Question Statement

The vertices of a triangle are $(3,-4)$, $(4,h)$, and $(2,6)$. Find $h$ if the area of the triangle is 10 square units.

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Background and Explanation

The area of a triangle given its three vertices can be found using the determinant formula (shoelace formula). This method uses a 3×3 determinant where the columns contain the x-coordinates, y-coordinates, and ones respectively.

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Solution

We use the determinant formula for the area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$:

$\text{Area}=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Given that the area is 10 square units, we set up the equation:

$\frac{1}{2}\left|\begin{array}{rrr}3& -4& 1\\ 4& h& 1\\ 2& 6& 1\end{array}\right|=10$

Step 1: Expand the determinant

We expand along the first row, computing the three 2×2 minors:

$\frac{1}{2}\left[3\left|\begin{array}{ll}h& 1\\ 6& 1\end{array}\right|+4\left|\begin{array}{ll}4& 1\\ 2& 1\end{array}\right|+1\left|\begin{array}{ll}4& h\\ 2& 6\end{array}\right|\right]=10$

Step 2: Evaluate each 2×2 determinant

• First minor: $\left|\begin{array}{ll}h& 1\\ 6& 1\end{array}\right|=h\cdot 1-1\cdot 6=h-6$

• Second minor: $\left|\begin{array}{ll}4& 1\\ 2& 1\end{array}\right|=4\cdot 1-1\cdot 2=4-2=2$

• Third minor: $\left|\begin{array}{ll}4& h\\ 2& 6\end{array}\right|=4\cdot 6-h\cdot 2=24-2h$

Substituting these back:

$\frac{1}{2}[3(h-6)+4(4-2)+1(24-2h)]=10$

Step 3: Simplify the expression

$\frac{1}{2}[3h-18+8+24-2h]=10$

Combining like terms inside the brackets:

• $3h-2h=h$
• $-18+8+24=14$

So we have:

$\frac{1}{2}(h+14)=10$

Step 4: Solve for $h$

Multiply both sides by 2:

$h+14=20$

Therefore:

$\boxed{h=6}$

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Key Formulas or Methods Used

• Determinant formula for triangle area: $\text{Area}=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$
• Expansion of 3×3 determinant: Along the first row using cofactors
• 2×2 determinant: $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$

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Summary of Steps

1. Set up the area formula using the determinant with the three given vertices
2. Expand the 3×3 determinant along the first row into three 2×2 determinants
3. Evaluate each 2×2 determinant and substitute back
4. Simplify the algebraic expression inside the brackets
5. Multiply both sides by 2 and solve the resulting linear equation for $h$