find the orthocentre of the triangle with the vertices (1,3),(0,-2) and (-3,1)
Answers
Answer:
we take a(1,3) b (0,-2) C(-3,1) and point of orthocenter o(h, k) then we find slope of lines bc, ac, ab, which are -1, 1/2, 5 and then find o by using m1*m2=-1 we have two equ k-h-2=0and 2h+k+2=0then we find value of h and k which are h=4/3and k =-14/3
Answer:
the orthocenter of the ΔABC is O( -4/3 , 2/3 )
Step-by-step explanation:
Let the vertices of the triangle be A,B, and C
such that A(1,3) , B(0,-2) ,C ( -3,1)
=> the sides of the triangle are AB, BC and AC
let BC , base of the triangle , take a point D such that AD⊥BC
similarly take E on AC such that BE⊥AC
and take F on AB such that CF ⊥ AB
so, we can say that the point of concurrency of AD,BE,and CF is the orthocentre, O of the given Δ ABC.
Slope of AB: A(1,3) , B(0,-2)
m(AB) = y₂-y₁ / x₂-x₁
= -2-3 / 0-1
= -5/-1
= 5
∴ slope of CF which perpendicular to AB is -1/ m(AB)
=>m(CF)= -1/5
∴ eqn. of CF with C(-3,1) and slope -1/5 can be written as
y-y₁ = m (x-x₁)
y-1 = -1/5 ( x - (-3))
5(y-1) = -1(x+3)
5y-5 = -x-3 => x+5y-2 = 0 ------------------------------- i
similarly we have to find slope of any other perpendicular,
let us take AD.
Slope of BC with B(0,-2) C(-3,1)
m(BC) = y₂-y₁ / x₂-x₁ => 1- (-2) / -3-0
= 1+2 / -3 => 3/-3
= -1
∴ slope of AD⊥BC can be given as -1/ m(BC)
=> m(AD) = -1/-1 = 1
then, equation of AD with A(1,3) and slope 1 can be written as
y-y₁ = m (x-x₁)
y-3 = 1(x-1)
x-y+2 = 0 ------------------------------------------- ii
now , we have eqns of the 2 perpendiculars AD and CF
the intersection point of these 2 lines gives the Orthocentre
so,
x+5y-2 = 0 ------------------------------- i
(-) x-1y+2 = 0 ------------------------------------------- ii
------------------------
6y - 4 = 0
=> y = 4/6
y = 2/3
substitute y value in eqn.i ,
=> x+5(2/3)-2 = 0
=> x+10/3 = 2
=> x = 2- (10/3) = 6-10 / 3
x = -4/3
so, the orthocenter of the ΔABC is O( -4/3 , 2/3 )