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The study of geometric figures using coordinate axes is known as coordinate geometry. Polygons, ellipses, straight lines, hyperbolas, curves, and circles can all be readily drawn and scaled in the coordinate axes. Moreover, using the coordinate system to work algebraically and examine the attributes of geometric figures is referred to as coordinate geometry.
Let’s discuss the steps to find the image of a point in a plane example.
Coordinates of a Point
A coordinate is a numerical value that aids in locating a point in space. The coordinates of a point in a two-dimensional space are (x, y). Let’s look at the terms associated with the steps to find an image of a point in a plane.
- Abscissa: It is the x value in the point (x, y) and the distance from the origin along the x-axis.
- Ordinate: It is the perpendicular distance of the point from the x-axis, which is parallel to the y-axis, and it is the ‘y’ value in the point (x, y).
- The dimensions of a point are used to calculate distance, determine the midpoint, calculate the slope of a line, and calculate the equation of a line, among other things.
The aim here is to find the image of a point P in the 2-D plane created by the mirror given a point P in the 2-D plane and the mirror equation.
Equation of mirror is in form of ax + by + c = 0
Let’s find out the solution.
Quest : P is equal to (1, 0), a is equal to -1, b is equal to 1, c is equal to zero.Find the mirror of P.
Ans : Q is equal to (0, 1)
Quest : P is equal to (3, 3), a is equal to 0, b is equal to 1, c is equal to -2.Find the mirror of P.
Ans: Q is equal to (3, 1)
R must be the midway point of P and Q since the item and image are equidistant from the mirror.
Since the mirror equation is given as ax + by + c = 0, the line going through P and Q has an equation perpendicular to the mirror. As a result, the equation for the line crossing between P and Q is ay – bx + d = 0. P also crosses through a line that passes through P and Q. Thus, we use P’s coordinate in the equation above.
a*y1 – b*x1 + d = 0
d = b*x1 – a*y1
R is also the intersection of the mirror and the line that passes through P and Q. As a result, we discover a solution.
ax + by + c = 0
ay -bx + d = 0
We can find x and y here because a, b, c, and d are all known. Also, R’s coordinates, namely x3 and y3, are now known.
A general formula of image of (x1,y1) about line ax+by+c=0 is
x-x1a=y-y1b=ax1+by1+ca2+b2
Let’s try out with some examples:
Example 1:
Determine the image of point (3,8) in relation to the line x+3y = 7, assuming that the line is a flat mirror.
Solution: Let the point be (h,k). So, h-3/1 = k-8/3 = -2[3 + 3(8) – 7]/1 + 9
We arrived at this result by solving these two equations.
h = -1 and k = -4
It only took about 15 seconds.
This method can assist you in quickly locating the image of a point to save time.
Note: This approach also works with 3D planes.
Example 2:
A line perpendicular to 5x = y + 7 is drawn. If the area of the triangle created by this line with coordinate axes is 10 sq. units, find the line’s equation.
5x – y – 7 is equal to 0
The line which is perpendicular to the above line is x + 5y + k is equal to 0
By applying A (x, 0), we get the value of x in terms of k.
x + 5(0) + k is equal to 0
x + k is equal to 0
x is equal to -k
By applying B (0, y), we get the value of y in terms of k.
1(0) + 5y + k = 0
5y + k = 0
y = -k/5
Here, x is equal to base of triangle, y = height of triangle
It could be said that the area of triangle is equal to 10 square units
(1/2) ⋅ Base ⋅ Height is equal to 10
(1/2) ⋅ (-k) ⋅ (-k/5) is equal to 10
k2 = 100 ⇒ k is equal to ± 10
x + 5y is equal to ± 10
Conclusion
Coordinate geometry aids in the definition of points in space. The x-axis and primary y-axis axes are defined first, and then the points are measured and labelled with relation to these points. In addition, numerous geometric forms such as a line, curve, circle, ellipse, and hyperbola can be displayed in the steps to find an image of a point in a plane.
