The image of point (1, 3) in the line x + y – 6 = 0 is:
To find the image of a point in a line, we use following conditions:
(i) l1 is perpendicular to l2 (ii) Mid point of AB lies on l1
solving (i) and (ii) we get: (x1, y1) ≡ (3, 5)
Given the triangle with vertices A (–4, 9), B (10, 4), C (–2, –1). Find the equation of the altitude through A.
Find the equation of perpendicular bisector of the line joining the points (1, 1) and (2, 3).
Find the coordinates of the foot of the perpendicular from the point (2, 3) on the line y = 3x + 4.
Find the equation of the line passing through (a, b) and parallel to px +qy + 1 = 0.
Find the equation of the line perpendicular to 3x + 4y + 1= 0 and passing through (1, 1).
Find the distance of line h (x + h) + k (y + k) = 0 from the origin.
Determine the distance between the lines : 6x + 8y – 45 = 0 and 3x + 4y – 5 = 0.
The algebraic sum of the perpendicular distances from A(x1, y1), B(x2,y2) and C(x3, y3) to a variable line is zero, then the line passes through:
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:
If t1, t2, t3 are distinct, then the points are collinear if: