First slide

Cartesian plane

Question

$The centroid of an equilateral triangle is (0, 0) . If two vertices of the triangle lie on x + y = 2 \sqrt{2}, then one of them will have its coordinates$

Easy

A

$(\sqrt{2} + \sqrt{6}, \sqrt{2} - \sqrt{6})$
B

$(\sqrt{2} + \sqrt{3}, \sqrt{2} - \sqrt{3})$
C

$(\sqrt{2} + \sqrt{5}, \sqrt{2} - \sqrt{5})$
D

None of these

Solution

Let the vertices B and C lie on the given line. Then, $O D = \frac{2 \sqrt{2}}{\sqrt{2}} = 2 (p e r p e n d i c u l a r d i s \tan c e f r o m o r i g i n = \frac{|c|}{\sqrt{a^{2} + b^{2}}})$
$L e t e q u a t i o n o d O D b e y = m x, \sin c e s l o p e i s p e r p e n d i c u l a r t o x + y = 2 \sqrt{2} i s 1, ∴ e q u a t i o n i s x = y s o l v i n g t h e t w o e q u a t i o n s w e g e t D (\sqrt{2}, \sqrt{2})$ =
$Also, B D = O D \times \tan 60^{\circ} = 2 \sqrt{3} for the coordinates of B and C .$
Slope of the given line is -1, slope of OD is 1 ,hence angle of inclination is $45^{0}$
Using parametric equation of line, we get
$\begin{array}{ll} \frac{x - \sqrt{2}}{1 / \sqrt{2}} = \frac{y - \sqrt{2}}{1 / \sqrt{2}} = \pm 2 \sqrt{3} \\ or C \equiv (\sqrt{2} + \sqrt{6}, \sqrt{2} - \sqrt{6}) \\ and B \equiv (\sqrt{2} - \sqrt{6}, \sqrt{2} + \sqrt{6}) \end{array}$

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