The diagonals of a parallelogram ABCD are along the lines $x+3 y=4$ and $6 x-2 y=7$. Then ABCD must be a

The diagonals of a parallelogram ABCD are along the lines $x+3 y=4$ and $6 x-2 y=7$. Then ABCD must be a
  1. rectangle.
  2. square.
  3. rhombus.
  4. cyclic quadrilateral

Solution

Slope of $x+3 y=4$ is $\mathrm{m}_1=-\frac{1}{3}$ Slope of $6 x-2 y=7$ is $m_2=3$ Here, $\mathrm{m}_1 \cdot \mathrm{~m}_2=-1$ $\therefore \quad$ The diagonals are perpendicular to each other. $\therefore \quad$ Parallelogram ABCD is a rhombus.

Asked in: MHT CET 2024 (02 May Shift 2)

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