What are the points on the y-axis whose distance from the line $$\dfrac{x}{3}+\dfrac{y}{4}$$ = 1 is 4 units.

Asked by Abhisek | 1 year ago |  66

##### Solution :-

Consider (0, b) as the point on the y-axis whose distance from line

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

It can be written as 4x + 3y – 12 = 0 ……. (1)

By comparing equation (1) to the general equation of line Ax + By + C = 0, we get

A = 4, B = 3 and C = – 12

We know that the perpendicular distance (d) of a line Ax + By + C = 0 from (x1, y1) is written as

$$4=|\dfrac{4(0)+3(b)-12}{\sqrt{4^2+3^2}}|$$

$$4=|\dfrac{3b-12}{5}|$$

By cross multiplication

20 = |3b – 12|

We get

20 = ± (3b – 12)

Here 20 = (3b – 12) or 20 = – (3b – 12)

It can be written as

3b = 20 + 12 or 3b = -20 + 12

So we get

b = $$\dfrac{32}{3}$$or b = $$- \dfrac{8}{3}$$

Hence, the required points are (0,$$\dfrac{32}{3}$$) and (0, $$- \dfrac{8}{3}$$).

Answered by Pragya Singh | 1 year ago

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