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(B) The given pair of straight lines is $ax^2 + 2hxy + by^2 = 0$.The lines bisecting the angle between the coordinate axes are $y = x$ and $y = -x$.If

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$(a + b)^2 = 4h^2$

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(B) The given pair of straight lines is $ax^2 + 2hxy + by^2 = 0$.The lines bisecting the angle between the coordinate axes are $y = x$ and $y = -x$.If $y = x$ is one of the lines,then substituting $y = x$ into the equation gives:$ax^2 + 2h(x)(x) + b(x)^2 = 0$$ax^2 + 2hx^2 + bx^2 = 0$$(a + 2h + b)x^2 = 0$This implies $a + b + 2h = 0$,or $a + b = -2h$.Squaring both sides,we get $(a + b)^2 = (-2h)^2 = 4h^2$.Similarly,if $y = -x$ is one of the lines,substituting $y = -x$ gives:$ax^2 + 2h(x)(-x) + b(-x)^2 = 0$$ax^2 - 2hx^2 + bx^2 = 0$$(a - 2h + b)x^2 = 0$This implies $a + b = 2h$,which also leads to $(a + b)^2 = 4h^2$ upon squaring.Thus,the condition is $(a + b)^2 = 4h^2$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

If one of the lines of the pair of straight lines $ax^2 + 2hxy + by^2 = 0$ bisects the angle between the coordinate axes,then:

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