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(B) The given pair of straight lines is $ax^2 + 2hxy + by^2 = 0$. Since one of the lines bisects the angle between the coordinate axes,its slope $m$ m

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

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(B) The given pair of straight lines is $ax^2 + 2hxy + by^2 = 0$. Since one of the lines bisects the angle between the coordinate axes,its slope $m$ must be $	an(45^{\circ}) = 1$ or $	an(135^{\circ}) = -1$. If $m = 1$,the line is $y = x$. Substituting $y = x$ into the equation $ax^2 + 2hxy + by^2 = 0$,we get: $ax^2 + 2hx(x) + b(x)^2 = 0$ $ax^2 + 2hx^2 + bx^2 = 0$ $(a + 2h + b)x^2 = 0$ This implies $a + b + 2h = 0$,or $a + b = -2h$. Squaring both sides,we get $(a + b)^2 = (-2h)^2 = 4h^2$. If $m = -1$,the line is $y = -x$. Substituting $y = -x$ into the equation,we get: $ax^2 + 2hx(-x) + b(-x)^2 = 0$ $ax^2 - 2hx^2 + bx^2 = 0$ $(a - 2h + b)x^2 = 0$ This implies $a + b = 2h$. Squaring both sides,we get $(a + b)^2 = (2h)^2 = 4h^2$. In both cases,the result is $(a + b)^2 = 4h^2$.

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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