Question

The circumcentre of the triangle formed by the lines \(x + y + 2 = 0\), \(2x + y + 8 = 0\) and \(x - y - 2 = 0\) is

• A. \((-5, 1)\)
• B. \((-4, 0)\)
• C. \((0, -2)\)
• D. \(\left(-\frac{8}{3}, -\frac{2}{3}\right)\)

Hint:

To find the circumcentre of a triangle, find the intersection points of the sides, calculate the midpoints, and use the perpendicular bisector method to find the point equidistant from all three vertices.

Answer 1

The Correct Option is B

Step 1: Find the intersection points of each pair of lines to identify the vertices of the triangle. - Solve \(x + y + 2 = 0\) and \(2x + y + 8 = 0\): \[ \text{From } x + y + 2 = 0 \Rightarrow y = -x - 2. \] Substituting into \(2x + y + 8 = 0\): \[ 2x + (-x - 2) + 8 = 0 \Rightarrow x = -6, \quad y = 4. \] So, the intersection point is \((-6, 4)\). - Solve \(2x + y + 8 = 0\) and \(x - y - 2 = 0\): \[ \text{From } x - y - 2 = 0 \Rightarrow y = x - 2. \] Substituting into \(2x + y + 8 = 0\): \[ 2x + (x - 2) + 8 = 0 \Rightarrow 3x = -6 \Rightarrow x = -2, \quad y = -4. \] So, the intersection point is \((-2, -4)\). - Solve \(x + y + 2 = 0\) and \(x - y - 2 = 0\): \[ \text{From } x - y - 2 = 0 \Rightarrow y = x - 2. \] Substituting into \(x + y + 2 = 0\): \[ x + (x - 2) + 2 = 0 \Rightarrow 2x = 0 \Rightarrow x = 0, \quad y = -2. \] So, the intersection point is \((0, -2)\). 
Step 2: Now, we use the perpendicular bisector method to find the circumcenter. - First, find the midpoints of the sides: - Midpoint of \((-6, 4)\) and \((-2, -4)\) is \((-4, 0)\). - Midpoint of \((-2, -4)\) and \((0, -2)\) is \((-1, -3)\). - Midpoint of \((0, -2)\) and \((-6, 4)\) is \((-3, 1)\). - The perpendicular bisector of any side is the line passing through the midpoint and perpendicular to the line segment. - We see that the midpoint of \((-6, 4)\) and \((-2, -4)\) is \((-4, 0)\), and since this coincides with the intersection of the perpendicular bisectors of the sides, this is the circumcenter.