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Question 103 of 129

Q.Find the nearest point on the line x−2y=5x-2y=5 from the origin.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019Subjective· 3mImportance★★★★★
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Using the foot-of-perpendicular formula from the origin to the line x−2y−5=0x-2y-5=0 gives the nearest point (1,−2)(1,-2), at distance 5\sqrt5.

Write the line as x−2y−5=0x-2y-5=0, so a=1a=1, b=−2b=-2, c=−5c=-5.

The foot of the perpendicular from (x0,y0)=(0,0)(x_0,y_0)=(0,0) to ax+by+c=0ax+by+c=0 is given by:

x=x0−a(ax0+by0+c)a2+b2,y=y0−b(ax0+by0+c)a2+b2x = x_0 - \dfrac{a(ax_0+by_0+c)}{a^2+b^2}, \qquad y = y_0 - \dfrac{b(ax_0+by_0+c)}{a^2+b^2}

Here ax0+by0+c=0+0−5=−5ax_0+by_0+c = 0+0-5=-5, and a2+b2=1+4=5a^2+b^2=1+4=5.

x=0−1×(−5)5=0+1=1x = 0 - \dfrac{1\times(-5)}{5} = 0+1 = 1.

y=0−(−2)×(−5)5=0−2=−2y = 0 - \dfrac{(-2)\times(-5)}{5} = 0 - 2 = -2.

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