ML Aggarwal Solutions Class 6 Mathematics Solutions for Basic Geometrical Concept Exercise 10 in Chapter 10 - Basic Geometrical Concept
Question 2 Basic Geometrical Concept Exercise 10
How many lines can be drawn through three collinear points?
Answer:
A set of three or more points on the same straight line is known as a collinear point. Collinear points cannot exist on different lines but they can exist on different planes. Collinearity is the property of points being collinear. The Slope formula can be applied to identify if the points are collinear or not.
Only one line can be drawn through three collinear points.
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There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points.
Number of straight lines formed joining the 10 points, taking 2 points at a time =
\[{}^{10} C_2 = \frac{10}{2} \times \frac{9}{1} = 45\]
Number of straight lines formed joining the 4 points, taking 2 points at a time =\[{}^4 C_2 = \frac{4}{2} \times \frac{3}{1} = 6\]
But, when 4 collinear points are joined pair wise, they give only one line.
∴ Required number of straight lines =\[45 - 6 + 1 = 40\]
Concept: Combination
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Solution : There are ten points `P_[1], P_[2],.., P_[10]`.
For one line two points are required.
Through point `P_[1]` there will be 9 lines when `P_[1]` is joined with any of the nine other points.
Similarly, there will be nine lines passing through each point.
So, number of lines is `9xx10` or 90.
But there is double counting in above answer. Why ?
One of the nine lines passing through point
`P_[1] " is" P_[1] P_[2]. "But" P_[1] P_[2]` is also one of the lines passing through point `P_[2]`.
Thus, line `P_[1]P_[2]` and similarly each line is counted twice. Therefore, actual number of lines is `[90]/[2]=45`.
There are 10 points in a place, no three of these points are in a straight line. What is the total number of straight lines which can be formed by joining the points?
This question was previously asked in
NDA [Held On: 21 Apr 2019] Maths Previous Year paper
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- 90
- 45
- 40
- 30
Answer [Detailed Solution Below]
Option 2 : 45
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Electric charges and coulomb's law [Basic]
10 Questions 10 Marks 10 Mins
Concept
If there are n points in a plane and no three points lie on a straight line, then the total number of straight line segments that can be formed is given by: \[C\left[ {n,\;2} \right] = \frac{{{\rm{n}}!}}{{2!\left[ {{\rm{n\;}} - {\rm{\;}}2} \right]!}}\]
Calculation
Given: n = 10.
\[\Rightarrow \;C\left[ {10,\;2} \right] = \frac{{10!}}{{2!\left[ {10 - {\rm{\;}}2} \right]!}} = \frac{{10 \times 9}}{2} = 45\]
Hence, the number of straight lines that can be formed is 45.
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Solution
The correct option is D
25
Explanation for the correct option:
We have been given that there are 10 points of which 7 are collinear.
We know that to make a line we need at least 2 points. So, the number of ways to select 2 points from 10 points to make straight lines is C210 .
As it is given that 7 points are collinear which means that we can’t make lines for all the 7 points. So, the number of ways to select 2 points from these 7 points which will not give any distinct line is C27
So, we will subtract7C2 from C 210. But 7 collinear points are on a straight line. Therefore we have to add +1 in C2 10
∴Total number of lines =C210-C2 7+1
=10!2!8!-7!2!5 !+1∵Crn=n!r !n-r!=10×9×8!2×1×8!-7× 6×5!2×1×5!+1=45-21+1=46-21 =25
Thus there are 25 straight lines that can be drawn out of 10 points of which 7 are collinear.
Hence, option D is the correct answer.