Let A be the set of all points (α, β) such that the area of triangle formed by the points (5, 6), (3, 2) and (α, β) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :

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Straight Lines Ncert Solutions Maths class 11th Maths Class 11th

Let A be the set of all points (α, β) such that the area of triangle formed by the points (5, 6), (3, 2) and (α, β) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :

Option 1 -

$\frac{16}{\sqrt[]{5}}$

Option 2 -

$\frac{4}{\sqrt[]{5}}$

Option 3 -

$\frac{12}{\sqrt[]{5}}$

Option 4 -

$\frac{8}{\sqrt[]{5}}$

6 Views | Posted a month ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

$| \begin{matrix} α & β & 1 \\ 5 & 6 & 1 \\ 3 & 2 & 1 \end{matrix} | = 24$

$4 α - 2 β - 8 = \pm 24$

$4 α - 2 β = 24 + 8, 4 α - 2 β = - 24 + 8$

$2 (2 α - β) = 32 2 α - β = - 8$

Distance of origin

$D = \sqrt[]{α^{2} + {(2 α + 8)}^{2}}$

$α = - \frac{16}{5}$

$D = \sqrt[]{{(\frac{- 16}{5})}^{2} + {(\frac{8}{5})}^{2}}$

if 2 - = 16

$D = \frac{16}{\sqrt[]{5}}$

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Let A be the set of all points (α, β) such that the area of triangle formed by the points (5, 6), (3, 2) and (α, β) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :

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