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
Distance between two points
Let
Then,
The distance of a point from the origin is |z - 0| = |z|
Three points A(z1), B(z2) and C(z3) are collinear, then AB + BC = AC
i.e.
Perpendicular bisector
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
Equation of Circle
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
Equation of Circle in second form
This equation also represents a circle. This can be verified by putting
Equation of Ellipse
$
\left|z-z_1\right|+\left|z-z_2\right|=k \quad\left(k>\left|z_1-z_2\right|\right)
$$
|This represents an ellipse as the sum of distances of point
Equation of Hyperbola
$
This represents a hyperbola as the difference of distances of point
Section Formula
The complex number
And
The complex number z dividing
Centroid of Triangle
Centroid of the triangle with vertices z1, z2 and z3 is given by (z1 + z2+ z3)/3
Solved Examples Based On the Distance between two points and the Perpendicular bisectors:
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.
Summary
Distance between two points helps to find the distance by using a mathematical formula that plays an important role in measuring distance. The perpendicular bisector equation involves either finding the line with the appropriate slope passing through the midpoint or ensuring any point on it is equidistant from the endpoints of the segment.
Complex numbers are the numbers in which complex or imaginary parts exist. It is represented as a+ib.
击
Let $z_1=x+$ iy and $z_2=x+i y$
Then, $\left|\mathrm{z}_1-\mathrm{z}_2\right|=\left|\left(\mathrm{x}_1-\mathrm{x}_2\right)+\mathrm{i}\left(\mathrm{y}_1-\mathrm{y}_2\right)\right|$ $=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}$
$=$ distance between points $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and
$\left(\mathrm{x}_2, \mathrm{y}_2\right)=$ distance between between $\mathrm{z}_1$ and $\mathrm{z}_2$
where $\mathrm{z}_1=\mathrm{x}_1+\mathrm{i} \mathrm{y}_1$ and $\mathrm{z}_2=\mathrm{x}_2+\mathrm{iy}_2$
$\left|z-z_1\right|=\left|z-z_2\right|$
This is the equation of perpendicular to the bisector of $A B$, where $A\left(z_1\right)$ and $B\left(z_2\right)$.
The equation of the circle whose center is at the point $z_0$ and has a radius of $r$ is given by
$$
\left|z-z_0\right|=r
$$
Centroid of the triangle with vertices $z_1, z_2$, and $z_3$ is given by $\left(z_1+z_2+z_3\right) / 3$.
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