In this page orthocentre of triangle question2 we are going to see
solution of first problem in the quiz orthocentre of triangle.

**Definition:**

It can be shown that the altitudes of a triangle are concurrent and the point of concurrence is called the orthocentre of the triangle.The orthocentre is denoted by O.

Let ABC be the triangle AD,BE and CF are three altitudes from A,B and C to BC,CA and AB respectively.Find the slopes of the altitudes AD,BE and CF.Now we have to find the equation of AD,BE and CF using the slope point form.By solving any two altitudes we can find the orthocentre.

**Question 2:**

**Find the co ordinates of the orthocentre of a triangle whose **

**vertices are ****(3,4) (2,-1) and (4,-6).**

**Solution:**

Let the given points be A (3,4) B (2,-1) and C (4,-6)

Now we need to find the slope of AC.From that we have to find the slope of the perpendicular line through B.

Slope of AC = [(y₂ - y₁)/(x₂ - x₁)]

A (3,4) and C (4,-6)

here x₁ = 3,x₂ = 4,y₁ = 4 and y₂ = -6

= [(-6-4)/(4-3)]

= (-10)/1

= -10

Slope of the altitude BE = -1/ slope of AC

= -1/(-10)

= 1/10

Equation of the altitude BE:

(y - y₁) = m (x -x₁)

Here B (2,-1) m = 1/10

(y - (-1)) = 1/10 (x - 2)

10 (y + 1) = (x - 2)

10 y +10 = x - 2

x - 10 y - 2 - 10 = 0

x - 10 y - 12 = 0

x - 10 y - 12 = 0 --------(1)

Now we need to find the slope of BC.From that we have to find the slope of the perpendicular line through D.

Slope of BC = [(y₂ - y₁)(x₂ - x₁)]

B (2,-1) and C (4,-6)

here x₁ = 2,x₂ = 4,y₁ = -1 and y₂ = -6

= [(-6-(-1))/(4-2)]

= (-6+1)/2

= -5/2

Slope of the altitude AD = -1/ slope of AC

= -1/(-5/2)

= 2/5

Equation of the altitude AD:

(y - y₁) = m (x -x₁)

Here A(3,4) m = 2/5

(y - 4)) = 2/5(x - 3)

5 (y - 4) = 2 (x - 3)

5 y - 20 = 2x - 6

2 x - 5 y - 6 + 20 = 0

2 x - 5 y + 14 = 0 --------(2)

x - 10 y - 12 = 0 --------(1)

2 x - 5 y + 14 = 0 --------(2)

x - 10 y - 12 = 0

(2) x 2=> 4 x - 10 y + 28 = 0

(-) (+) (-)

--------------------

- 3 x - 39 = 0

- 3 x = 39

x = 39/(-3)

x = - 13

Substitute the value of x in the first equation

x - 10 y - 12 = 0

-13 - 10 y = 12

-10 y = 12 + 13

-10 y = 25

y = 25/(-10)

y = 5/2

So the orthocentre is (-13,5/2) Orthocentre of triangle question2

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