Find the orthocen

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).