## Question

### Solution

Correct option is

t1 + t2 + t3 = 0   Points ABC are collinear. It means area formed by the triangle ABC = 0    ⇒     t1 + t2 + t3 = 0     [âˆµ t1t2 and t3 distinct real number]

#### SIMILAR QUESTIONS

Q1

Find the equation of perpendicular bisector of the line joining the points (1, 1) and (2, 3).

Q2

Find the coordinates of the foot of the perpendicular from the point (2, 3) on the line y = 3x + 4.

Q3

Find the equation of the line passing through (ab) and parallel to px +qy + 1 = 0.

Q4

Find the equation of the line perpendicular to 3x + 4y + 1= 0 and passing through (1, 1).

Q5

Find the distance of line h (x + h) + k (y + k) = 0 from the origin.

Q6

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

Q7

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:

Q8

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:

Q9

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

Q10

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: