## Question

### Solution

Correct option is

x – 2y + 8 = 0

The family of the lines

(x + y – 1) + λ (2x + 3y – 5) = 0

Passes through intersection of

x + y – 1 = 0                               …(i)

2x + 3y – 5 = 0                             … (ii)

Solving (i) and (ii), we get (x1y1) ≡ (–2, 3)

Family of the line

(3x + 2y – 4) + µ(x + 2y – 6) = 0

Solving (iii) and (iv), we get: Equation of line belonging to both the families will pass through (x1y1) and (x2y2) ⇒      x – 2y + 8 = 0 belongs to both the families.

#### SIMILAR QUESTIONS

Q1

Determine the distance between the lines : 6x + 8y – 45 = 0 and 3x + 4y – 5 = 0.

Q2

The algebraic sum of the perpendicular distances from A(x1y1), B(x2,y2) and C(x3y3) to a variable line is zero, then the line passes through:

Q3

If A (cos α, sin α), B (sin α, –cos α), C (1, 2) are the vertices of a ΔABC, then as α varies the locus of its centroid is:

Q4

The image of point (1, 3) in the line x + y – 6 = 0 is:

Q5

If t1t2t3 are distinct, then the points are collinear if:

Q6

A and B are two fixed points. The vertex C of a Δ ABC moves such that cot A + cot B = constant. Locus of C is a straight line:

Q7

The number of integer values of m, for which the x-co-ordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer, is:

Q8

The vertices of a triangles are A (x1x1 tan α), B (x2x2 tan β) and C (x3,x3 tan γ). If the circumcentre of Δ ABC coincides with the origin and H (ab) be its orthocenter, then a/b is equal to:

Q9

A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinates axes at points P and Q. As L varies, the absolute minimum values of OP + OQ is (O is origin)

Q10

If p1p2p3 be the length perpendicular from the points

(m2, 2m), (mm’m + m’) and (m’2, 2m’)

respectively on the line 