# M.C.Q : Straight Line (1 to 10)

LEVEL – I

1. Let P and Q be the points on the line joining A(–2 , 5) and B(3 , 1) such that AP = PQ = QB , then the mid point of PQ is

(A) (1/2 , 3)

(B) (-1/4 ,4)

(C) (2, 3)

(D) (– 1, 4)

2. P is a point lying on line y = x then maximum value of |PA – PB| , ( where A ≡ (1 , 3) , B ≡(5 , 2)) is

(A) √5

(B) 2√2

(C) √17

(D) 3/ √2

3. Locus of point of intersection of the perpendicular lines one belonging to (x + y – 2) + λ(2x + 3y – 5) = 0 and other to (2x + y – 11) + λ(x + 2y – 13) = 0 is a

(A) circle

(B) straight line

(C) pair of lines

(D) None of these

4. A(-3, 4), B(5, 4), C and D form a rectangle. If x – 4y + 7 = 0 is a diameter of the circumcircle of the rectangle ABCD then area of rectangle ABCD is

(A) 8

(B) 16

(C) 32

(D) 64

5. The triangle ABC has medians AD, BE, CF. AD lies along the line y = x + 3 , BE lies along the line y = 2x + 4, AB has length 60 and angle C = 90°, then the area of ΔABC is

(A) 400

(B) 200

(C) 100

(D) none of these

6. A line passes through (1, 0). The slope of the line, for which its intercept between y = x – 2 and y = -x + 2 subtends a right angle at the origin, is

(A) ± 2/3,

(B) ± 3/2

(C) ± 1

(D) none of these

7. The locus of the image of origin in line rotating about the point (1 , 1) is

(A) x2 + y2 = 2(x + y)

(B) x2 + y2 = (x + y)

(C) x2 + y2 = 2(x – y)

(D) x2 + y2 = (x – y)

8. A line through the point (–a, 0) cuts from the second quadrant a triangular region with area T. The equation for the line is

(A) 2Tx + a2 y + 2aT = 0

(B) 2Tx – a2y + 2aT = 0

(C) 2Tx + a2y – 2aT = 0

(D) 2Tx – a2y – 2aT = 0

9. If A1 , A2 , A3 , … , An are n points in a plane whose coordinates are (xi , yi), i = 1 , 2 , … , n respectively. A1A2 is bisected by the point G1 ; G1A3 is divided by G2 in ratio 1 : 2 and G2A4 is divided by G3 in the ratio 1 : 3 , G3A5 at G4 in the ratio 1 : 4 and so on until all the points are exhausted, then the coordinates of fixed point Gn – 1 so obtained will be

(A)

(B)

(C)

(D)

10. The vertex A of a triangle ABC is the point (-2, 3) whereas the line perpendicular to the sides AB and AC are x – y – 4 = 0 and 2x – y – 5 = 0 respectively. The right bisectors of sides meet at P(3/2 , 5/2) . Then the equation of the median of side BC is

(A) 5x + 2y = 10

(B) 5x – 2y = 16

(C) 2x – 5y = 10

(D) none of these