Question

If $3x^2-11xy+10y^2-7x+13y+k=0$ denotes a pair of straight lines, then the point of intersection of the lines is

• A. (1, 3)
• B. (3, 1)
• C. (-3, 1)
• D. (1, -3)

Answer 1

Answer: (B)

The pair of straight lines is
$3x^2-11xy+10y^2-7x+13y+k=0$
then on compairing with
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
$\Rightarrow <br/><br/>\{\begin{array}{l}<br/><br/>a=3,h=-\frac{11}{2},b=10\\ <br/><br/>g=-\frac{7}{2},f=\frac{13}{2},c=k<br/><br/>\end{array}$
Then, the point of intersection of the lines is
$\left[\frac{hf-bg}{ab-h^2},\frac{gh-af}{ab-h^2}\right]$
$=\left[\frac{\frac{-143}{4}+\frac{70}{2}}{30-\frac{121}{4}},\frac{\frac{77}{4}-\frac{39}{2}}{30-\frac{121}{4}}\right]$
$=\left[\frac{-3}{120-121},\frac{77-78}{120-121}\right]$
$=\left(\frac{-3}{-1},\frac{-1}{-1}\right)=(3,1)$