You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Slope Form a Line

If 3-z1/2-z...

Question

If $(\frac{3 - z_{1}}{2 - z_{1}}) (\frac{2 - z_{2}}{3 - z_{2}}) = k (k > 0)$ .
Then prove that the points $A (z_{1}), B (z_{2}), C (3)$ and $D (2)$ (taken in clockwise sense) are concyclic.

Open in App

Solution

$(\frac{3 - z_{1}}{2 - z_{1}}) (\frac{2 - z_{2}}{3 - z_{2}}) = k$
$∴ a r g (\frac{3 - z_{1}}{2 - z_{1}} \cdot \frac{2 - z_{2}}{3 - z_{2}}) = a r g (k)$
Now if $k > 0$ , $a r g (k) = 0$
$∴ a r g (\frac{3 - z_{1}}{2 - z_{1}}) + a r g (\frac{2 - z_{2}}{3 - z_{2}}) = 0$
or $a r g (\frac{3 - z_{1}}{2 - z_{1}}) = a r g (\frac{3 - z_{2}}{2 - z_{2}})$
Hence, chord DC subtends same angle at A and B.
Therefore, point A, B, C, D are concyclic.
Ans: 1

Suggest Corrections

Similar questions

If $⎧ ⎩ \frac{3 - z_{1}}{2 - z_{1}} ⎫ ⎭ ⎧ ⎩ \frac{2 - z_{2}}{3 - z_{2}} ⎫ ⎭$ = k, then points $A (z_{1})$ , $B (z_{2})$ , $C (3, 0)$ and $D (2, 0)$ (taken in clockwise sense) will

Z_{1} \neq Z_{2}

are two points in an argand plane. If

a | Z_{1} | = b | Z_{1} |

Prove that

\frac{a Z_{1} - b Z_{2}}{a Z_{1} + b Z_{2}}

is purely imaginary.

Q. Let

z_{1}, z_{2}

be any two complex numbers and a, b two real numbers such that

a^{2} + b^{2} \neq 0

, then

| z_{1} |^{2} + | z_{2} |^{2} - | z_{1}^{2} + z_{2}^{2} | \leq \frac{2 | a z_{1} + b z_{2} |^{2}}{a^{2} + b^{2}} \leq | z_{1} |^{2} + | z_{2} |^{2} + | z_{1}^{2} + z_{2}^{2} |

.
If this is true enter 1, else enter 0.

| z_{1} |

and

| z_{2} |

are two distinct points in an Argand plane. If

a | z_{1} | = b | z_{2} |

(where

a, b \in R

), then the point

(a z_{1} / b z_{2}) + (b z_{2} / a z_{1})

is a point on the

Q. Consider an elipse having its foci at

A (z_{1}) and B (z_{2})

in the argand plane. If the eccentricity of the ellipse is

e

and it is known that origin is an interior point of the ellipse, then

e

lies in the interval

Join BYJU'S Learning Program

If (3−z12−z1)(2−z23−z2)=k(k>0).Then prove that the points A(z1),B(z2),C(3) and D(2) (taken in clockwise sense) are concyclic.

(3−z12−z1)(2−z23−z2)=k∴arg(3−z12−z1⋅2−z23−z2)=arg(k)Now if k>0, arg(k)=0∴arg(3−z12−z1)+arg(2−z23−z2)=0 or arg(3−z12−z1)=arg(3−z22−z2)Hence, chord DC subtends same angle at A and B. Therefore, point A, B, C, D are concyclic.Ans: 1

If $(\frac{3 - z_{1}}{2 - z_{1}}) (\frac{2 - z_{2}}{3 - z_{2}}) = k (k > 0)$ .
Then prove that the points $A (z_{1}), B (z_{2}), C (3)$ and $D (2)$ (taken in clockwise sense) are concyclic.