If $\lambda x^{2}-5 x y+6 y^{2}+x-3 y=0$ represents a pair of straight lines, then their point of intersection is
1) $(1,3)$
2) $(-1,-3)$
3) $(3,1)$
4) $(-3,-1)$

Hint: Here we first need to equate the given expression with the general form. So that we can find the unknown values. And for finding the point of intersection, we have to substitute the values in the straight-line equation. The resultant values represent the points of intersection.

Formula Used: The junction of two lines is referred to as the point of intersection. The equations $a_{1} x+b_{1} y+c_{1}=0$and $a_{2} x+b_{2} y+c_{2}=0$correspondingly depict these two lines.

Complete step by step Solution:
Given $\lambda x^{2}-5 x y+6 y^{2}+x-3 y=0$
 $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$
We get $a=\lambda, h=-5 / 2, b=6, g=1 / 2, f=-3 / 2, c=0$
For a straight line,
$a b c+2 f g h-a f^{2}-b g^{2}-c h^{2}=0$
Then substituting the values
We get
$=>\lambda \times 6 \times 0+2 \times-3 / 2 \times 1 / 2 \times-5 / 2-\lambda \times 9 / 4-6 \times 1 / 4-0=0$
$=>0+15 / 4-9 \lambda / 4-3 / 2=0$
$=>9 / 4=9 \lambda / 4$
$=>\lambda=1$
So the equation becomes $x^{2}-5 x y+6 y^{2}+x-3 y=0$
$=>x^{2}-3 x y-2 x y+6 y^{2}+x-3 y=0$
$=>x(x-3 y)-2 y(x-3 y)+x-3 y=0$
Simplify the terms
$=>(x-3 y)(x-2 y+1)=0$
So $x-3 y=0$……...(i)
$x-2y+1=0$…………..(2)
Solving (i) and (ii)
We get$x=-3, y=-1$
The point of intersection of the pair of straight lines $x^{2}-5 x y+6 y^{2}+x-3 y=0$ is $(-3,-1)$


Hence, the correct option is 4.

Note: Students should keep in mind that when two lines are said to intersect when they have exactly one point in common. There is a point at which the intersecting lines meet. The point of intersection is the same location that appears on all intersecting lines. There will be a place where the two coplanar, non-parallel straight lines intersect.
Finding the intersection of three or more lines is achievable. We can determine the answer for the location of the intersection of two lines by resolving the two equations.