What are the points on y-axis whose distance from the line x 3 + y 4 =1 is 4 units?

Let the required point be (0, a) Given, distnace of (0, a) from line 4x + 3y - 12 = 0 is 4units. D= | ax_1+by+c a^2+b^2 | 4= | 4(0)+3(a)-12 4^2+3^2 | 4= | -3a+12 5 | 4= -3a+12 5 -3a=20-12 a=-8 3 And 4= 3a-12 5 3a=20+12 =32 3 So, Required points are (0,32 3 ), (0,-8 3 )

What are the points on y-axis whose distance from the line x 3 + …

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Question

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

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Answer

Let the required point be (0, a)
Given, distnace of (0, a) from line 4x + 3y - 12 = 0 is 4units.
$\text{D}=\Big|\frac{\text{ax}_1+\text{by}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$
$4=\Big|\frac{4(0)+3(\text{a})-12}{\sqrt{4^2+3^2}}\Big|$
$4=\Big|\frac{-3\text{a}+12}{5}\Big|$
$\Rightarrow4=\frac{-3\text{a}+12}{5}$
$\Rightarrow-3\text{a}=20-12$
$\text{a}=-\frac{8}{3}$
And $4=\frac{3\text{a}-12}{5}$
$\Rightarrow3\text{a}=20+12$
$\Rightarrow\text{a}=\frac{32}{3}$
So, Required points are
$\Big(0,\frac{32}{3}\Big),\Big(0,\frac{-8}{3}\Big)$

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