Problem 13
Question
There are \(n\) points in a plane, in which no three are in a straight line except ' \(m\) ' which are all in a straight line. Find the number of (a) different straight lines, (b) different triangles, (c) different quadrilaterals that can be formed with the given points as vertices.
Step-by-Step Solution
Verified Answer
(a) \(\binom{n}{2} - \binom{m}{2} + 1\), (b) \(\binom{n}{3} - \binom{m}{3}\), (c) \(\binom{n}{4} - \binom{m}{4}\).
1Step 1: Calculation of Total Possible Lines
To calculate the total number of possible lines that can be formed with \(n\) points, we use the combination formula for selecting 2 points out of \(n\): \(\binom{n}{2}\). This gives the total number of two-point combinations which determine lines.
2Step 2: Excluding Collinear Points in Line Calculation
Since \(m\) points are collinear (meaning they are on the same straight line), they account for only one line despite being chosen in different pairs. Thus, we must subtract the extra lines counted which are only internal to this group. Therefore, the number of overlapping lines is \(\binom{m}{2}\).
3Step 3: Total Number of Different Lines
The number of different lines is the difference between the total possible lines and the overlapping lines formed by the collinear points. Hence, the number of different lines = \(\binom{n}{2} - \binom{m}{2} + 1\).
4Step 4: Total Possible Triangles
To find the number of triangles, we need to select any 3 points from \(n\). This can be calculated using the combination \(\binom{n}{3}\).
5Step 5: Excluding Degenerate Triangles with Collinear Points
Any triangle using all 3 of the collinear \(m\) points would be degenerate (they lie on a straight line). Therefore, we must exclude these cases: \(\binom{m}{3}\).
6Step 6: Total Number of Different Triangles
The number of different triangles that can be formed is therefore given by \(\binom{n}{3} - \binom{m}{3}\).
7Step 7: Total Possible Quadrilaterals
To find the number of quadrilaterals, we need to select any 4 points from \(n\). This can be calculated using the combination \(\binom{n}{4}\).
8Step 8: Excluding Quadrilaterals with Collinear Points
Any quadrilateral that includes 4 of the collinear \(m\) points will itself be degenerate. Therefore, we must exclude these cases: \(\binom{m}{4}\).
9Step 9: Total Number of Different Quadrilaterals
The number of different quadrilaterals that can be formed is given by \(\binom{n}{4} - \binom{m}{4}\).
Key Concepts
Collinear PointsCombinationsPlane Geometry
Collinear Points
In geometry, collinear points refer to points that lie on the same straight line. Understanding collinearity is crucial when dealing with problems in plane geometry that involve determining possible lines or shapes, like triangles and quadrilaterals.
There are a few key characteristics of collinear points:
There are a few key characteristics of collinear points:
- If you can draw one straight line through two or more points, they are collinear.
- For some calculations, collinear points simplify the situation because they limit the number of forms or arrangements you must consider. For example, if three or more points are collinear, they cannot form a triangle.
- Identifying collinear points is critical when subtracting possible degenerate (not actually forming the intended shape) forms from total possible combinations.
Combinations
Combinations are a fundamental concept in combinatorics, a branch of mathematics focused on counting, arranging, and analyzing possible configurations within a given set. They are used to determine how many ways you can choose a group of items from a larger collection, without regard to order.
In the formula for combinations, denoted as \(\binom{n}{k}\), the 'n' represents the total number of elements to choose from, and 'k' represents the number of elements to choose. The formula is given by:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]This calculation allows one to compute:
In the formula for combinations, denoted as \(\binom{n}{k}\), the 'n' represents the total number of elements to choose from, and 'k' represents the number of elements to choose. The formula is given by:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]This calculation allows one to compute:
- Lines that can be drawn by selecting any two out of 'n' points.
- Triangles by choosing any three points from 'n'.
- Quadrilaterals by selecting any four points from 'n'.
Plane Geometry
Plane geometry is the study of shapes and figures in a flat, two-dimensional world. It mostly involves studying points, lines, angles, polygons (like triangles and quadrilaterals), and circles. Understanding how these geometric figures interact is crucial when solving problems related to them.
Some important features of plane geometry include:
Some important features of plane geometry include:
- All points lie in a single plane, meaning any two points can always be connected by a straight line.
- It is possible to create various shapes by connecting points through lines.
- Analyzing properties of shapes, such as angles and side lengths, is typical for figuring out relationships and measurements.
Other exercises in this chapter
Problem 12
Total number of six-digit numbers that can be formed, having the property that every succeeding digit is greater than the preceding digit, is equal to a. \({ }^
View solution Problem 12
Number of points of intersection of \(n\) straight lines if \(n\) satisfies \({ }^{n+5} P_{n+1}=\frac{11(n-1)}{2} \times{ }^{n+3} P_{n}\) is a. 15 b. 28 c. 21 d
View solution Problem 13
Numbers greater than 1000 but not greater than 4000 , which can be formed with the digits \(0,1,2,3,4\) (repetition of digits is allowed), are a. 350 b. 375 c.
View solution Problem 13
Number of shortest ways in which we can reach from the point \((0,0,0)\) to point \((3,7,11)\) in space where the movement is possible only along the \(x\)-axis
View solution