Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Let P(x, y) be the required point.
Using the section formula, we obtain
\(x=\)\(\frac{2\times4+3\times(-1)}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)
\(y=\frac{2\times(-3)+3\times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Therefore, the point is (1, 3).
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:
If the circles \( x^2 + y^2 - 8x - 8y + 28 = 0 \) and \( x^2 + y^2 - 8x - 6y + 25 - a^2 = 0 \) have only one common tangent, then \( a \) is:
Let \( a \) be an integer multiple of 8. If \( S \) is the set of all possible values of \( a \) such that the line \( 6x + 8y + a = 0 \) intersects the circle \( x^2 + y^2 - 4x - 6y + 9 = 0 \) at two distinct points, then the number of elements in \( S \) is:
Assertion (A): The sum of the first fifteen terms of the AP $ 21, 18, 15, 12, \dots $ is zero.
Reason (R): The sum of the first $ n $ terms of an AP with first term $ a $ and common difference $ d $ is given by: $ S_n = \frac{n}{2} \left[ a + (n - 1) d \right]. $
Assertion (A): The sum of the first fifteen terms of the AP $21, 18, 15, 12, \dots$ is zero.
Reason (R): The sum of the first $n$ terms of an AP with first term $a$ and common difference $d$ is given by: $S_n = \frac{n}{2} \left[ a + (n - 1) d \right].$