
Find the value of x so that the inclination of the line joining the points (x, -3) and (2, 5) is \[{{135}^{\circ }}\].
Answer
571.8k+ views
Hint: We know that the slope of a line joining the two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is equal to the tangent of the angle made by the line with x-axis in anticlockwise direction given by as follows:
\[slope=\tan \theta =\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
Complete step-by-step solution:
We have been given a line joining the points (x, -3) and (2, 5) which makes an angle of \[{{135}^{\circ }}\] with the x-axis.
We know that the slope of a line joining the two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is equal to the tangent of the angle made by the line with x-axis in anticlockwise direction given by as follows:
\[slope=\tan \theta =\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
So we have \[\theta ={{135}^{\circ }},{{x}_{1}}=2,{{x}_{2}}=x,{{y}_{1}}=5,{{y}_{2}}=-3\]
\[\Rightarrow \tan {{135}^{\circ }}=\dfrac{-3-5}{x-2}\]
Since we know that \[\tan {{135}^{\circ }}=\tan \left( {{90}^{\circ }}+{{45}^{\circ }} \right)=-\cot {{45}^{\circ }}=-1\] as in second quadrant tangent function is negative.
\[\begin{align}
& \Rightarrow -1=\dfrac{-3-5}{x-2} \\
& \Rightarrow -1=\dfrac{-8}{x-2} \\
\end{align}\]
On cross multiplication, we get as follows:
\[\Rightarrow -x+2=-8\]
On adding - 2 to both the sides of the equation, we get as follows:
\[\begin{align}
& \Rightarrow -x-2+2=-8-2 \\
& \Rightarrow -x=-10 \\
& \Rightarrow x=10 \\
\end{align}\]
Therefore the value of x is equal to 10.
Note: Use the value of \[\tan {{135}^{\circ }}\] very carefully as sometimes we used the value of \[\tan {{135}^{\circ }}\] as equal to 1 which is which is wrong and thus gives the wrong answer. Also, substitute the values of \[{{x}_{1}},{{x}_{2}},{{y}_{1}},{{y}_{2}}\] carefully in the formula of slope of the line. Also, remember that in \[\tan \theta \], \[\theta \] is the angle between the line and the x-axis in an anticlockwise direction with the x-axis.
\[slope=\tan \theta =\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
Complete step-by-step solution:
We have been given a line joining the points (x, -3) and (2, 5) which makes an angle of \[{{135}^{\circ }}\] with the x-axis.
We know that the slope of a line joining the two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is equal to the tangent of the angle made by the line with x-axis in anticlockwise direction given by as follows:
\[slope=\tan \theta =\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
So we have \[\theta ={{135}^{\circ }},{{x}_{1}}=2,{{x}_{2}}=x,{{y}_{1}}=5,{{y}_{2}}=-3\]
\[\Rightarrow \tan {{135}^{\circ }}=\dfrac{-3-5}{x-2}\]
Since we know that \[\tan {{135}^{\circ }}=\tan \left( {{90}^{\circ }}+{{45}^{\circ }} \right)=-\cot {{45}^{\circ }}=-1\] as in second quadrant tangent function is negative.
\[\begin{align}
& \Rightarrow -1=\dfrac{-3-5}{x-2} \\
& \Rightarrow -1=\dfrac{-8}{x-2} \\
\end{align}\]
On cross multiplication, we get as follows:
\[\Rightarrow -x+2=-8\]
On adding - 2 to both the sides of the equation, we get as follows:
\[\begin{align}
& \Rightarrow -x-2+2=-8-2 \\
& \Rightarrow -x=-10 \\
& \Rightarrow x=10 \\
\end{align}\]
Therefore the value of x is equal to 10.
Note: Use the value of \[\tan {{135}^{\circ }}\] very carefully as sometimes we used the value of \[\tan {{135}^{\circ }}\] as equal to 1 which is which is wrong and thus gives the wrong answer. Also, substitute the values of \[{{x}_{1}},{{x}_{2}},{{y}_{1}},{{y}_{2}}\] carefully in the formula of slope of the line. Also, remember that in \[\tan \theta \], \[\theta \] is the angle between the line and the x-axis in an anticlockwise direction with the x-axis.
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