Q.Find area of the triangle with vertices at the point given in each of the following :

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: $|2x-y-2| = |x+y-1|$.
4. This splits as $(2x-y-2)-(x+y-1)=0$ or $(2x-y-2)+(x+y-1)=0$, i.e. $x-2y-1=0$ or $3x-3=0$ (i.e. $x=1$).
5. Combined pair-of-lines equation: $(x-2y-1)(x-1)=0$.
6. Expand: $x^2-x-2xy+2y-x+1 = x^2-2xy-2x+2y+1=0$.

Common Mistakes

• Dropping the modulus and equating the two linear expressions directly, which loses the second branch of the locus and gives only a straight line, not the quadratic asked for.
• Sign slips in the $2\times2$ area determinant.

ANSWER: C

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 "$-$": $-x-3y+11=-12 \Rightarrow x+3y-23=0$.
6. The locus is the union of these two lines; combine via product: $(x+3y+1)(x+3y-23)=0$. Let $u=x+3y$: $(u+1)(u-23)=u^2-22u-23=0$.
7. Expand $u^2 = (x+3y)^2 = x^2+6xy+9y^2$, so the combined equation is $x^2+6xy+9y^2 -22x-66y-23=0$.

Common Mistakes

• Only reporting one of the two lines (forgetting P could be on either side of AB for the same area).
• Sign errors while resolving the absolute value into the $\pm12$ cases.

ANSWER: C

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. $B=P-4\left(\dfrac12,\dfrac{\sqrt3}{2}\right)=(1-2,\,2-2\sqrt3)=(-1,\,2-2\sqrt3)$.
4. Area of $\triangle OAB=\dfrac12|x_Ay_B-x_By_A|=\dfrac12\left|3(2-2\sqrt3)-(-1)(2+2\sqrt3)\right|=\dfrac12|8-4\sqrt3|$.
5. Since $4\sqrt3\approx6.93<8$, this is $\dfrac12(8-4\sqrt3)=4-2\sqrt3$.

Common Mistakes

• Placing both A and B on the same side of P (using $r=+4$ twice) instead of opposite sides.
• Sign error in the cross-product/determinant area formula.

ANSWER: A

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 $O=(0,0)$, $P=(x,y)$, $A=(6,0)$: $\text{Area}=\dfrac12|6(0-y)|=3|y|$.
3. Given condition: Area$(POB) = 2\times$Area$(POA)$: $2|x| = 2(3|y|) = 6|y|$.
4. So $|x|=3|y| \Rightarrow x^2 = 9y^2 \Rightarrow x^2-9y^2=0$.

Common Mistakes

• Mixing up which triangle corresponds to which fixed point (using $A$'s coordinates for the $B$-triangle or vice versa).
• Dropping the factor of 2 in the given ratio condition.

ANSWER: C

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

• Using the wrong vertex order and getting a $y$-dependent expression.
• Dropping the absolute value and losing one branch of the locus.

ANSWER: D

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 $30^\circ$.
5. Area $= \dfrac12 \cdot CA \cdot CB \cdot \sin(\angle ACB) = \dfrac12(3)(4)\sin30^\circ = \dfrac12(12)\left(\dfrac12\right) = 3$.

Common Mistakes

• Confusing which two sides the given angle $\angle ACB$ lies between (it's between $CA$ and $CB$, opposite $AB$ — not between $AB$ and one of the others).
• Forgetting to solve for the actual scale factor $t$ from the given $AB$ value, and instead just using the ratio $3:4$ directly in the area formula (which would give the wrong absolute area).

ANSWER: B