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Q.Find the distance from the point (1,2)(1, 2) to the line 5x+12y−3=05x+12y-3=0.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2022Subjective· 3mImportance★★★★★
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The distance from (1,2)(1,2) to 5x+12y−3=05x+12y-3=0 is 22 units.

The distance from point (x1,y1)(x_1,y_1) to line ax+by+c=0ax+by+c=0 is d=∣ax1+by1+c∣a2+b2d=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}.

Here a=5,b=12,c=−3a=5,b=12,c=-3, (x1,y1)=(1,2)(x_1,y_1)=(1,2):

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