NDA_Math_2025_Q84

Question:

Let A(3, -1) j and B(1, 1) be the end points of line segment AB. Let P be the middle point of the line segment AB. Let Q be the point situated at a distance of 2 units from P on the perpendicular bisector line of AB. What are the possible coordinates of Q?

Let A(3, -1) j and B(1, 1) be the end points of line segment AB. Let P be the middle point of the line segment AB. Let Q be the point situated at a distance of 2 units from P on the perpendicular bisector line of AB. What are the possible coordinates of Q?

Options:

  • a) (2, 1)
  • b) (3, 1)
  • c) (2, 2)
  • d) (1, 3)
  • a) (2, 1)
  • b) (3, 1)
  • c) (2, 2)
  • d) (1, 3)

correct answer : b)

Explanation:

Coordinate of P = 31
2
11
2
+- +
 
,
= P(2, 0)
Slope of AB = 11
13
+

= -1
Equation of AB is
y – 1 = -1 (x – 1) Þ x + y – 2 = 0
Distance from Q(x, y)
xy+- 2
2
= 2
x + y – 2 = 2 Þ x + y = 4 …(i)
Slope of PQ = y
x- 2
Þ PQ ^ AB
\ (-1) y
x –

 
2 = -1 Þ y = x – 2
From (i)
x + x – 2 = 4 Þ x = 3 and y = 1
Q(3, 1)
Oswaal NDA/NA Year-wise Solved Papers

Coordinate of P = 31
2
11
2
+- +
 
,
= P(2, 0)
Slope of AB = 11
13
+

= -1
Equation of AB is
y – 1 = -1 (x – 1) Þ x + y – 2 = 0
Distance from Q(x, y)
xy+- 2
2
= 2
x + y – 2 = 2 Þ x + y = 4 …(i)
Slope of PQ = y
x- 2
Þ PQ ^ AB
\ (-1) y
x –

 
2 = -1 Þ y = x – 2
From (i)
x + x – 2 = 4 Þ x = 3 and y = 1
Q(3, 1)
Oswaal NDA/NA Year-wise Solved Papers

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