Question

What is the slope of the line which is bisector of third quadrant??

Answer 1

Answer:

Plz mark brainliest ❤️❤️❤️

Step-by-step explanation:

Answer 2

$x^2 + y^2 - 2x + 2y = 0$ is the Slope of the Line

Step-by-step explanation:

• Let us assume a circle to be touching the bisector of the third quadrant.  
• Since it's a circle, it will be touching the bisector of the first quadrant too.

The bisector of the First and Third Quadrant is a tangent y = x center of the line. So (2,0) will be the chord of the circle.

The intersection of the line y = −x and the line x = 1 is at the (1,−1), (1,−1) will be its center.

$r = \sqrt 2$

$(x-1)^2 + (y+1)^2 = (\sqrt 2)^2$

⇒ $x^2 + 1 -2x + y^2 + 1 + 2y = 2$

⇒$x^2 + y^2 - 2x + 2y = 0$

$x^2 + y^2 - 2x + 2y = 0$ can be called as the slope of the line or Diameter of circle.