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Question: IIf the parametric equation of a line is given by \(x=4+\dfrac{t}{\sqrt{2}}\) and \(y=- 1+\sqrt{2}...

IIf the parametric equation of a line is given by x=4+t2x=4+\dfrac{t}{\sqrt{2}} and y=1+2ty=- 1+\sqrt{2}t where tt is the parameter, then
(a) Slope of the line is tan1(2){{\tan }^{-1}}\left( 2 \right)
(b) Slope of the line is tan1(12){{\tan }^{-1}}\left( \dfrac{1}{2} \right)
(c) Intercept made by the line on the x-axis =92=\dfrac{9}{2}
(d) Intercept made by the line on the y-axis =9=-9

Explanation

Solution

Hint: Simplify the given line equation and substitute it into the parametric equation.

The given equations are,
x=4+t2x=4+\dfrac{t}{\sqrt{2}} and y=1+2ty=-1+\sqrt{2}t
We have to rearrange these such that we can formulate an equation in xx and yy terms. To
change the parametric form of the equation, multiply the equation x=4+t2x=4+\dfrac{t}{\sqrt{2}} by 22,
2x=8+2t22x=8+\dfrac{2t}{\sqrt{2}}
2x=8+2t2x=8+\sqrt{2}t
From this we can write,
2t=2x8\sqrt{2}t=2x-8
Now, we can substitute this in the equation y=1+2ty=-1+\sqrt{2}t,
y=1+(2x8)y=-1+\left( 2x-8 \right)
y=2x9y=2x-9
The options indicate that we need to compute the slope and the intercepts of the line y=2x9y=2x-9. It is
in the form of y=mx+cy=mx+c, where mm is the slope and cc is the y-intercept.
On comparing the equation y=2x9y=2x-9 with the general form, we get the slope as 22 and the y-
intercept as 9-9.
The x-intercept can be computed by taking y=0y=0,
0=2x90=2x-9
x=92x=\dfrac{9}{2}
Looking at the options, we get that both option (c) and (d) are true.

Note: The slope of a line is given by tanθ\tan \theta or by yx\dfrac{y}{x}. We get the slope as 22 for
the line in the question. It means that tanθ=2\tan \theta =2 is the slope of the line. The angle of
inclination of a line is represented by tan1θ{{\tan }^{-1}}\theta . So, the options (a) and (b) do not
represent the slope but the angle of inclination of the line.