If alpha represents the square of the distanc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation of the pair of lines is $x^2-y^2-x+3y-2=0$. Factoring the expression,we get $(x-y+1)(x+y-2)=0$. Thus,the lines are $L_1: x-y+1=

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(A) The given equation of the pair of lines is $x^2-y^2-x+3y-2=0$. Factoring the expression,we get $(x-y+1)(x+y-2)=0$. Thus,the lines are $L_1: x-y+1=0$ and $L_2: x+y-2=0$. Solving for the intersection point $P$: Adding the equations: $(x-y+1) + (x+y-2) = 2x-1=0 \implies x=\frac{1}{2}$. Substituting $x=\frac{1}{2}$ into $x+y-2=0$,we get $y=\frac{3}{2}$. So,$P = (\frac{1}{2}, \frac{3}{2})$. $\alpha$ is the square of the distance from the origin $(0,0)$ to $P$: $\alpha = (\frac{1}{2})^2 + (\frac{3}{2})^2 = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2}$. $\beta$ is the product of the perpendicular distances from the origin to $L_1$ and $L_2$: $d_1 = \frac{|0-0+1|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$. $d_2 = \frac{|0+0-2|}{\sqrt{1^2+1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. $\beta = d_1 	imes d_2 = \frac{1}{\sqrt{2}} 	imes \sqrt{2} = 1$. Therefore,$\alpha \beta = \frac{5}{2} 	imes 1 = \frac{5}{2}$.

If $\alpha$ represents the square of the distance between the origin and the point of intersection of the lines $x^2-y^2-x+3y-2=0$ and $\beta$ represents the product of the perpendicular distances from the origin to the pair of lines,then $\alpha \beta=$

Similar Questions

One bisector of the angle between the lines given by $a(x - 1)^2 + 2h(x - 1)y + by^2 = 0$ is $2x + y - 2 = 0$. The other bisector is

The point of intersection of the lines represented by the equation $3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$ is

If pairs of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 q x y-y^2=0$ are such that each pair bisects the angle between the other pair,then:

The equation of the bisectors of the angles between the lines represented by ${x^2} + 2xy \cot \theta + {y^2} = 0$ is

The joint equation of the pair of lines passing through the point of intersection of the lines represented by $2x^{2}-xy-15y^{2}-7x+32y-9=0$ and parallel to the coordinate axes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module