Find the points on the x-axis, whose distances from the line $$\dfrac{x}{3}+\dfrac{y}{4}=1$$ are 4 units.

Asked by Pragya Singh | 1 year ago |  108

##### Solution :-

The equation of line is

$$\dfrac{x}{3}+\dfrac{y}{4}=1$$

4x + 3y = 12

4x + 3y – 12 = 0 …. (1)

Now, compare equation (1) with general equation of line Ax + By + C = 0, where A = 4, B = 3, and C = -12

Let (a, 0) be the point on the x-axis, whose distance from the given line is 4 units.

So, the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

$$d= \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$$

$$4= \dfrac{|4a+3\times 0-12|}{\sqrt{4^2+3^2}}$$

$$4= \dfrac{|4a-12|}{\sqrt{16+9}}$$

$$4= \dfrac{|4a-12|}{5}$$

|4a – 12| = 4 × 5

± (4a – 12) = 20

4a – 12 = 20 or – (4a – 12) = 20

4a = 20 + 12 or 4a = -20 + 12

a = $$\dfrac{32}{4}$$ or a = $$- \dfrac{8}{4}$$

a = 8 or a = -2

The required points on the x – axis are (-2, 0) and (8, 0)

Answered by Abhisek | 1 year ago

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