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Point P(a, b) lie on the line y=x+1 . P is shifted in the direction perpendicular to given line so that it meets x-axis at (\alpha, 0), then

 

Option: 1

\alpha=2 a+1

 


Option: 2

\alpha=a+1

 


Option: 3

\alpha=a-1

 


Option: 4

\alpha=a+2


Answers (1)

best_answer

Parametric equation of line passing through (a, b) and perpendicular to y=x+1  is
                    \frac{x-a}{-1 / \sqrt{2}}=\frac{y-b}{1 / \sqrt{2}}=\alpha
Any point on this line is given by
                    \left(a+\frac{\alpha}{\sqrt{2}}, b+\frac{\alpha}{\sqrt{2}}\right)
If this point is lying on the x-axis, theh
                    b+\frac{\alpha}{\sqrt{2}}=0 \Rightarrow \alpha=-\sqrt{2} b
Hence, the point is (a-b, 0).

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Ritika Kankaria

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