Top answers

The line L passes through the points (-2,3) and (6,9). How do I find the equation of the line that is parallel to L and passes through the point (5,-1)?

Find the gradient of line L. The gradient can be defined as the "change in y" divided by the "change in x". For this situation, the "change in y" would be 9 - 3 = 6. The "c...
JC
Answered by Justin C. • Maths tutor
15237 Views

Prove that an angle subtended by an arc is double at the centre then at the perimeter.

After drawing, bisect the perimeter angle in two from the centre. We now have two isosceles triangles with the sides the same length (1 radius from the centre). We know that 2 angles in each triangle add up ...
MP
Answered by Matthew P. • Maths tutor
7889 Views

The complex conjugate of 2-3i is also a root of z^3+pz^2+qz-13p=0. Find a quadratic factor of z^3+pz^2+qz-13p=0 with real coefficients and thus find the real root of the equation.

z-2+3i times z-2-3i = z 2 -4z+13. z 3 -2z 2 +5z+26 divided by z 2 -4z+13 = z+2. Therefore the real root is z=-2.
WN
Answered by William N. • Maths tutor
6264 Views

Given that 2-3i is a root to the equation z^3+pz^2+qz-13p=0, show that p=-2 and q=5.

Substitute 2-3i into equation using part i (2-3i) 3 =-46-9i. -46-9i+p(-5-12i)+q(2-3i)-13p=0. -46-18p+2q-9i-12pi-3iq=0. Real: -46-18p+2q=0 and Imaginary: -9-12p-3q=0. p=-2, q=5
WN
Answered by William N. • Maths tutor
11569 Views

Show (2-3i)^3 can be expressed in the form a+bi where a and b are negative integers.

(2-3i) x (2-3i) = -5-12i. -5-12i x (2-3i) = -46-9i. a=-46, b=-9
WN
Answered by William N. • Maths tutor
4551 Views