Let the equations of two sides of a triangle be 3x - 2y + 6 = 0 and 4x + 5y - 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is :

Question

Let the equations of two sides of a triangle be 3x - 2y + 6 = 0 and 4x + 5y - 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is : (A) 122y - 26x - 1675 = 0 (B) 26x + 61y + 1675 = 0 (C) 122y + 26x + 1675 = 0 (D) 26x - 122y - 1675 = 0

Tardigrade Admin · Accepted Answer

Equation of AB is 3x - 2y + 6 = 0 Equation of AC is 4x + 5y - 20 = 0 Equation of BE is 2x + 3y - 5 = 0 Equation of CF is 5x - 4y - 1 = 0 ⇒ Equation of BC is 26x - 122y = 1675

Q. Let the equations of two sides of a triangle be $3 x - 2 y + 6 = 0$ and $4 x + 5 y - 20 = 0$ . If the orthocentre of this triangle is at $(1, 1)$ , then the equation of its third side is :

$122 y - 26 x - 1675 = 0$

$26 x + 61 y + 1675 = 0$

$122 y + 26 x + 1675 = 0$

$26 x - 122 y - 1675 = 0$

Equation of $A B$ is $3 x - 2 y + 6 = 0$
Equation of $A C$ is $4 x + 5 y - 20 = 0$
Equation of $BE$ is $2 x + 3 y - 5 = 0$
Equation of $CF$ is $5 x - 4 y - 1 = 0$
$\Rightarrow$ Equation of BC is $26 x - 122 y = 1675$

Q. Let the equations of two sides of a triangle be 3x−2y+6=0 and 4x+5y−20=0. If the orthocentre of this triangle is at (1,1), then the equation of its third side is :

122y−26x−1675=0

26x+61y+1675=0

122y+26x+1675=0

26x−122y−1675=0