If |8 + z| + |z - 8| = 16,where z is a comple | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation is $|z - (-8)| + |z - 8| = 16$. This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = -8$ and $z_2 = 8$. The distance b

Vedclass · Accepted Answer

$C$ $A$ straight line

Vedclass · Answer

(C) The given equation is $|z - (-8)| + |z - 8| = 16$. This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = -8$ and $z_2 = 8$. The distance between the two fixed points $z_1$ and $z_2$ is $|8 - (-8)| = |16| = 16$. Since the sum of the distances from $z$ to the two fixed points is equal to the distance between the points themselves $(|z - z_1| + |z - z_2| = |z_1 - z_2|)$,the locus of $z$ is the line segment joining the points $-8$ and $8$. However,in the context of standard geometry of complex numbers,if the sum of distances equals the distance between foci,it represents a degenerate ellipse (a line segment). Given the options provided,it is a straight line segment.

If $|8 + z| + |z - 8| = 16$,where $z$ is a complex number,then the point $z$ will lie on

Similar Questions

If $z=x+iy$,where $x, y \in \mathbb{R}$,$(x, y) \neq (0, -4)$ and $\text{Arg}\left(\frac{2z-3}{z+4i}\right)=\frac{\pi}{4}$,then the locus of $z$ is

$P$ is a point denoting $z$ in the Argand diagram. If $\frac{z-i}{z-1}$ is always purely imaginary,then the locus of $P$ is

Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as:
$S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$
$S_{2} = \{ z \in C : \operatorname{Re}((1 - i)z) \geq 1 \}$
$S_{3} = \{ z \in C : \operatorname{Im}(z) \leq 1 \}$
Then the set $S_{1} \cap S_{2} \cap S_{3}$

$\sinh(ix)$ is equal to

If the vertices $A, B$ and $C$ of an isosceles $\triangle ABC$ are respectively $z_1, z_2$ and $z_3$ and if $\angle C=90^{\circ}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module