Q.Find values of $k$ if area of triangle is 4 sq. units and vertices are

Concept: Area of a triangle using coordinates — the absolute value of half the determinant of the vertices.

For vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$:

(i) Vertices $(k,0), (4,0), (0,2)$:

So $|-2k+8| = 8$. This gives $-2k+8 = 8$ or $-2k+8 = -8$, hence $k = 0$ or $k = 8$.

(ii) Vertices $(-2,0), (0,4), (0,k)$: …

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ can be found using the determinant formula:

Why does this work? The expression inside the absolute value is actually twice the signed area of the triangle. The sign tells you the orientation (clockwise or anticlockwise), but we only care about the magnitude. The formula comes from the cross product of two side vectors in the plane — the area of a parallelogram formed by those vectors is the determinant, and the triangle is half of that.

This is the same as the expression above, just written in determinant form. We'll use the expanded version for clarity.

(i) Vertices: $(k, 0), (4, 0), (0, 2)$

1. Plug into the formula. Let $(x_1, y_1) = (k, 0)$, $(x_2, y_2) = (4, 0)$, $(x_3, y_3) = (0, 2)$. The area is:

2. Simplify inside the absolute value.

So:

3. Set the area equal to 4.

Multiply both sides by 2:

4. Solve the absolute value equation. This gives two cases:

• First case: $-2k + 8 = 8 \implies -2k = 0 \implies k = 0$
• Second case: $-2k + 8 = -8 \implies -2k = -16 \implies k = 8$

So for part (i), $k = 0$ or $k = 8$.

(ii) Vertices: $(-2, 0), (0, 4), (0, k)$

1. Plug into the formula. Let $(x_1, y_1) = (-2, 0)$, $(x_2, y_2) = (0, 4)$, $(x_3, y_3) = (0, k)$. The area is: …

Method: Finding an Unknown Coordinate Given a Fixed Triangle Area

Use this whenever one vertex has an unknown coordinate and the triangle's area is specified.

Steps

Step 1: Substitute the known and unknown coordinates into the area formula

leaving the unknown (say $k$) inside the expression.

Step 2: Simplify to a single linear expression in the unknown

Combine the known terms and collect the coefficient of the unknown.

Step 3: Set the expression equal to twice the given area (clearing the $\tfrac12$) …

Common Mistakes

Mistake 1: Solving only one of the two absolute-value cases and reporting a single value

Why it's wrong: an equation of the form $|E|=B$ always splits into $E=B$ and $E=-B$ — reporting just one root misses a second, equally valid position of the vertex that gives the same triangle area. Correct approach: always write out and solve both cases before finalising the answer.

Mistake 2: Forgetting to multiply both sides by $2$ before removing the modulus …