Match the following :
Column -I | Column-II |
(i) The ratio in which the line 3x – 2y + 5 = 0 divides the join of (6, – 7) and (– 2, 3) is | [A] |
(ii) The ratio in which the line 3x + 4y + 2 = 0 divides the distance between 3x + 4y + 5 = 0, and 3x + 4y – 5 = 0 , is | [B] 91 |
(iii) If the extremities of the base of an isosceles triangle are the points (2a,0) and (0, a) and the equation of one of the sides is x = 2a, then the area of the triangle is | [C] 37 : 7 |
(iv) The line 3x + 2y = 24 meets y- axis at A and x- axis at B. The perpendicular bisector of AB meets the line through (0, –1) parallel to x- axis at C. The area of the triangle ABC is | [D] 3 : 7 |
[E] |
Text Solution
Verified by Experts(i) [C]; (ii) [D]; (iii) [A]; (iv) [B]
Ans.
(i) [C]
(ii) [D]
(iii) [A], [E]
(iv) [B]
Sol. (i) Do yourself
(ii) Lines 3x + 4y + 2 = 0 and 3x + 4y + 5= 0 are on the same side of the origin. The distance between these lines is d1= 
= 
Lines 3x + 4y + 2 =0 and 3x + 4y –5 = 0 are on the opposite sides of the origin. The distance between these
lines is d2 = 
= 
Thus, 3x + 4y +2 = 0 divides the distance between
3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 in the ratio d1: d2
i.e. 3 : 7
(iii) Let the coordinates of the third vertex be (2a,t).
AC = BC ⇒ t = 
⇒ t = 
So, the coordinates of the third vertex C are 
So, Area of triangle = ± 
= ± 
(iv) The coordinates of A and B are (0,12) and (8,0) respectively. The equation of the perpendicular bisectors of AB is
y – 6 = 
(x – 4) or 2x – 3y + 10 = 0 ...(1)
Equation of a line passing through (0, –1) and parallel to x- axis is y = –1. This meets (1) at C, therefore the coordinates of C are (–13/2, –1). Hence, the area of the triangle ABC is
Δ = 
= 91 sq. units.
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