Q.If the co-ordinates of the vertices of an equilateral triangle with sides of length $a$ are $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, then $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2 = \dfrac{3a^4}{4}$.

Concept and Intuition

The signed area of a polygon with ordered vertices is given by the shoelace formula. With three fixed vertices and one variable point $P$, the area becomes a linear expression in the coordinates of $P$, and fixing the area gives the locus.

Step-by-Step Solution

1. Vertices in order: $A(1,2), B(2,1), C(1,-2), P(x,y)$.
2. Shoelace sum $\sum (x_i y_{i+1} - x_{i+1} y_i)$:
   • $A\to B: 1\cdot1 - 2\cdot2 = -3$
   • $B\to C: 2\cdot(-2) - 1\cdot1 = -5$
   • $C\to P: 1\cdot y - x\cdot(-2) = y+2x$
   • $P\to A: x\cdot2 - 1\cdot y = 2x-y$
3. Total $= -3-5+(y+2x)+(2x-y) = 4x-8$.
4. Area $= \tfrac12|4x-8| = 9 \Rightarrow |4x-8| = 18 \Rightarrow 4x = 26$ or $4x=-10 \Rightarrow x = 6.5$ or $x = -2.5$.
5. As a single equation: $(x-6.5)(x+2.5)=0 \Rightarrow x^2 - 4x - 16.25 = 0 \Rightarrow 4x^2 - 16x - 65 = 0$.

Common Mistakes …

Concept and Intuition

Any point on a line through $P(1,2)$ making angle $\theta$ with the positive $x$-axis, at signed distance $r$ from $P$, is $(1+r\cos\theta,\ 2+r\sin\theta)$. Taking $r=+4$ and $r=-4$ gives the two points A, B (on opposite sides of P, since both are stated to be 4 units from P). The area of a triangle with one vertex at the origin is the half the absolute cross product of the other two vertices' position vectors.

Step-by-Step Solution

1. Direction cosines for $60°$: $(\cos60°,\sin60°)=\left(\dfrac12,\dfrac{\sqrt3}{2}\right)$.
2. $A=P+4\left(\dfrac12,\dfrac{\sqrt3}{2}\right)=(1+2,\,2+2\sqrt3)=(3,\,2+2\sqrt3)$.
3. $B=P-4\left(\dfrac12,\dfrac{\sqrt3}{2}\right)=(1-2,\,2-2\sqrt3)=(-1,\,2-2\sqrt3)$. …

Concept and Intuition

For a triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $P(x,y)$, twice the area is the determinant $x_1(y_2-y)+x_2(y-y_1)+x(y_1-y_2)$. Since area is a magnitude, equating two areas means equating the absolute values of two linear expressions in $x,y$ — and $|L_1|=|L_2|$ is equivalent to $(L_1-L_2)(L_1+L_2)=0$, a pair of straight lines (hence a quadratic locus, matching the quadratic options given).

Step-by-Step Solution

1. With $A(1,0)$, $B(0,-2)$, $P(x,y)$: $2\,\text{Area}(PAB) = |1(-2-y)+0(y-0)+x(0-(-2))| = |2x-y-2|$.
2. With $A(1,0)$, $C(2,-1)$, $P(x,y)$: $2\,\text{Area}(PAC) = |1(-1-y)+2(y-0)+x(0-(-1))| = |x+y-1|$.
3. Setting the areas equal: $|2x-y-2| = |x+y-1|$. …

Concept and Intuition

In the Argand plane, $|z_i-z_j|$ is just the Euclidean distance between the two points — i.e. an ordinary side length of the triangle. So this problem is really plane geometry: we're told the ratio of two sides meeting at $C$ (i.e. $CA:CB = 3:4$), the included angle at $C$, and the length of the side opposite $C$ (i.e. $AB$). The law of cosines connects all of these, letting us solve for the actual side lengths (not just their ratio), after which the area formula $\tfrac12 ab\sin C$ finishes it.

Step-by-Step Solution

1. Let $CA = 3t$ and $CB = 4t$ (respecting the given ratio $AC:BC=3:4$).
2. Law of cosines at vertex $C$ (angle between $CA$ and $CB$, opposite side $AB$):

3. Given $AB^2 = \left(\sqrt{25-12\sqrt3}\right)^2 = 25-12\sqrt3$. Equating: $t^2(25-12\sqrt3) = 25-12\sqrt3 \Rightarrow t^2=1 \Rightarrow t=1$.
4. So $CA=3,\ CB=4$, with included angle $30^\circ$. …

Concept and Intuition

The locus of a point maintaining a fixed triangular area with two fixed points is always a pair of straight lines parallel to the line joining the two fixed points (one on each side). The area formula, once the absolute value is resolved into $+12$ and $-12$ cases, yields two linear equations; since the locus is the union of both lines, the combined single equation is their product.

Step-by-Step Solution

1. Area of $\triangle PAB = \dfrac12\left| x_A(y_B-y) + x_B(y-y_A) + x(y_A-y_B) \right|$ with $A=(2,3)$, $B=(-4,5)$, $P=(x,y)$.
2. Substitute: $\dfrac12\left|2(5-y) + (-4)(y-3) + x(3-5)\right| = \dfrac12\left|10-2y-4y+12-2x\right| = \dfrac12\left|22-6y-2x\right| = \left|{-x-3y+11}\right|$.
3. Set equal to 12: $|-x-3y+11| = 12 \Rightarrow -x-3y+11 = \pm12$.
4. Case "+": $-x-3y+11=12 \Rightarrow x+3y+1=0$.
5. Case "$-$": $-x-3y+11=-12 \Rightarrow x+3y-23=0$. …

Concept and Intuition

For a triangle with one vertex at the origin and the other two at fixed points, the area formula simplifies nicely — it becomes proportional to just one coordinate of the moving point. Setting up both areas in terms of $P=(x,y)$ and equating per the given ratio gives the locus directly.

Step-by-Step Solution

1. Area of $\triangle POB$ with $O=(0,0)$, $P=(x,y)$, $B=(0,4)$: $\text{Area}=\dfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| = \dfrac12|x(0-4)| = 2|x|$.
2. Area of $\triangle POA$ with $O=(0,0)$, $P=(x,y)$, $A=(6,0)$: $\text{Area}=\dfrac12|6(0-y)|=3|y|$. …