Question

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Find the circumcentre of the triangle formed by the straight lines x+y=0, 2x+y+5=0 and x-y=2​

Answer 1

Given lines :

• x + y = 0 ------ (1)

• 2x + y + 5 = 0 ------ (2)

• x - y = 2 ------ (3)

To Find :

• Circumcentre of the triangle

Solution :

Clearly, lines one and three are perpendicular.

∴ Given lines forms a right- angled triangle

➺ In a right angled triangle circumcentre is mid point of the hypotenuse

Equation - ➌

➺ x - y = 2

➠ x = 2 + y

By substituting value 'x' in eq-➋

$\implies{\sf{2(2 + y) +y + 5 = 0}}$

$\implies{\sf{4 + 2y + y + 5 = 0}}$

$\implies{\sf{9 + 3y = 0}}$

$\implies{\sf{3y = -9}}$

$\implies{\sf{y = \dfrac{-9}{3}}}$

$\implies{\sf{\pink{y = -3}}}$

Now we will find 'x' by substituting value 'y'

$\implies{\sf{x = 2 - y}}$

$\implies{\sf{x = 2 + (-3)}}$

$\implies{\sf{x = 2 - 3}}$

$\implies{\sf{\pink{x = -1}}}$

Hence, B( x,y ) = (-1, -3)

Equation - ➊

➺ x + y = 0

➠ x = -y

By substituting value 'x' in eq- ➋

$\implies{\sf{2(-y) + y + 5 = 0}}$

$\implies{\sf{-2y + y + 5 = 0}}$

$\implies{\sf{-y + 5 = 0}}$

$\implies{\sf{-y = -5}}$

$\implies{\sf{\pink{y = 5}}}$

Now we will find 'x' by substituting value 'y'

$\implies{\sf{x = -y}}$

$\implies{\sf{x = -(5)}}$

$\implies{\sf{\pink{x = 5}}}$

Hence, C( x, y ) = (-5 ,5 )

$\sf{Mid point = {\dfrac{-1-5}{2}, {\dfrac{-3+5}{2}}}}$

$\implies{\sf{\dfrac{-6}{2},{\dfrac{2}{2}}}}$

$\implies{\bf{\red{(-3, 1)}}}$

Hence, The circumcentre of the triangle formed by the straight lines is (-3, 1 )

Answer 2

Answer:

Step-by-step explanation:

The given lines

(n-2y + 5) - (x-y + 4)

-2y + y + 5-4 = 0

- 9+1=0

y = 1 - 3