The coordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC$ $(AC = BC)$ are $(-2, 3)$ and $(2, 0)$ respectively. $A$ line parallel to $AB$ and having a $y$-intercept equal to $\frac{43}{12}$ passes through $C$. Then the coordinates of $C$ are:

  • A
    $\left(-\frac{3}{4}, 1\right)$
  • B
    $\left(1, \frac{17}{6}\right)$
  • C
    $\left(\frac{2}{3}, \frac{4}{5}\right)$
  • D
    $(1, 0)$

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Match the following:
List-$I$List-$II$
$A$. The equation of line passing through $(4,3)$ whose $X$-intercept is twice its $Y$-intercept$I$. $x+y-2\sqrt{2}=0$
$B$. The equation of the line passing through the centroid and circumcentre of $\triangle ABC$ with vertices $A(1,1), B(3,3), C(6,-6)$$II$. $7x+23y-8=0$
$C$. The equation of the line whose $X$-intercept is $(-3/5)$ and is perpendicular to $x-y+2=0$$III$. $x+2y+\sqrt{2}=0$
$D$. The equation of the line whose distance from the origin is $2$ and the normal drawn from the origin makes an angle $45^{\circ}$ with the positive direction of $X$-axis$IV$. $x+2y-10=0$
$V$. $5x+5y+3=0$

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$A$ line passing through $P(2,3)$ and making an angle of $30^{\circ}$ with the positive direction of the $x$-axis meets the curve $x^2-2xy-y^2=0$ at $A$ and $B$. Then the value of $PA \cdot PB$ is

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