Q52 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
The line $2x+y=4$ intersects X and Y axes at points A and B respectively. Let $C(p, q)$ be the point such that the lines AC and BC are perpendicular. The distance of point C from the midpoint of segment AB is...
- A$\sqrt{2}$
- Correct answerB$\sqrt{5}$
- C$2\sqrt{2}$
- D$2\sqrt{5}$
Solution
The line $2x+y=4$ intersects the X-axis at $A(2, 0)$ and the Y-axis at $B(0, 4)$.
The midpoint of segment AB is $M\left(\dfrac{2+0}{2}, \dfrac{0+4}{2}\right) = (1, 2)$.
Since the lines AC and BC are perpendicular, $\angle ACB = 90^{\circ}$.
This implies that the point C lies on a circle with AB as its diameter.
The distance of point C from the midpoint of segment AB is the radius of this circle.
The length of the diameter AB is $\sqrt{(2-0)^2 + (0-4)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.
The radius of the circle is $\dfrac{AB}{2} = \dfrac{2\sqrt{5}}{2} = \sqrt{5}$.
Thus, the distance of point C from the midpoint of segment AB is $\sqrt{5}$.
The midpoint of segment AB is $M\left(\dfrac{2+0}{2}, \dfrac{0+4}{2}\right) = (1, 2)$.
Since the lines AC and BC are perpendicular, $\angle ACB = 90^{\circ}$.
This implies that the point C lies on a circle with AB as its diameter.
The distance of point C from the midpoint of segment AB is the radius of this circle.
The length of the diameter AB is $\sqrt{(2-0)^2 + (0-4)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.
The radius of the circle is $\dfrac{AB}{2} = \dfrac{2\sqrt{5}}{2} = \sqrt{5}$.
Thus, the distance of point C from the midpoint of segment AB is $\sqrt{5}$.
Hence, the correct answer is option: 'B'.