Q. 114.5( 4 Votes )
Find the orthocenter of the triangle the equations of whose sides are x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0.
Answer :
Given: Sides of triangle are are x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0.
Assuming: AB, BC and AC be the sides of triangle whose equation is are x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0.
To find:
Orthocenter of triangle.
Concept Used:
Point of intersection of two lines.
Explanation:
x + y – 1 = 0 …… (i)
2x + 3y – 6 = 0 …… (ii)
4x – y + 4 = 0. …… (iii)
By solving equation (i) and (ii) By cross multiplication
⇒ x = - 3 ,y = 4
B( - 3, 4)
By Solving equation (i) and (iii) By cross multiplication
⇒ x 
, y
A
Equation of BC is 2x + 3y = 6
Altitude AD is perpendicular to BC,
Therefore, equation of AD is x + y + k = 0
AD is passing through A
⇒ k = - 1
Equation of AD is x + y – 1 = 0 …… (iv)
Altitude BE is perpendicular to AC.
⇒ Let the equation of DE be x – 2y = k
BE is passing through D( - 3, 4)
⇒ - 3 – 8 = k
⇒ k = - 11
Equation of BE is x – 2y = - 11 …… (v)
By solving equation (iv) and (v),
We get, x = - 3 and y = 4
Hence, the orthocenter of triangle is ( - 3, 4).
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