The value of k so that x2 + y2 + kx + 4y + 2 = 0 and 2(x2 + y) - 4x - 3y + k = 0 cut orthogonally is
The two lines x = my + n, z = py + q and x = m'y + n', z =p'y + q' are perpendicular to each other, if
mm' + pp' = 1
mm' + pp' = - 1
The angle between lines joining the origin to the point of intersection of the line x + y = 2 and the curve y2 - x2 = 4 is
The line joining two points A(2, 0), B(3, 1) is rotated about A in anti-clockwise direction through an angle of 15°. The equation of the line in the now position, is
√3x - y - 2√3 = 0
x - 3√y - 2 = 0
√3x + y - 2√3 = 0
x - √3y - 2 = 0
A.
√3x - y - 2√3 = 0
Here, slope of AB = 1/1
⇒ tan(0) = m1 = 1 or θ = 45°
Thus, slope of new line is tan(45° + 15°)
= tan(60°) = √3
[∵ it is rotated anti-clockwise, so the angle will be 45° + 15° = 60°]
Hence, the equation is y = √3x + c, but it still passes through (2, 0), hence c = - 2√3.
Thus, required equation is
y = √3x - 2√3
A straight line through the point A (3, 4) is such that its intercept between the axes is bisected at A. Its equation is
3x - 4y + 7 = 0
4x + 3y = 24
3x + 4y = 25
x + y = 7