The points divide the line segment into three equal parts, so the ratio of division is \(1 : 2\) for the first point and \(2 : 1\) for the second point.
For the first point, using the section formula, the coordinates dividing the segment \((-2, 2)\) and \((7, -4)\) in the ratio \(1 : 2\) are:
\[ x = \frac{1 \times 7 + 2 \times (-2)}{1 + 2} = \frac{7 - 4}{3} = 1, \quad y = \frac{1 \times (-4) + 2 \times 2}{1 + 2} = \frac{-4 + 4}{3} = 0 \]
For the second point, dividing the segment in the ratio \(2 : 1\), we get:
\[ x = \frac{2 \times 7 + 1 \times (-2)}{2 + 1} = \frac{14 - 2}{3} = 4, \quad y = \frac{2 \times (-4) + 1 \times 2}{2 + 1} = \frac{-8 + 2}{3} = -2 \]
So, the coordinates of the trisection points are:
\((1, 0) \quad \text{and} \quad (4, -2)\)
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: