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Misc-II · Q131

Q.Find points on the X-axis whose distance from the line x/3 + y/4 = 1 is 4 unit.

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The line x3+y4=1\dfrac{x}{3}+\dfrac{y}{4}=1 is 4x+3y−12=04x+3y-12=0. For a point (x,0)(x,0) on the X-axis, distance =∣4x−12∣5=4⇒∣4x−12∣=20=\dfrac{|4x-12|}{5}=4 \Rightarrow |4x-12|=20. So $4x-12=20 \Rightarrow …

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