Question

find the orthocentreof a triangle formed by the lines x+2y=0,4x+3y-5=0,3x+y=0​

Answer 1

Answer:

Given equations are:

x+2y=0......(1)

4x+3y−5=0.......(2)

3x+y=0..........(3)

Solving (1) and (2), vertex A=(0,0)

Solving (1) and (3), vertex B=(2,−1)

Equation of BC is 4x+3y−5=0

AB is perpendicular to BC and passes through A=(0,0)

Equation of AB is 3x−4y=0........(4)

BE is perpendicular to AC

Therefore equation of BE is x−3y=k

BE passes through B=(2,−1)

2+3=k⇒k=5

Equation of BE is x−3y=5........(5)

Solving (4) and (5)

Orthocenter is O(−4,−3)

Answer 2

Answer:

(-4,-3)

Step-by-step explanation:

Let AB, BC and CA be represented by the lines

x+2y=0       --(1)

4x+3y=5     --(2)

3x+y=0       --(3)

Solving (1) and (3), A=(0,0)

∴Equation to AD ⊥(2) is 3x-4y=0  --(i)

Solving (1) and (2) ; B=(2,-1)

∴Equation to BE⊥(3) is (x-2)-3(y+1) = 0

⇒x-3y-5 = 0  --(ii)

Solving (i) and (ii), the orthocenter = (-4,-3).