Question

The equation of the pair of straight lines through the point $(1,1)$ and perpendicular to the pair of straight lines $3x^2-8xy+5y^2=0$ is

• A. $5x^2+8xy+3y^2-14x-18y+16=0$
• B. $5x^2+8xy+3y^2-18x-14y+16=0$
• C. $5x^2-8xy+3y^2-18x-14y+16=0$
• D. $5x^2-8xy+3y^2-14x-18y+16=0$

Answer 1

The correct option is B $5x^2+8xy+3y^2-18x-14y+16=0$

$3x^2-8xy+5y^2=0$

$3-8\frac{y}{x}+5\frac{y^2}{x^2}=0$

$3-8m+5m^2=0$

$5m^2-8m+3=0$

$m_1=1$ and $m_2=\frac{3}{5}$

Let $m_1^{{}^{\prime }}$ and $m_2^{{}^{\prime }}$ be slopes of lines perpendicular to given pair of straight lines.

Then,

$m_1^{{}^{\prime }}=\frac{-1}{m}$ and $m_2^{{}^{\prime }}=\frac{-1}{\frac{3}{5}}$

$m_1^{{}^{\prime }}=-1$ and $m_2^{{}^{\prime }}=\frac{-5}{3}$

Equation of line with slope $m_1^{{}^{\prime }}=-1$ and passing through $(1,1)$ is

$(y-1)=(-1)(x-1)$

$⟹x+y-2=0$

Equation of line with slope $m_2^{{}^{\prime }}=\frac{-5}{3}$ and passing through $(1,1)$ is

$(y-1)=\left(\frac{-5}{3}\right)(x-1)$

$⟹5x+3y-8=0$

Combines equation of above two lines is

$(x+y-2)(5x+3y-8)=0$

$5x^2+3y^2+8xy-18x-14y+16=0$

Hence, the answer is option (B)