The area bounded by the angle bisectors of th | vedclass

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

Question

(A) The given equation is ${x^2} - ({y^2} - 2y + 1) = 0$,which simplifies to ${x^2} - {(y - 1)^2} = 0$.This represents the pair of lines $(x - (y - 1)

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The given equation is ${x^2} - ({y^2} - 2y + 1) = 0$,which simplifies to ${x^2} - {(y - 1)^2} = 0$.This represents the pair of lines $(x - (y - 1))(x + (y - 1)) = 0$,or $(x - y + 1)(x + y - 1) = 0$.The lines are $x - y + 1 = 0$ and $x + y - 1 = 0$.The angle bisectors of these lines are given by $\frac{x - y + 1}{\sqrt{1^2 + (-1)^2}} = \pm \frac{x + y - 1}{\sqrt{1^2 + 1^2}}$.This simplifies to $x - y + 1 = \pm (x + y - 1)$.Case $1$: $x - y + 1 = x + y - 1 \implies 2y = 2 \implies y = 1$.Case $2$: $x - y + 1 = -(x + y - 1) \implies x - y + 1 = -x - y + 1 \implies 2x = 0 \implies x = 0$.Thus,the angle bisectors are the lines $x = 0$ and $y = 1$.The third line is $x + y = 3$.The vertices of the triangle formed by $x = 0$,$y = 1$,and $x + y = 3$ are found by intersection:Intersection of $x = 0$ and $y = 1$ is $(0, 1)$.Intersection of $x = 0$ and $x + y = 3$ is $(0, 3)$.Intersection of $y = 1$ and $x + y = 3$ is $(2, 1)$.The triangle is a right-angled triangle with vertices $(0, 1)$,$(0, 3)$,and $(2, 1)$.The base length is $2 - 0 = 2$ and the height is $3 - 1 = 2$.Area = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 2 = 2$.

The area bounded by the angle bisectors of the lines ${x^2} - {y^2} + 2y = 1$ and the line $x + y = 3$ is:

Similar Questions

The area (in sq units) of the quadrilateral formed by two pairs of lines $\lambda^2 x^2 - m^2 y^2 - n(\lambda x + my) = 0$ and $\lambda^2 x^2 - m^2 y^2 + n(\lambda x + my) = 0$ is:

Area of the triangle formed by the lines $3x^2-4xy+y^2=0$ and $2x-y=6$ is

From the point $(3,-4)$,perpendicular lines $L_1$ and $L_2$ are drawn to each of the lines represented by $S \equiv 2x^2+3xy-2y^2-7x+y+3=0$. The area of the quadrilateral formed by the pair of lines $S=0$,$L_1$,and $L_2$ is (in square units):

The area of the region enclosed between the pair of lines $xy = 0$ and the lines $xy + 5x - 4y - 20 = 0$ is .....

If the pair of straight lines given by $Ax^2+2Hxy+By^2=0$ $(H^2>AB)$ forms an equilateral triangle with the line $ax+by+c=0$,then $(A+3B)(3A+B)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module