Distance of Origin From Centroid — Two Sides and Orthocenter Given

JEE Main 2022 — 29 June, Evening Shift. Previous Year Question.

Problem

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 $\left(\dfrac{7}{3},\dfrac{7}{3}\right)$, is:

Key insight. The two given lines meet at one vertex of the triangle. The other two vertices can each be found as the intersection of one given side with the altitude through the orthocenter perpendicular to the other given side — since an altitude from a vertex is perpendicular to the opposite side and passes through both that vertex and the orthocenter.

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Approach

Find vertex $A$ as the intersection of the two given sides. For the other two vertices, construct the altitude from each remaining vertex — perpendicular to the opposite side, passing through the given orthocenter — and intersect each altitude with the appropriate given side. Once all three vertices are known, compute the centroid and its distance from the origin.

Solution

Step 1 — Find vertex A as the intersection of the two given sides

Solving $x-2y+1=0$ and $2x-y-1=0$ simultaneously: from the first, $x=2y-1$. Substituting into the second:

$$2(2y-1)-y-1=0 \implies 4y-2-y-1=0 \implies 3y=3 \implies y=1, \quad x=1$$

So $A=(1,1)$.

Step 2 — Find vertex B using the altitude from B

Say side $AB$ is $x-2y+1=0$ and side $AC$ is $2x-y-1=0$ (slope $2$). The altitude from $B$ is perpendicular to $AC$, so its slope is $-\dfrac{1}{2}$, and it passes through the orthocenter $H=\left(\dfrac{7}{3},\dfrac{7}{3}\right)$:

$$y-\frac{7}{3} = -\frac{1}{2}\left(x-\frac{7}{3}\right)$$

$B$ lies on $AB$ ($x=2y-1$). Substituting:

$$y-\frac{7}{3} = -\frac{1}{2}\left(2y-1-\frac{7}{3}\right) = -\frac{1}{2}\left(2y-\frac{10}{3}\right) = -y+\frac{5}{3}$$

$$2y = \frac{5}{3}+\frac{7}{3} = 4 \implies y=2, \quad x=2(2)-1=3$$

So $B=(3,2)$.

Step 3 — Find vertex C using the altitude from C

The altitude from $C$ is perpendicular to $AB$ (slope $\dfrac{1}{2}$), so its slope is $-2$, passing through $H$:

$$y-\frac{7}{3} = -2\left(x-\frac{7}{3}\right)$$

$C$ lies on $AC$ ($y=2x-1$). Substituting:

$$2x-1-\frac{7}{3} = -2\left(x-\frac{7}{3}\right) = -2x+\frac{14}{3}$$

$$4x = 1+\frac{7}{3}+\frac{14}{3} = 1+7 = 8 \implies x=2, \quad y=2(2)-1=3$$

So $C=(2,3)$.

Step 4 — Verify with the third altitude (optional check)

$BC$ has slope $\dfrac{3-2}{2-3}=-1$, so the altitude from $A$ has slope $1$: $y-1=1(x-1) \implies y=x$. This passes through $H=\left(\dfrac{7}{3},\dfrac{7}{3}\right)$ ✓, confirming the vertices are correct.

Step 5 — Compute the centroid

$$G = \left(\frac{1+3+2}{3}, \frac{1+2+3}{3}\right) = (2,2)$$

Step 6 — Compute the distance from the origin

$$OG = \sqrt{2^2+2^2} = \sqrt8 = 2\sqrt2$$

Answer

Common mistakes

• Assigning the wrong given line to the wrong side. Since the labelling of which line is $AB$ and which is $AC$ is a free choice (both lead to the same final triangle, just with $B$ and $C$ swapped), the key is being consistent — the altitude from a vertex must be perpendicular to the side opposite that vertex, not the side through it.
• Skipping the verification step. Confirming the third altitude (from $A$) also passes through the given orthocenter is a valuable check that catches sign or slope errors made in finding $B$ or $C$.

Practise next

• A triangle has two sides $x-y+1=0$ and $x+2y-4=0$, with orthocenter $(1,2)$. Find the distance from the origin to the centroid, using the same altitude-construction method.