Locus of its orthocentre is (x+y-7)^2 + (x - y - 1)² = 100 .The correct choice is option B.
Let H be the orthocenter of triangle ABC. Then, the equations of the altitudes of triangle ABC can be found using the slope-intercept form:
Altitude from A to BC:
Slope = -(5 sin θ - 3)/(5 cos θ - 5) = -sin θ + 1/cos θ
y - 4 = (-sin θ + 1/cos θ)(x - 3)
Altitude from B to AC:
Slope = (5 sin θ + 4)/(5 cos θ - 3) = sin θ + 4/cos θ
y - 5 sin θ = (sin θ + 4/cos θ)(x - 5 cos θ)
Altitude from C to AB:
Slope = -(5 cos θ + 4)/(5 sin θ - 3) = -cos θ - 4/sin θ
y + 5 cos θ = (-cos θ - 4/sin θ)(x - 5 sin θ)
To find the point of intersection of these three altitudes, we can solve the system of equations:
y - 4 = (-sin θ + 1/cos θ)(x - 3)
y - 5 sin θ = (sin θ + 4/cos θ)(x - 5 cos θ)
y + 5 cos θ = (-cos θ - 4/sin θ)(x - 5 sin θ)
Solving this system of equations, we get the coordinates of the orthocenter H:
x = (3 + 5 sin θ + 5 cos θ)/3
y = (4 + 5 sin θ - 5 cos θ)/3
Substituting these values of x and y into the equation of the circle:
(x + y - 7)^2 + (x - y - 1)^2 = 100
We can simplify this equation to:
(2x - 6)^2 + (2y - 8)^2 = 100
This is the equation of a circle with center (3, 4) and radius 10. Therefore, the locus of the orthocenter of triangle ABC is a circle with center (3, 4) and radius 10.
So the answer is option B (x+y-7)^2 + (x - y - 1)^2 = 100.
Question:-
Vertices of a variable triangle are (3, 4), (5 cos 0, 5 sin 0) and (5 sin 0, -5 cos 0), where 0 ∈ R. Locus of its orthocentre is
A (x+y-1)^2 + (x - y - 7)² = 100
B (x+y-7)^2 + (x - y - 1)² = 100
C (x+y-7)^2 + (x + y - 1)² = 100
D (x+y-7)^2 + (x - y + 1)² = 100