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Question

find the value of $$k$$. if the following equation represent a pair of lines $$3{x^2} + 10xy + 3{y^2} + 16y + k = 0$$.


Solution

Given equation is $$3x^2+10xy+3y^2+16y+k=0$$
Comparing it with $$ax^2+2hxy+by^2+2gx+2fy+c=0$$
We get, $$a=3,\,h=5,\,b=3,\,g=0,\,f=8,\,c=k$$
The equation represents pair of straight lines.
$$\therefore$$  $$abc+2fgh-af^2-bg^2-ch^2=0$$
$$\Rightarrow$$  $$3\times 3\times k+2\times 8\times 0\times 5-3(8)^2-3(0)^2-k(5)^2=0$$
$$\Rightarrow$$  $$9k+0-192-0-25k=0$$
$$\Rightarrow$$  $$-16k-192=0$$
$$\Rightarrow$$  $$-16k=192$$
$$\therefore$$  $$k=-12$$

Maths

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