First, solve the system of equations:
1. \(y^2 = 12x\)
2. \(3y = 4x + 6\)
From equation (2), express \(x\) in terms of \(y\):
\(x = \frac{3y - 6}{4}\)
Substitute \(x\) in equation (1):
\(y^2 = 12 \left( \frac{3y - 6}{4} \right)\)
Simplify:
\(y^2 = 9y - 18\)
Rearrange to form a quadratic equation:
\(y^2 - 9y + 18 = 0\)
Factor the quadratic:
\((y - 3)(y - 6) = 0\)
Thus, \(y = 3\) or \(y = 6\).
Find corresponding \(x\) values:
For \(y = 3\):
\(x = \frac{3(3) - 6}{4} = \frac{9 - 6}{4} = \frac{3}{4}\)
For \(y = 6\):
\(x = \frac{3(6) - 6}{4} = \frac{18 - 6}{4} = 3\)
The points of intersection are \(\left( \frac{3}{4}, 3 \right)\) and \((3, 6)\).
Calculate the distance between the points:
\(\text{Distance} = \sqrt{\left( 3 - \frac{3}{4} \right)^2 + (6 - 3)^2}\)
\(= \sqrt{\left( \frac{9}{4} \right)^2 + 3^2}\)
\(= \sqrt{\frac{81}{16} + 9}\)
\(= \sqrt{\frac{81}{16} + \frac{144}{16}}\)
\(= \sqrt{\frac{225}{16}}\)
\(= \frac{15}{4}\)
\(= 3.75\)