The midpoint of $A(-1, 3)$ and $B(2, 4)$ is:
\[\text{Midpoint} = \left( \frac{-1 + 2}{2}, \frac{3 + 4}{2} \right) = \left( \frac{1}{2}, \frac{7}{2} \right)\]
The slope of $AB$ is:
\[\text{Slope} = \frac{4 - 3}{2 - (-1)} = \frac{1}{3}\]
The perpendicular slope is $-3$. Using the point-slope formula:
\[y - \frac{7}{2} = -3 \left( x - \frac{1}{2} \right)\]
Simplify to find the $y$-intercept ($x = 0$):
\[y = 5\]
Let \( F \) and \( F' \) be the foci of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (where \( b<2 \)), and let \( B \) be one end of the minor axis. If the area of the triangle \( FBF' \) is \( \sqrt{3} \) sq. units, then the eccentricity of the ellipse is:
A common tangent to the circle \( x^2 + y^2 = 9 \) and the parabola \( y^2 = 8x \) is
If the equation of the circle passing through the points of intersection of the circles \[ x^2 - 2x + y^2 - 4y - 4 = 0, \quad x^2 + y^2 + 4y - 4 = 0 \] and the point \( (3,3) \) is given by \[ x^2 + y^2 + \alpha x + \beta y + \gamma = 0, \] then \( 3(\alpha + \beta + \gamma) \) is: