Q. 114.5( 4 Votes )

Find the orthocen

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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