The midpoint of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\(\text{Mid-point} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Substitute the given points \((-1, 3)\) and \((8, 3)\):
\(\text{Mid-point} = \left( \frac{-1 + 8}{2}, \frac{3 + 3}{2} \right) = \left( \frac{7}{2}, \frac{6}{2} \right)\)
Simplify:
\(\text{Mid-point} = \left( \frac{7}{2}, 3 \right)\)
Thus, the correct answer is:
\(Option\space d)\ \left( \frac{7}{2}, \frac{9}{4} \right)\)
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: