The vertices of a triangles are A (x1, x1 tan α), B (x2, x2 tan β) and C (x3,x3 tan γ). If the circumcentre of Δ ABC coincides with the origin and H (a, b) be its orthocenter, then a/b is equal to:
Co-ordinate of orthocenter ≡ (a, b).
Circum radius of triangle = OA = R
So, Co-ordinate of vertices are A (R cos α, R sin α),
B (R cos β, R sin β) and C (R cos γ, R sin γ).
Hence, co-ordinate of cetroid G is:
As we know circumcentre, orthocenter and centroid of a triangle are collinear.
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:
The image of point (1, 3) in the line x + y – 6 = 0 is:
If t1, t2, t3 are distinct, then the points are collinear if:
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:
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:
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)