One vertex of the equilateral triangle with centroid at the origin and one side as x+y-2=0 is (A) (-1,-1) (B)

NCERT Exemplar · Q41

x+y-2=0

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Concept understanding — Distance From Point To Line

Distance from a Point to a Line

The distance from a point to a line is the shortest distance — the length of the perpendicular dropped from the point onto the line. In 3D we compute it with vectors and the cross product.

Let the line be $\vec{r}=\vec{a}+\lambda\vec{b}$ (a point $A$ with position vector $\vec{a}$ , direction $\vec{b}$ ), and let $P$ be the given point with position vector $\vec{p}$ .

The idea

Look at the triangle formed by $A$ , $P$ and the foot of the perpendicular $M$ . The segment $\overrightarrow{AP}=\vec{p}-\vec{a}$ is the hypotenuse, and the perpendicular distance $d=PM$ is the side opposite the angle $\theta$ between $\overrightarrow{AP}$ and the line:

d = |\overrightarrow{AP}|\sin\theta.

But the cross product already contains $\sin\theta$ : $|\overrightarrow{AP}\times\vec{b}| = |\overrightarrow{AP}|\,|\vec{b}|\sin\theta$ . Dividing by $|\vec{b}|$ isolates the distance.

d = \frac{|(\vec{p}-\vec{a})\times\vec{b}|}{|\vec{b}|}

Example

Distance of $P(1,2,3)$ from the line $\vec{r}=(\hat{i}+\hat{j})+\lambda(2\hat{i}-\hat{j}+2\hat{k})$ .

Here $\vec{a}=(1,1,0)$ , $\vec{b}=(2,-1,2)$ , and $\overrightarrow{AP}=\vec{p}-\vec{a}=(0,1,3)$ .

$\overrightarrow{AP}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&3\\2&-1&2\end{vmatrix} = (2+3)\hat{i} - (0-6)\hat{j} + (0-2)\hat{k} = 5\hat{i}+6\hat{j}-2\hat{k}.$ …

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The centroid is at distance $\sqrt2$ from the given side, so the triangle's altitude is $3\sqrt2$ and its circumradius is $2\sqrt2$ . The vertex opposite the side lies along the perpendicular through the centroid, on the far side away from the given side — this gives $(-2,-2)$ , option (C).

Step 1: Distance from the centroid to the given side

Side: $x+y-2=0$ . Centroid at the origin:

d=\frac{|0+0-2|}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt2}=\sqrt2

Step 2: Altitude and circumradius

In an equilateral triangle the centroid divides every median in ratio $2:1$ (vertex side : side-midpoint side), so:

distance from centroid to a side $=\tfrac13$ of the altitude $h$ , giving $h=3d=3\sqrt2$ ;
distance from centroid to a vertex (the circumradius $R$ ) $=\tfrac23$ of the altitude, giving $R=\tfrac23h=2\sqrt2$ .

Step 3: Direction of the opposite vertex

The side $x+y-2=0$ has slope $-1$ , so the perpendicular through the centroid has slope $1$ — it lies along the line $y=x$ (direction $(1,1)$ ).

The foot of the perpendicular from the origin to the side is found by solving $t+t-2=0 \Rightarrow t=1$ , i.e. the foot is at $(1,1)$ , which is in the $+(1,1)$ direction from the origin. Since the centroid lies between the side and the vertex opposite it, that vertex must lie in the opposite direction, $-(1,1)$ , at distance $R$ from the origin: …

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Q.One vertex of the equilateral triangle with centroid at the origin and one side as x+y−2=0x+y-2=0x+y−2=0 is (A) (−1,−1)(-1,-1)(−1,−1) (B) (2,2)(2,2)(2,2) (C) (−2,−2)(-2,-2)(−2,−2) (D) (2,−2)(2,-2)(2,−2)

Step 1: Distance from the centroid to the given side

Step 2: Altitude and circumradius

Step 3: Direction of the opposite vertex

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Q.One vertex of the equilateral triangle with centroid at the origin and one side as $x+y-2=0$ is
(A) $(-1,-1)$
(B) $(2,2)$
(C) $(-2,-2)$
(D) $(2,-2)$

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