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  • Point  lie on the line <img alt="\mathrm{y=x+1 . P}" src="https://entranceco

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

\mathrm{\frac{x-a}{-\frac{1}{\sqrt{2}}}=\frac{\frac{y-b}{1}}{\sqrt{2}}=\mathrm{\alpha}}
any point on this line is given by \mathrm{\left(a+\frac{\alpha}{\sqrt{2}}, b+\frac{\alpha}{\sqrt{2}}\right)}
If this point is lying on the X-axis, then \mathrm{\mathrm{b}+\frac{\frac{\alpha}{\sqrt{2}}}{}=0 \Rightarrow \alpha=-\sqrt{2} b}
Hence the point is (a – b, 0)

Hence (A) is the correct answer.

Posted by

Pankaj Sanodiya

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