Let the equation 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
Answers
26x - 122y - 1675 = 0 is the equation of its third side
Step-by-step explanation:
equation of two sides of a triangle be
AB 3x - 2y + 6 = 0 and AC 4x + 5y - 20 = 0
Corresponding Equation of altitudes
CF y = -2x/3 + c & BE y = 5x/4 + c
(as slope of AB * CF = - 1 , Slope of AC * BE = -1)
Passes through 1 , 1
=> c = 5/3 & c = -1/4
y = -2x/3 + 5/3 & y = 5x/4 - 1/4
=> 3y + 2x = 5 & 5x - 4y = 1
CF & AC intersect at C
3y + 2x = 5 & 4x + 5y - 20 = 0 intersection
=> y = -10 , x = 35/2
=> C = ( 35/2 , - 10)
BE & AB intersect at B
5x - 4y = 1 & 3x - 2y + 6 = 0
=> x = - 13 , y = -33/2
=> B = (-13 , -33/2)
C (35/2 , - 10) & B ( -13 , - 33/2)
BC Slope = (-33/2 + 10)/(-13 - 35/2)
= -13/(-61)
= 13/61
Equation of line BC
y = 13x/61 + c
Using B coordiante (-13 , -33/2)
-33/2 = 13(-13)/61 + c
=> c = -1675/122
Equation of line BC
y = 13x/61 - 1675/122
=> 122y = 26x - 1675
=> 26x - 122y - 1675 = 0
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