If , then the distance of the point from the line is _________. [2025]

Question

If $α = 1 + \sum_{r = 1}^{6} {(- 3)}^{r - 1} {}^{12}C_{2 r - 1}$ , then the distance of the point $(12, \sqrt{3})$ from the line $α x - \sqrt{3} y + 1 = 0$ is _________. [2025]

Smart Achievers · Accepted Answer

(5)

Given, $α = 1 + \sum_{r = 1}^{6} {(- 3)}^{r - 1} {}^{12}C_{2 r - 1}$

$= 1 + \sum_{r = 1}^{6} {}^{12}C_{2 r - 1} \frac{{(\sqrt{3} i)}^{2 r - 1}}{\sqrt{3} i}$

Let $t = \sqrt{3} i$ , then we have

$α = 1 + \frac{1}{\sqrt{3} i} [{}^{12}C_{1} t + {}^{12}C_{3} t^{3} + . . . + {}^{12}C_{11} t^{11}]$

$= 1 + \frac{1}{\sqrt{3} i} [\frac{{(1 + \sqrt{3} i)}^{12} - {(1 - \sqrt{3} i)}^{12}}{2}]$

$= 1 + \frac{1}{\sqrt{3} i} [\frac{{(- 2 ω^{2})}^{12} - {(2 ω)}^{12}}{2}] = 1$ [ $∵ ω^{3} = 0$ ]

$∴$ Distance of $(12, \sqrt{3})$ from $x - \sqrt{3} y + 1 = 0$ is

$| \frac{12 - \sqrt{3} (\sqrt{3}) + 1}{\sqrt{{(1)}^{2} + {(- \sqrt{3})}^{2}}} |$ $= \frac{12 - 3 + 1}{\sqrt{1 + 3}} = 5$ .

Solution

Q.
If $α = 1 + \sum_{r = 1}^{6} {(- 3)}^{r - 1} {}^{12}C_{2 r - 1}$ , then the distance of the point $(12, \sqrt{3})$ from the line $α x - \sqrt{3} y + 1 = 0$ is _________. [2025]

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Solution

Q. If α=1+∑r=16(–3)r–1 C2r–112, then the distance of the point (12,3) from the line αx–3y+1=0 is _________. [2025]

Ans. (5) Given, α=1+∑r=16(–3)r–1 C2r–112 =1+∑r=16C2r–112(3i)2r–13i Let t=3i, then we have α=1+13i[C112t+C312t3+...+C1112t11] =1+13i[(1+3i)12–(1–3i)122] =1+13i[(–2ω2)12–(2ω)122]=1 [∵ ω3=0] ∴ Distance of (12,3) from x–3y+1=0 is |12–3(3)+1(1)2+(–3)2|=12–3+11+3=5.

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Q.
If $α = 1 + \sum_{r = 1}^{6} {(- 3)}^{r - 1} {}^{12}C_{2 r - 1}$ , then the distance of the point $(12, \sqrt{3})$ from the line $α x - \sqrt{3} y + 1 = 0$ is _________. [2025]