One-vertex-of-the-equalateral-triangle-with-centroid-at-the-origin-and-one-side-as-x-y-2-0-is-a-1-1-b-2-2-c-2-2-d-2-2-

Question Number 77000 by vishalbhardwaj last updated on 02/Jan/20

One vertex of the equalateral triangle

with centroid at the origin and one

side as x + y - 2 = 0 is

(a) (- 1, - 1) (b) (2, 2) (c) (- 2, - 2)

(d) (2, - 2)

Commented by vishalbhardwaj last updated on 03/Jan/20

please tell the correct answer with complete

explanation \dots .

Answered by mr W last updated on 04/Jan/20

d i s t a n c e f r o m o r i g i n t o g i v e n l i n e :

\frac{∣ 0 \times 1 + 0 \times 1 - 2 ∣}{\sqrt{1^{2} + 1^{2}}} = \sqrt{2}

p e r p e n d i c u l a r l i n e t h r o u g h o r i g i n :

y = x

t h e v e r t e x m u s t l i e o n t h i s l i n e,

t h e r e f o r e (a), (b), (c) a r e p o s s i b l e .

d i s t a n c e f r o m v e r t e x t o t h e o r i g i n

i s 2 t i m e s o f t h e d i s t a n c e f r o m o r i g i n

t o t h e g i v e n l i n e, i . e . 2 \sqrt{2}, t h e r e f o r e

(b) a n d (c) a r e p o s s i b l e .

d i s t a n c e f r o m v e r t e x t o g i v e n l i n e

i s 3 t i m e s o f t h a t f r o m o r i g i n,

i . e . 3 \sqrt{2},

(b) : \frac{∣ 2 \times 1 + 2 \times 1 - 2 ∣}{\sqrt{1^{2} + 1^{2}}} = \sqrt{2} \neq 3 \sqrt{2}

(c) : \frac{∣ - 2 \times 1 - 2 \times 1 - 2 ∣}{\sqrt{1^{2} + 1^{2}}} = 3 \sqrt{2}

\Rightarrow o n l y (c) i s c o r r e c t .

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