
How do you find the slope of $( - 4,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1)$, $( - 2,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 5)$?
Answer
534k+ views
Hint: We will get a straight line joining the given two points. The slope of the line is the change in the value of $y$ with respect to $x$. For a straight line, if two points $A({x_1},{\kern 1pt} {\kern 1pt} {\kern 1pt} {y_1})$ and $B({x_2},{\kern 1pt} {\kern 1pt} {\kern 1pt} {y_2})$ are situated on the line, then by using the slope formula we can calculate the slope (m) as, $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.
Complete step by step solution:
We have to find the slope of the line passing through the points $( - 4,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1)$, $( - 2,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 5)$.
As we already know two points on the line, we will use the slope formula to find the slope of the line.
The slope formula is given by $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
where, $A({x_1},{\kern 1pt} {\kern 1pt} {\kern 1pt} {y_1})$ and $B({x_2},{\kern 1pt} {\kern 1pt} {\kern 1pt} {y_2})$ are the two points on the line
$m$ is the slope of the line
From the given points we can write,
${x_1} = - 4$, ${y_1} = - 1$, ${x_2} = - 2$ and ${y_2} = - 5$
Putting the values in the formula we get,
$
\Rightarrow m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \\
\Rightarrow m = \dfrac{{( - 5) - ( - 1)}}{{( - 2) - ( - 4)}} \\
\Rightarrow m = \dfrac{{ - 5 + 1}}{{ - 2 + 4}} \\
\Rightarrow m = \dfrac{{ - 4}}{2} = - 2 \\
$
Thus, the value of $m$ is $ - 2$.
Hence, the slope of the line passing through $( - 4,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1)$ and $( - 2,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 5)$ is $ - 2$.
Note: For a line making obtuse angle with the x-axis, the slope is negative as the behavior of $y$ is opposite to that of $x$, i.e. the value of $y$ decreases for increase in the value of $x$ and the value of $y$ increases for decrease in the value of $x$. We can also find the slope of the line by first calculating the equation of the line passing through the given points and then using the slope-intercept formula $y = mx + c$, where $m$ is the slope of the line and $c$ is the y-intercept. The choice of the method depends on the information given in the question and the ease of solution.
Complete step by step solution:
We have to find the slope of the line passing through the points $( - 4,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1)$, $( - 2,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 5)$.
As we already know two points on the line, we will use the slope formula to find the slope of the line.
The slope formula is given by $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
where, $A({x_1},{\kern 1pt} {\kern 1pt} {\kern 1pt} {y_1})$ and $B({x_2},{\kern 1pt} {\kern 1pt} {\kern 1pt} {y_2})$ are the two points on the line
$m$ is the slope of the line
From the given points we can write,
${x_1} = - 4$, ${y_1} = - 1$, ${x_2} = - 2$ and ${y_2} = - 5$
Putting the values in the formula we get,
$
\Rightarrow m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \\
\Rightarrow m = \dfrac{{( - 5) - ( - 1)}}{{( - 2) - ( - 4)}} \\
\Rightarrow m = \dfrac{{ - 5 + 1}}{{ - 2 + 4}} \\
\Rightarrow m = \dfrac{{ - 4}}{2} = - 2 \\
$
Thus, the value of $m$ is $ - 2$.
Hence, the slope of the line passing through $( - 4,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1)$ and $( - 2,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 5)$ is $ - 2$.
Note: For a line making obtuse angle with the x-axis, the slope is negative as the behavior of $y$ is opposite to that of $x$, i.e. the value of $y$ decreases for increase in the value of $x$ and the value of $y$ increases for decrease in the value of $x$. We can also find the slope of the line by first calculating the equation of the line passing through the given points and then using the slope-intercept formula $y = mx + c$, where $m$ is the slope of the line and $c$ is the y-intercept. The choice of the method depends on the information given in the question and the ease of solution.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

