2}}=-9$
Now, we will put the value of ${{x}_{1}},{{x}_{2}},{{y}_{1}},\text{and }{{y}_{2}}$ in $\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$to get the slope.
$\Rightarrow m=\dfrac{\left( -9-3 \right)}{\left( 1-2.5 \right)}$
$\Rightarrow m=\dfrac{-12}{-1.5}$
Since, we can write 1.5 as $\dfrac{15}{10}$, so will put $\dfrac{15}{10}$ in place of 1.5 and we will also cancel the minus(-) at the numerator and denominator.
$\Rightarrow m=\dfrac{12}{\dfrac{15}{10}}$
$\Rightarrow m=\dfrac{12\times 10}{15}$
Now, we will divide numerator and denominator both by 3 as 12 and 15 both are multiple of 3.
$\Rightarrow m=\dfrac{4\times 10}{5}$
Now, we will divide 10 by 5, then we will get:
$\Rightarrow m=2\times 4$
$\therefore m=8$

Hence, slope of the line passing through $\left( 2.5,3 \right),\left( 1,-9 \right)$ is 8.

Note: Student are required to note that when we have general equation of the line as $ax+by+c=0$ , then slope of the line is equal to $-\dfrac{a}{b}$ and y-intercept is equal to $-\dfrac{c}{a}$ . We can also find the slope of the line by equating the first derivative of the line equation to 0 i.e. $\dfrac{dy}{dx}=0$ and when we put x = 0, we will get the y-intercept of the line.