The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are y–3=2±3x–2 .

3 Views|Posted a year ago
Asked by Shiksha User
1 Answer
V
a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

Let  ABC  be  an  equilateral  triangle  with  vertex(2,3)  and  the  opposite  side  is  x+y=2with  slope  −1.  Suppose  slope  of  line  AB  is  m.  each  angle  of  equilateral  triangle  is  600.∴  Angle  between  AB  and  BC              tan600=|−1−m1+(−1)m|⇒                  3=|1+m1−m|⇒                  3=±(1+m1−m)Taking  (+)  sign⇒                  3=1+m1−m      ⇒3−3m=1+m⇒      3m+m=3−1     ⇒m(3+1)=3−1⇒                   m=3−13+1     ⇒m=3−13+1*3−13−1⇒                   m=3+1−233−1=2−3Taking  (−)  sign⇒                  3=−(1+m1−m)      ⇒3−3m=−1−m⇒      −3m+m=−1−3          ⇒m(−3+1)=−1−3⇒                     m=−1−31−3         ⇒m=−1−31−3*1+31+3⇒                     m=−1−3−3−31−3=−4−23−2=−2(2+3)−2=2+3So,  the  equations  of  other  two  lines  are                 y−3=(2±3)(x−2)Hence,  the  given  statement  is  True.

Thumbs Up IconUpvote Thumbs Down Icon

Similar Questions for you

  | 1 2 − 2 i + 1 | = α ( 1 2 − 2 i ) + β ( 1 + i )  

9 4 + 4 = α ( 1 2 − 2 i ) + β ( 1 + i )

5 2 = α ( 1 2 ) + β + i ( − 2 α + β )             

α 2 + β = 5 2      ...(1)

 –2α + β = 0                    …(2)

Solving (1) and (2)

α 2 + 2 α = 5 2

5 2 α = 5 2            

a =

...Read more

Variance = ∑ x 2 n − ( x ¯ ) 2  

6 0 2 + 6 0 2 + 4 4 2 + 5 8 2 + 6 8 2 + α 2 + β 2 + 5 6 2 8 = ( 5 8 ) 2 = 6 6 . 2            

7 2 0 0 + 1 9 3 6 + 3 3 6 4 + 4 6 2 4 + 3 1 3 6 + α 2 + β 2 8 = 3 3 6 4 = 6 6 . 2             

2 5 3 2 . 5 + α 2 + β 2 8 − 3 3 6 4 = 6 6 . 2            

α2 + β2 = 897.7 × 8

= 7181.6

...Read more

Start with

(1) E ¯ : 6 ! 2 ! = 3 6 0  

(2)    G E ¯ : 5 ! 2 ! , G N ¯ : 5 ! 2 !  

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

...Read more

( 1 + x ) 1 1 =   1 1 C 0 +   1 1 C 1 x +   1 1 C 2 x 2 +   . . . . .   + 1 1 C 1 1 x 1 1

= 2 1 2 − 2 − 2 4 1 2
= 2 1 2 − 2 6 1 2 = 4 0 7 0 1 2 = 2 0 3 5 6 = m n
m + n = 2035 + 6 = 2041

 

f ( x ) = { 2 + 2 x , x ∈ ( − 1 ,   0 ) 1 − x 3 , x ∈ [ 0 ,   3 )

g ( x ) = { x , x ∈ [ 0 ,   1 ) − x , x ∈ ( − 3 ,   0 )   ->g(x) = |x|, x Î (–3, 1)

f ( g ( x ) ) = { 2 + 2 | x | , | x | ∈ ( − 1 ,   0 ) ⇒ x ∈ ? 1 − | x | 3 , | x | ∈ [ 0 ,   3 ) ⇒ x ∈ ( − 3 ,   1 )            

f ( g ( x ) ) = { 1 − x 3 , x ∈ [ 0 ,   1 ) 1 + x 3 , x ∈ ( − 3 ,   0 )

Range of fog(x) is [0, 1]

            

            Range of fog(x) is [0, 1]

...Read more

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

67K
Colleges
|
1.2K
Exams
|
7.2L
Reviews
|
1.9M
Answers

Learn more about...

Maths NCERT Exemplar Solutions Class 11th Chapter Ten 2025

Maths NCERT Exemplar Solutions Class 11th Chapter Ten 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering