The joint equation of two altitudes of an equilateral triangle is (√3 x-y+8-4 √3)(-√3 x-y+12+4 √3)=0 The third altitude has the equation

Question

The joint equation of two altitudes of an equilateral triangle is (√3 x-y+8-4 √3)(-√3 x-y+12+4 √3)=0 The third altitude has the equation (A) √9 x=2-4 √3 (B) y-10=0 (C) √3 x=2+4 √3 (D) y+10=0

Tardigrade Admin · Accepted Answer

Clearly, the third altitude will be the bisector of obtuse angle between the given altitudes.
Now given two altitudes are

3 x - y + 8 - 43 = 0

.... (i)
and

- 3 x - y + 12 + 43 = 0

.....(ii)

a_{1} a_{2} + b_{1} b_{2} = - 33 + (- 1) (- 1) = - 2 < 0

∴

Obtuse angle bisector is:

\frac{3 x - y + 8 - 4 3}{3 + 1} = - \frac{- 3 x - y + 12 + 4 3}{3 + 1}

or

y = 10

Q. The joint equation of two altitudes of an equilateral triangle is $(3 x - y + 8 - 43) (- 3 x - y + 12 + 43) = 0$ The third altitude has the equation

$9 x = 2 - 43$

$y - 10 = 0$

$3 x = 2 + 43$

$y + 10 = 0$

Q. The joint equation of two altitudes of an equilateral triangle is (3​x−y+8−43​)(−3​x−y+12+43​)=0 The third altitude has the equation

9​x=2−43​

y−10=0

3​x=2+43​

y+10=0

Q. The joint equation of two altitudes of an equilateral triangle is $(3 x - y + 8 - 43) (- 3 x - y + 12 + 43) = 0$ The third altitude has the equation

$9 x = 2 - 43$

$y - 10 = 0$

$3 x = 2 + 43$

$y + 10 = 0$