Question Statement

Using the formula of area, find $x$ if the points $P(3,2)$, $Q(-1,x)$ and $R(7,3)$ are collinear.

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Background and Explanation

Three points are collinear if they lie on the same straight line, which geometrically means the area of the triangle formed by these points is zero. We use the determinant formula for the area of a triangle given its vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$.

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Solution

Setting Up the Condition for Collinearity

For three points to be collinear, the area of the triangle formed by them must equal zero. The area of triangle $PQR$ with vertices $P(3,2)$, $Q(-1,x)$, and $R(7,3)$ is given by:

$\text{Area}=\frac{1}{2}\left|\begin{array}{rrr}3& 2& 1\\ -1& x& 1\\ 7& 3& 1\end{array}\right|$

Since the points are collinear:

$\frac{1}{2}\left|\begin{array}{rrr}3& 2& 1\\ -1& x& 1\\ 7& 3& 1\end{array}\right|=0$

Multiplying both sides by 2:

$\left|\begin{array}{rrr}3& 2& 1\\ -1& x& 1\\ 7& 3& 1\end{array}\right|=0$

Expanding the Determinant

We expand the $3\times 3$ determinant along the first row using cofactor expansion:

$3\left|\begin{array}{cc}x& 1\\ 3& 1\end{array}\right|-2\left|\begin{array}{rr}-1& 1\\ 7& 1\end{array}\right|+1\left|\begin{array}{rr}-1& x\\ 7& 3\end{array}\right|=0$

Now we calculate each $2\times 2$ determinant:

• First minor: $\left|\begin{array}{cc}x& 1\\ 3& 1\end{array}\right|=x(1)-1(3)=x-3$
• Second minor: $\left|\begin{array}{rr}-1& 1\\ 7& 1\end{array}\right|=(-1)(1)-1(7)=-1-7=-8$
• Third minor: $\left|\begin{array}{rr}-1& x\\ 7& 3\end{array}\right|=(-1)(3)-x(7)=-3-7x$

Substituting these back:

$3(x-3)-2(-8)+1(-3-7x)=0$

Simplifying and Solving for $x$

Expanding the terms:

$3x-9+16-3-7x=0$

Combining like terms:

• For $x$: $3x-7x=-4x$
• For constants: $-9+16-3=4$

This gives us:

$-4x+4=0$

Solving for $x$:

$-4x=-4$

$x=1$

Therefore, the value of $x$ is 1.

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Key Formulas or Methods Used

• Collinearity Condition: Three points are collinear if the area of the triangle formed by them equals zero
• Area of Triangle (Determinant Form): $\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$
• Determinant Expansion: Expansion of $3\times 3$ determinant along first row: $a_{11}{M}_{11}-a_{12}{M}_{12}+a_{13}{M}_{13}$ where $M_{ij}$ are the minors
• $2\times 2$ 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: Write the determinant expression for the area of triangle $PQR$ using coordinates $P(3,2)$, $Q(-1,x)$, $R(7,3)$
2. Apply collinearity condition: Set the area equal to zero (removing the $\frac{1}{2}$ factor by multiplying by 2)
3. Expand the determinant: Use cofactor expansion along the first row to reduce to $2\times 2$ determinants
4. Calculate minors: Evaluate each $2\times 2$ determinant using the rule $ad-bc$
5. Simplify the equation: Combine like terms to get a linear equation in $x$
6. Solve: Isolate $x$ to find $x=1$