Question

Match the items of Column I with those of Column II.

| Column I | Column II |
| (A) Equation of the line through the point (2, 3), such that its x-intercept is twice its y intercept, is | (p) 2x + 33y - 46 = 0 |
| (B) If A (-5,6), B(-1, -4) and C(3,2) are the vertices of a triangle, then the equation of the line passing through the centroid and the circumcentre of ΔABC is | (r) 28x - 21y + 12 = 0 |
| (C) The equation of the line passing through the point (1,-6) , whose product of the intercepts on the axes is 1, is | (*) x + 2y - 8 |
| (D) The equation of the line whose x-intercept is -3/7 and is perpendicular to the line 3x + 4y – 10 = 0, is | (t) 4x + y + 2 = 0 |

Answer 1

(A) Let the line be

x/2a + y/a = 1

Since it passes through the point (2, 3), we have

2/2a + 3/a = 1

2 + 6 = 2a

a = 4

Hence the required line equation is

x/8 + y/4 = 1

x + 2y - 8 = 0

Answer: (A) → (*)

(B) We have G as the cenrtroid of ΔABC which is given by (-1,4/3). Equation of the perpendicular bisector of the side BC is

2x + 3y + 1 = 0 ....(1)

Equation of the perpendicular bisector of the side AB is

2x - 5y + 11 = 0 ....(2)

Solving Eqs. (1) and (2), the circumcentre of ΔABC is given by (-19/8,5/4)

or * = (-19/8,5/4) and G = (-1, 4/3)

Then the slope of the line SG is

Hence, the equation of the line SG is

y - 4/3 = 2/33(x + 1)

2x - 33y+ 46 = 0

Answer: (B) → (p)

(C) Let the equation of the line be

x/a + ay = 1

This passes through (1, 6). This implies that

Therefore, the required lines are

Answer: (C) → (q), (t)

(D) Let the required line be

x/(-3/7) + y/b = 1 ⇒ -7x/3 + y/b = 1

Since this line is perpendicular to the line 3x + 4y - 10 = 0 , we have

(+7b/3(-3/4) = -1) ⇒ b = +4/7

Therefore, the required line is

-7x/3 + 7y/4 = 1

28x - 21y + 12 = 0

Answer: (D) → (r)