What are the points on the yaxis whose distance from the line 3x+4y=1is 4 units.

Solution: Given equation is ,3x+4y=1⇒4x+3y−12=0Let the required point is A(0,p).Then, distance from (0,p)=∣∣42+324(0)+3(p)−12∣∣Here, d=4,∴4=±53p−123p−12=20and3p−12=−20⇒p=332andp=−38So, required points will be (0,332) and (0,−38).

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What are the points on the yaxis whose distance from the line

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

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What are the points on the yaxis whose distance from the line

\frac{x}{3} + \frac{y}{4} = 1

is 4 units.

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Text solutionVerified

Solution: Given equation is ,

\frac{x}{3} + \frac{y}{4} = 1

\Rightarrow 4 x + 3 y - 12 = 0

Let the required point is

A (0, p)

.
Then, distance from

(0, p) = ∣ ∣ \frac{4 ( 0 ) + 3 ( p ) - 12}{4 ^{2} + 3 ^{2}} ∣ ∣

Here,

d = 4

∴ 4 = \pm \frac{3 p - 12}{5}

3 p - 12 = 20 and 3 p - 12 = - 20

\Rightarrow p = \frac{32}{3} and p = - \frac{8}{3}

So, required points will be

(0, \frac{32}{3})

and

(0, - \frac{8}{3})

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Question Text	What are the points on the yaxis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.
Answer Type	Text solution:1
Upvotes	150

What are the points on the yaxis whose distance from the line 3x​+4y​=1is 4 units.

Text solutionVerified

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