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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:
Option 1)

$122x+26y+1675=0$
Option 2)

$26x+61y+1675=0$
Option 3)

$122x-26y-1675=0$
Option 4)

$26x-122y-1675=0$

Answers (1)

best_answer

Equation of a line perpendicular to a given line -

$Bx-Ay+\lambda =0$ is the line perpendicular to $Ax+By+C =0$ .

- wherein

$\lambda$ is some other constant than $C$ .

General form of the equation of a line -

In $Ax+By+C=0$ , slope of a line = $\frac{-A}{B}$ .

- wherein

$a,b,c$ are the constants.

Let the equation of . $AB = 3x -2y +6 = 0$ . The equation of $AC = 4x +5y -20 = 0$ . Equation of $BE = 2x + 3y -5 = 0$ $(\because BE \perp AC)$

Equation of $CF = 5x -4y -1 =0$

$\Rightarrow$ Equation of $BC = 26x -122y -1675 = 0$

Option 1)

$122x+26y+1675=0$

Option 2)

$26x+61y+1675=0$

Option 3)

$122x-26y-1675=0$

Option 4)

$26x-122y-1675=0$

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