The distance of the point $(1, 2)$ from the line $x + y = 0$ measured parallel to the line $3x - y = 2$ is

  • A
    $\frac{3 \sqrt{10}}{4}$ units
  • B
    $\frac{3 \sqrt{2}}{8}$ units
  • C
    $10$ units
  • D
    $5 \sqrt{5}$ units

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If the equation of the straight line passing through the point of intersection of $x+2y-19=0$ and $x-2y-3=0$ and which is at a perpendicular distance of $5$ units from the point $(-2,4)$ is $5x+by+c=0$,then a possible value of $5+b+c$ is

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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$IV$. $x+2y-10=0$
$V$. $5x+5y+3=0$

Consider the fourteen lines in the plane given by $y=x+r$ and $y=-x+r$,where $r \in \{0, 1, 2, 3, 4, 5, 6\}$. The number of squares formed by these lines,whose sides are of length $\sqrt{2}$,is:

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