Question

Passage: Consider the straight line 3x + y + 4 = 0. Answer the following questions.

(i) The point on the line 3x + y + 4 = 0 which is equidistant from the points ( 5, 6) and (3, 2) is

(A) ( 1, 1)

(B) ( 2, 2)

(C) ( 3, 5)

(D) ( 1 3 , 3)

(ii) Equation of the line passing through the point (1, 1) and perpendicular to the given line is

(A) x - 3y + 4 = 0

(B) x - 3y + 5 = 0

(C) x - 3y - 4 = 0

(D) x - 3y + 2 = 0

(iii) If the line y + 5 = k(x - 3) is parallel to the given line then the area of the triangle formed by this line and the coordinate axes (in sq. units) is

(A) 8/3

(B) 16/3

(C) 4

(D) 5

Answer 1

Correct option (i) (B),(ii) (D),(iii) (A)

Explanation :

(i) Let A = (5 ,6) and B = (3, 2). The slope of AB is

and the midpoint of AB = (-1, 4). Hence, the perpendicular bisector of the segment (bar)AB is y - 4 = 2(x + 1) or 2x - y + 6 = 0 . Solving this equation and the given line equations, we have x = -2 and y = 2. Thus, (-2, 2) is the point on the given line which is equidistant from both A(-5 6) and B(3, 2).

(ii) Line perpendicular to the given line is of the form

y = 1/3x + c

This line passes through (1, 1). It implies that

1 = 1/3 + c ⇒ c = 2/3

Thus, the required line is

y = x/3 + 2/3 or x - 3y + 2 = 0

(iii) The line y + 5 = k(n - 3) is parallel to the given line ⇒ k = -3. That is,

3x + y = 4

or x/(4/3) + y/4 = 1

Hence, the area of the triangle is

1/2(4/3)(4) = 8/3