The circumcenter of the triangle formed by th | vedclass

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

Question

(C) First,factorize the equation $xy + 2x + 2y + 4 = 0$. $x(y + 2) + 2(y + 2) = 0$ $(x + 2)(y + 2) = 0$. This represents two lines: $x = -2$ and $y =

Vedclass · Accepted Answer

$(-1, -1)$

Vedclass · Answer

(C) First,factorize the equation $xy + 2x + 2y + 4 = 0$. $x(y + 2) + 2(y + 2) = 0$ $(x + 2)(y + 2) = 0$. This represents two lines: $x = -2$ and $y = -2$. The third line is given as $x + y + 2 = 0$. These three lines form a triangle with vertices at the intersection points: $1$. Intersection of $x = -2$ and $y = -2$ is $(-2, -2)$. $2$. Intersection of $x = -2$ and $x + y + 2 = 0$ is $(-2, 0)$. $3$. Intersection of $y = -2$ and $x + y + 2 = 0$ is $(0, -2)$. Since the triangle is a right-angled triangle with the right angle at $(-2, -2)$,the circumcenter is the midpoint of the hypotenuse. The hypotenuse connects $(-2, 0)$ and $(0, -2)$. Midpoint = $(\frac{-2 + 0}{2}, \frac{0 - 2}{2}) = (-1, -1)$.

The circumcenter of the triangle formed by the lines $xy + 2x + 2y + 4 = 0$ and $x + y + 2 = 0$ is

Similar Questions

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations of the sides $AB$ and $AC$ are $px + qy = 1$ and $qx + py = 1$. The equation of the median through $A$ is:

The centroid of the triangle formed by the lines $x+y-1=0$,$x-y-1=0$,and $x-3y+3=0$ is

Find the incenter of the triangle formed by the axes and the line $\frac{x}{a} + \frac{y}{b} = 1$.

If the mid-points of the sides $BC$,$CA$ and $AB$ of a triangle $ABC$ are respectively $(2,1)$,$(-1,-2)$ and $(3,3)$,then the equation of the side $BC$ is

The point $(-4, 5)$ is a vertex of a square and one of its diagonals lies along the line $7x - y + 8 = 0$. The equation of the other diagonal is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module