Correct answer: (0, -1)
Explanation:
In a parallelogram, the diagonals bisect each other.
So, the point of intersection of the diagonals is the midpoint of both diagonals. Let the vertices of the parallelogram in order be:
\( A(2, -3),\ B(6, 5),\ C(-2, 1),\ D(-6, -7) \) The diagonals are \( AC \) and \( BD \) Find the midpoint of diagonal \( AC \): \[ \text{Midpoint of } AC = \left( \frac{2 + (-2)}{2},\ \frac{-3 + 1}{2} \right) = (0, -1) \] Find the midpoint of diagonal \( BD \): \[ \text{Midpoint of } BD = \left( \frac{6 + (-6)}{2},\ \frac{5 + (-7)}{2} \right) = (0, -1) \] Since both diagonals intersect at the same midpoint, the point of intersection is (0, -1).
In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).
If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).