Question

# If $$(a+b)^2=4h^2$$, prove that one of the lines given by the equation $$ax^2+2hxy+by^2=0$$ will bisect the angle between the co-ordinate axes.

Solution

## Solution:-The equation of the bisectors of the angles between the coordinate axes are $$y = x$$ and $$y = -x$$.Let $$y = x$$ belong to $$a{x}^{2} + 2hxy + b{y}^{2} = 0$$$$\therefore a{x}^{2} + 2h {x}^{2} + b{x}^{2} = 0$$$$\Rightarrow a + b = -2h ..... \left( 1 \right)$$If $$y = -x$$ belongs to $$a{x}^{2} + 2hxy + b{y}^{2} = 0$$.$$\therefore a{x}^{2} - 2h{x}^{2} + b{x}^{2} = 0$$$$\Rightarrow a + b = 2h ..... \left( 2 \right)$$From $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have$$\left( a + b \right) = \pm 2h$$Squaring both sides, we get$${\left( a + b \right)}^{2} = 4{h}^{2}$$Hence proved.Maths

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