The circumcentre of the triangle formed by th | vedclass

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

Question

(C) The given equation $xy + 2x + 2y + 4 = 0$ can be factored as $(x + 2)(y + 2) = 0$,which represents two lines: $x = -2$ and $y = -2$.The third line

Vedclass · Accepted Answer

$(-1, -1)$

Vedclass · Answer

(C) The given equation $xy + 2x + 2y + 4 = 0$ can be factored as $(x + 2)(y + 2) = 0$,which represents two lines: $x = -2$ and $y = -2$.The third line is $x + y + 2 = 0$.To find the vertices of the triangle,we find the intersection points of these lines:$1$. Intersection of $x = -2$ and $y = -2$ is $C(-2, -2)$.$2$. Intersection of $x = -2$ and $x + y + 2 = 0$ is $A(-2, 0)$.$3$. Intersection of $y = -2$ and $x + y + 2 = 0$ is $B(0, -2)$.Since the lines $x = -2$ and $y = -2$ are perpendicular,the triangle is a right-angled triangle with the right angle at $C(-2, -2)$.The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $AB$.The midpoint of $AB$ is $(\frac{-2 + 0}{2}, \frac{0 - 2}{2}) = (-1, -1)$.

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

Similar Questions

The potato spindle tuber disease is caused by which of the following?

The ratio between kinetic energy and the total energy of the electrons of a hydrogen atom according to Bohr's model is

$A$ square is inscribed in the circle $x^{2} + y^{2} - 2x + 4y - 93 = 0$ such that its sides are parallel to the coordinate axes. The coordinates of its vertices are:

On a temperature scale $Y$,water freezes at $-160^{\circ} Y$ and boils at $-50^{\circ} Y$. On this $Y$ scale,a temperature of $340 \ K$ is (in $^{\circ} Y$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module