Find the area of the triangle formed by the line x+y=3 and the angle bisectors of pair of straight lines x2-y2+2 y=1

Question

Find the area of the triangle formed by the line x+y=3 and the angle bisectors of pair of straight lines x2-y2+2 y=1 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

x2-(y-1)2=0 x-y+1=0, x+y-1=0 ⇒ Bisector are y-axis and line parallel to x-axis, i.e. y =1. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/2bc620e99608fd1e185dba715bb68dff-.png" /> Area of triangle =(1/2) ⋅ 2 × 2=2 sq. units

Q. Find the area of the triangle formed by the line $x + y = 3$ and the angle bisectors of pair of straight lines $x^{2} - y^{2} + 2 y = 1$

$x^{2} - (y - 1)^{2} = 0$
$x - y + 1 = 0, x + y - 1 = 0$
$\Rightarrow$ Bisector are $y$ -axis and line parallel to $x$ -axis, i.e. $y = 1$ .

Area of triangle $= \frac{1}{2} \cdot 2 \times 2 = 2$ sq. units

Q. Find the area of the triangle formed by the line x+y=3 and the angle bisectors of pair of straight lines x2−y2+2y=1

x2−(y−1)2=0 x−y+1=0,x+y−1=0 ⇒ Bisector are y-axis and line parallel to x-axis, i.e. y=1. Area of triangle =21​⋅2×2=2 sq. units

Q. Find the area of the triangle formed by the line $x + y = 3$ and the angle bisectors of pair of straight lines $x^{2} - y^{2} + 2 y = 1$

$x^{2} - (y - 1)^{2} = 0$
$x - y + 1 = 0, x + y - 1 = 0$
$\Rightarrow$ Bisector are $y$ -axis and line parallel to $x$ -axis, i.e. $y = 1$ .

Area of triangle $= \frac{1}{2} \cdot 2 \times 2 = 2$ sq. units