Distance between two points and perpendicular bisectors are the two important terms in coordinate geometry as well as in complex numbers. It helps in finding the distance between two complex numbers. Measurement of distance is a very important aspect of our day-to-day life.This formula comes from the knowledge of basic distance formula and it's application..
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In this article, we will cover the concept of the distance between two points and perpendicular bisectors. This concept falls under the broader category of complex numbers and quadratic equations, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.
Distance between two points
Let
Then,
The distance of a point from the origin is
Three points
i.e.
We can use the distance formula to find the equation of perpendicular bisector
Let two fixed points
As any point on the perpendicular bisector of
This is the equation of perpendicular of the bisector of
The equation of the circle whose center is at the point
If the center is the origin then,
Interior of the circle is represented by
The exterior is represented by
Here z can be represented as
This equation also represents a circle. This can be verified by putting
|This represents an ellipse as the sum of distances of point
This represents a hyperbola as the difference of distances of point
The complex number
And
The complex number z dividing
Centroid of the triangle with vertices
Example 1: Distance (in units) between
Solution:
As we learned in
Distance between Z 1 and Z 2 -
- wherein
Distance between
Here,
Hence, the answer is 5.
Example 2: If
Solution:
As we learned in
Perpendicular bisector -
Locus of point equidistant from two given points.
z will lie on the perpendicular bisector of line joining
- wherein
Hence, the answer is
Example 3: If
1) Real axis
2) Imaginary axis
3) Argand plane
4) Circle
Solution:
As we learned in
Perpendicular bisector -
Locus of point equidistant from two given points.
z will lie on the perpendicular bisector of line joining
- wherein
i.e real axis.
Hence, the answer is the option 1.
Example 4: If the set
Solution:
Let
Letz
Hence, the answer is 30 .
Example 5: Let
1) hyperbola with the length of the transverse axis 7
2) hyperbola with eccentricity 2
3) a straight line with the sum of its intercepts on the coordinate axes equals -18
4) A straight line with the sum of its intercepts on the coordinate axes equals 14
Solution
a straight line with the sum of intercept on
Hence, the answer is the option 4.
Complex numbers are the numbers in which complex or imaginary parts exist. It is represented as a+ib.
击
Let
Then,
where
This is the equation of perpendicular to the bisector of
The equation of the circle whose center is at the point
Centroid of the triangle with vertices
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