
How do you find the slope given $\left( -2,-2 \right)$ and$\left( -2,4 \right)$? \[\]
Answer
493.8k+ views
Hint: We recall the definition of slope form slope of point equation. We use the formula for slope of line joining two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ that is $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ to find the slope of line joining the given points $\left( -2,-2 \right)$ and$\left( -2,4 \right)$ .\[\]
Complete step-by-step solution:
We know from the Cartesian coordinate system that every linear equation can be represented as a line. If the line is inclined with positive $x-$axis at an angle $\theta $ then its slope is given by $m=\tan \theta $ and if it cuts $y-$axis at a distance $c$ from the origin the intercept is given by $c$. The slope-intercept form of equation is given by
\[y=mx+c\]
The slope $m$ here means rise over run which means to what extent the line raised itself above the positive $x-$axis with respect to the extension in the $x-$axis. If $m=0$ we get a line parallel to $x-$axis .We know that if the slope which undefined which means $m=\infty $ we get a line perpendicular to $x-$axis and if $m=0$ we get a line parallel to $x-$axis.\[\]
We know that when we are given a line joining two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ in plane then we can find the slope of the line as the ratio of horizontal range ${{y}_{2}}-{{y}_{1}}$ and vertical range ${{x}_{2}}-{{x}_{1}}$ which means
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
We are given the coordinates of two points $\left( -2,-2 \right)$ and$\left( -2,4 \right)$. So we have ${{x}_{1}}=-2,{{y}_{1}}=-2,{{x}_{2}}=-2,{{y}_{2}}=4$. So the slope of the line is
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{4-\left( -2 \right)}{-2-\left( -2 \right)}=\dfrac{6}{0}=\infty \]
So the slope is undefined and hence the line perpendicular to $x-$axis.
Note: We note that line passing given two points has the equation $x=-2$.The slope determines the orientation and inclination of the line. We note that if the slope is $m > 0$ positive then we get a line increasing from left to right. If the slope is negative that is $m < 0$ we get a line decreasing from left to right.
Complete step-by-step solution:
We know from the Cartesian coordinate system that every linear equation can be represented as a line. If the line is inclined with positive $x-$axis at an angle $\theta $ then its slope is given by $m=\tan \theta $ and if it cuts $y-$axis at a distance $c$ from the origin the intercept is given by $c$. The slope-intercept form of equation is given by
\[y=mx+c\]
The slope $m$ here means rise over run which means to what extent the line raised itself above the positive $x-$axis with respect to the extension in the $x-$axis. If $m=0$ we get a line parallel to $x-$axis .We know that if the slope which undefined which means $m=\infty $ we get a line perpendicular to $x-$axis and if $m=0$ we get a line parallel to $x-$axis.\[\]
We know that when we are given a line joining two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ in plane then we can find the slope of the line as the ratio of horizontal range ${{y}_{2}}-{{y}_{1}}$ and vertical range ${{x}_{2}}-{{x}_{1}}$ which means
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
We are given the coordinates of two points $\left( -2,-2 \right)$ and$\left( -2,4 \right)$. So we have ${{x}_{1}}=-2,{{y}_{1}}=-2,{{x}_{2}}=-2,{{y}_{2}}=4$. So the slope of the line is
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{4-\left( -2 \right)}{-2-\left( -2 \right)}=\dfrac{6}{0}=\infty \]
So the slope is undefined and hence the line perpendicular to $x-$axis.

Note: We note that line passing given two points has the equation $x=-2$.The slope determines the orientation and inclination of the line. We note that if the slope is $m > 0$ positive then we get a line increasing from left to right. If the slope is negative that is $m < 0$ we get a line decreasing from left to right.
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