The distance of the origin from the centroid of the triangle whose two sides have the equations x -2 y +1=0 and 2 x - y -1=0 and whose orthocenter is ((7/3), (7/3)) is:

Question

The distance of the origin from the centroid of the triangle whose two sides have the equations x -2 y +1=0 and 2 x - y -1=0 and whose orthocenter is ((7/3), (7/3)) is: (A) √2 (B) 2 (C) 2 √2 (D) 4

Tardigrade Admin · Accepted Answer

AB ≡ x -2 y +1=0 AC ≡ 2 x - y -1=0 So A (1,1) Altitude from B is BH = x +2 y -7=0 ⇒ B (3,2) Altitude from C is CH =2 x + y -7=0 ⇒ C (2,3) Centroid of triangle ABC = E (2,2) OE =2 √2

Q. The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2 y + 1 = 0$ and $2 x - y - 1 = 0$ and whose orthocenter is $(\frac{7}{3}, \frac{7}{3})$ is:

$2$

2

$22$

4

$A B \equiv x - 2 y + 1 = 0$
$A C \equiv 2 x - y - 1 = 0$
So $A (1, 1)$
Altitude from $B$ is $B H = x + 2 y - 7 = 0 \Rightarrow B (3, 2)$
Altitude from $C$ is $C H = 2 x + y - 7 = 0 \Rightarrow C (2, 3)$
Centroid of $△ A BC = E (2, 2) OE = 22$

Q. The distance of the origin from the centroid of the triangle whose two sides have the equations x−2y+1=0 and 2x−y−1=0 and whose orthocenter is (37​,37​) is:

2​

2

22​

4

AB≡x−2y+1=0 AC≡2x−y−1=0 So A(1,1) Altitude from B is BH=x+2y−7=0⇒B(3,2) Altitude from C is CH=2x+y−7=0⇒C(2,3) Centroid of △ABC=E(2,2)OE=22​

Q. The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2 y + 1 = 0$ and $2 x - y - 1 = 0$ and whose orthocenter is $(\frac{7}{3}, \frac{7}{3})$ is:

$2$

$22$

$A B \equiv x - 2 y + 1 = 0$
$A C \equiv 2 x - y - 1 = 0$
So $A (1, 1)$
Altitude from $B$ is $B H = x + 2 y - 7 = 0 \Rightarrow B (3, 2)$
Altitude from $C$ is $C H = 2 x + y - 7 = 0 \Rightarrow C (2, 3)$
Centroid of $△ A BC = E (2, 2) OE = 22$