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Exercise 12.1 · Q8

Q.What are the points on the x-axis whose perpendicular distance from the line x3+y4=1\frac{x}{3} + \frac{y}{4} = 1 is 4 units.

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Convert the line to general form, then set the perpendicular-distance formula from a point (x0,0)(x_0,0) equal to 4 and solve for x0x_0.

Perpendicular distance of point (x1,y1)(x_1,y_1) from line Ax+By+C=0Ax+By+C=0: d=∣Ax1+By1+C∣A2+B2d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.

  1. Write x3+y4=1\dfrac{x}{3}+\dfrac{y}{4}=1 in general form: multiply by 12: 4x+3y−12=04x+3y-12=0, so A=4, B=3, C=−12A=4,\ B=3,\ C=-12.
  2. A point on the x-axis is (x0,0)(x_0,0). Its distance from the line is

d=∣4x0+3(0)−12∣42+32=∣4x0−12∣5d=\frac{|4x_0+3(0)-12|}{\sqrt{4^2+3^2}}=\frac{|4x_0-12|}{5}

  1. Set d=4d=4: ∣4x0−12∣5=4⇒∣4x0−12∣=20\dfrac{|4x_0-12|}{5}=4\Rightarrow |4x_0-12|=20.
  2. Case (a): 4x0−12=20⇒4x0=32⇒x0=84x_0-12=20\Rightarrow4x_0=32\Rightarrow x_0=8. …

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