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Exercise 6.5 · Q17

Q.The length of the perpendicular from the origin to the line x3−y4=1\dfrac{x}{3}-\dfrac{y}{4}=1 is

(1) 115\dfrac{11}{5}
(2) 512\dfrac{5}{12}
(3) 125\dfrac{12}{5}
(4) −512-\dfrac{5}{12}
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Clear denominators to get 4x−3y−12=04x-3y-12=0, then use the perpendicular-distance formula from the origin.

The perpendicular distance of a point (x1,y1)(x_1,y_1) from a 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}}. We first put the given intercept-form line into this general form.

Step 1. Clear the fractions. The line is x3−y4=1\dfrac{x}{3}-\dfrac{y}{4}=1. Multiply throughout by the LCM of 33 and 44, which is 1212:

12(x3)−12(y4)=12  ⟹  4x−3y=12  ⟹  4x−3y−12=012\left(\dfrac{x}{3}\right)-12\left(\dfrac{y}{4}\right)=12 \implies 4x-3y=12 \implies 4x-3y-12=0

Step 2. Apply the perpendicular-distance formula from the origin (0,0)(0,0). Here a=4, b=−3, c=−12a=4,\ b=-3,\ c=-12: …

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