To find the equation of the locus of point [tex]\( P(x, y) \)[/tex] such that the area of triangle [tex]\( PAB \)[/tex] is 5 square units, we can utilize the determinant method to express the area of a triangle formed by the coordinates of its vertices.
Given:
[tex]\( A(2, 3) \)[/tex] and [tex]\( B(-3, 4) \)[/tex]
We need to find the locus of point [tex]\( P(x, y) \)[/tex] such that the area of triangle [tex]\( PAB \)[/tex] is 5 square units.
The formula for the area of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] is:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
For triangle [tex]\( PAB \)[/tex]:
- [tex]\( A(2, 3) \)[/tex]
- [tex]\( B(-3, 4) \)[/tex]
- [tex]\( P(x, y) \)[/tex]
Substituting these coordinates into the area formula, we get:
[tex]\[ \text{Area} = \frac{1}{2} \left| 2(4 - y) + (-3)(y - 3) + x(3 - 4) \right| \][/tex]
Simplifying inside the absolute value, we obtain:
[tex]\[ \text{Area} = \frac{1}{2} \left| 2(4 - y) - 3(y - 3) + x(3 - 4) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| 8 - 2y - 3y + 9 + x(-1) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| 17 - 5y - x \right| \][/tex]
Given that the area of the triangle [tex]\( PAB \)[/tex] is 5 square units, we set up the following equation:
[tex]\[ \frac{1}{2} \left| 17 - 5y - x \right| = 5 \][/tex]
To remove the fraction, multiply both sides by 2:
[tex]\[ \left| 17 - 5y - x \right| = 10 \][/tex]
This absolute value equation can be split into two linear equations:
1. [tex]\( 17 - 5y - x = 10 \)[/tex]
2. [tex]\( 17 - 5y - x = -10 \)[/tex]
Solving these, we get:
1. [tex]\( 17 - 5y - x = 10 \)[/tex]
[tex]\[ 17 - 10 = 5y + x \][/tex]
[tex]\[ x + 5y = 7 \][/tex]
2. [tex]\( 17 - 5y - x = -10 \)[/tex]
[tex]\[ 17 + 10 = 5y + x \][/tex]
[tex]\[ x + 5y = 27 \][/tex]
Thus, the equation of the locus of point [tex]\( P(x, y) \)[/tex] such that the area of triangle [tex]\( PAB \)[/tex] is 5 square units is:
[tex]\[ \left| x + 5y - 17 \right| = 10 \][/tex]
