If the orthocentre of the triangle, whose vertices are (1,2),(2,3) and (3,1) is (α, β), then the quadratic equation whose roots are α+4 β and 4 α+β, is

Question

If the orthocentre of the triangle, whose vertices are (1,2),(2,3) and (3,1) is (α, β), then the quadratic equation whose roots are α+4 β and 4 α+β, is (A) x2-19 x+90=0 (B) x2-20 x+99=0 (C) x2-18 x+80=0 (D) x2-22 x+120=0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/2a37174d6f6c4a7ecea147933cdadd14-.png" /> ((β-3/α-2))((1/-2))=-1 β-3=2 α-4 β=2 α-1 m AH × m BC =-1 ⇒ ((β-2/α-1))(-2)=-1 ⇒ 2 β-4=α-1 ⇒ 2(2 α-1)=α+3 ⇒ 3 α=5 α=(5/3), β=(7/3) ⇒ H ((5/3), (7/3)) α+4 β=(5/3)+(28/3)=(33/3)=11 β+4 α=(7/3)+(20/3)=(27/3)=9 x2-20 x+99=0

Q. If the orthocentre of the triangle, whose vertices are $(1, 2), (2, 3)$ and $(3, 1)$ is $(α, β)$ , then the quadratic equation whose roots are $α + 4 β$ and $4 α + β$ , is

$x^{2} - 19 x + 90 = 0$

$x^{2} - 20 x + 99 = 0$

$x^{2} - 18 x + 80 = 0$

$x^{2} - 22 x + 120 = 0$

Q. If the orthocentre of the triangle, whose vertices are (1,2),(2,3) and (3,1) is (α,β), then the quadratic equation whose roots are α+4β and 4α+β, is

x2−19x+90=0

x2−20x+99=0

x2−18x+80=0

x2−22x+120=0

(α−2β−3​)(−21​)=−1 β−3=2α−4 β=2α−1 mAH​×mBC​=−1 ⇒(α−1β−2​)(−2)=−1 ⇒2β−4=α−1 ⇒2(2α−1)=α+3 ⇒3α=5 α=35​,β=37​ ⇒H(35​,37​) α+4β=35​+328​=333​=11 β+4α=37​+320​=327​=9 x2−20x+99=0

Q. If the orthocentre of the triangle, whose vertices are $(1, 2), (2, 3)$ and $(3, 1)$ is $(α, β)$ , then the quadratic equation whose roots are $α + 4 β$ and $4 α + β$ , is

$x^{2} - 19 x + 90 = 0$

$x^{2} - 20 x + 99 = 0$

$x^{2} - 18 x + 80 = 0$

$x^{2} - 22 x + 120 = 0$