Question

the difference of the slope of the lines 3x^2-4xy+y^2=0​

Answer 1

$Consider \: the \: given \: pair \: of \ \\ equations \\ </p><p></p><p> {3x}^{2} −4xy+ {y}^{2} =0</p><p></p><p>$

$We \: know \: that \\ \: the \: standard \: \\ joint \: equation \: of \: two \: line \: is \: \\ </p><p></p><p> {ax}^{2} +2hxy+ {by}^{2} =0</p><p></p><p>$

$Therefore, \\ </p><p></p><p>a \: = \: 3 \\ </p><p></p><p>2h \: = \: −4 \\ </p><p></p><p>b=1 \\ </p><p></p><p> </p><p></p><p>Let  \: m1  \: and  \: m2 \: \\  be \: the \: slopes \: of \: both \: the \: lines \\ Then,</p><p></p><p>$

$m1+m2=−2h \div b \\ </p><p></p><p>m1+m2=−(−4) \div 1=4</p><p></p><p>$

$And, </p><p></p><p>m1 m2=a \div b </p><p></p><p>m1m2=3 \div 1=3</p><p></p><p> </p><p></p><p>$

$we \: know \: that \:$

$({m1 - m2)}^{2} = {(m1 + m2)}^{2} - 4m1m2$

${(m1 - m2)}^{2} = {(4)}^{2} - 4(3) \\ = 16 - 12 \\ = 4 \\ m1 - m2 = \sqrt{4} \\ m1 - m2 = 2$