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Properties of Modulus

Let x,y,z ∈ C...

Question

Let $x, y, z \in C$ satisfy $| x | = 1, | y - 6 - 8 i | = 3$ and $| z + 1 - 7 i | = 5$ respectively, then the minimum value of $| x - z | + | y - z |$ is equal to

A

1

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B

2

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C

5

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D

6

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Solution

The correct option is D $6$
$| z_{1} + z_{2} | \leq | z_{1} | + | z_{2} | \forall z_{1}, z_{2} \in C$

$| x - z | + | y - z | \geq | (x - z) + (z - y) | = | x - y |$

Equality holds when $x, y, z$ are collinear points. (The points are collinear on the line $3 y = 4 x$ in the Argand plane)

$| x - y |_{m i n} = C_{1} C_{2} - (r_{1} + r_{2}) = 10 - (1 + 3) = 6$

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Similar questions

x, y, z \in C

| x | = 1, | y - 6 - 8 i | = 3

| z + 1 - 7 i | = 5

respectively, then the minimum value of

| x - z | + | y - z |

Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} y^{5} z^{6} (z^{3} - y^{3}) & x^{4} z^{6} (x^{3} - z^{3}) & x^{4} y^{5} (y^{3} - x^{3}) y^{2} z^{3} (y^{6} - z^{6}) & x z^{3} (z^{6} - x^{6}) & x y^{2} (x^{6} - y^{6}) y^{2} z^{3} (z^{3} - y^{3}) & x z^{3} (x^{3} - z^{3}) & x y^{2} (y^{3} - x^{3}) \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y^{2} & z^{3} x^{4} & y^{5} & z^{6} x^{7} & y^{8} & z^{9} \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ_{1}

(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that,

\frac{y - z}{a (b - c)} = \frac{z - x}{b (c - a)} = \frac{x - y}{c (a - b)}

\frac{x}{3 x - y - z} = \frac{y}{3 y - z - x} = \frac{z}{3 z - x - y}

x + y + z \neq 0

then show that the value of each ratio is equal to 1.

(iii) If

\frac{a x + b y}{x + y} = \frac{b x + a z}{x + z} = \frac{a y + b z}{y + z}

x + y + z \neq 0

\frac{a + b}{2}

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

\frac{3 x - 5 y}{5 z + 3 y} = \frac{x + 5 z}{y - 5 x} = \frac{y - z}{x - z}

then show that every ratio =

\frac{x}{y}

z = x + i y

a m p (z - 1) = a m p (z + 3)

, then the value of

(x - 1) : y

x + y \propto z + \frac{1}{z}

x - y \propto z - \frac{1}{z}

z = 2

y = 1, x = 3

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