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A straight line has various real-world applications, such as in architecture, engineering, art, navigation, and even in video games. In the Class 11 maths exercise 9.1, you will learn about the basic coordinate geometry, like measuring the slope of the line, conditions for parallelism and perpendicularity of lines, angle between two lines, colinearity of three points, etc. Understanding straight lines helps you analyze motion and direction, interpret graphs and data visually, and design and build real-world structures. Also, check NCERT to learn more about a straight line.
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Our experienced subject experts designed the NCERT solutions straight line exercise 9.1 to offer a systematic and structured approach to these important concepts and help students to develop a clear understanding of critical concepts by a series of solved questions and conceptual explanations. These NCERT Solutions follow the CBSE pattern and also help in gaining knowledge about all the natural processes happening around them. These solutions provide a valuable resource to the students to enhance their performance in board exams.
Question 1: Draw a quadrilateral in the Cartesian plane, whose vertices are
Answer:
Area of ABCD = Area of ABC + Area of ACD
Now, we know that the area of a triangle with vertices
Therefore,
Area of triangle ABC
Similarly,
Area of triangle ACD
Now,
Area of ABCD = Area of ABC + Area of ACD
Answer:
it is given that it is an equilateral triangle and length of all sides is 2a
The base of the triangle lies on y-axis such origin is the midpoint
Therefore,
Coordinates of point A and B are
Now,
Apply Pythagoras theorem in triangle AOC
Therefore, coordinates of vertices of the triangle are
Question 3:(i) Find the distance between
PQ is parallel to the
Answer:
When PQ is parallel to the y-axis
then, x coordinates are equal i.e.
Now, we know that the distance between two points is given by
Now, in this case
Therefore,
Therefore, the distance between
Question 3:(ii) Find the distance between
PQ is parallel to the
Answer:
When PQ is parallel to the x-axis
then, x coordinates are equal i.e.
Now, we know that the distance between two points is given by
Now, in this case
Therefore,
Therefore, the distance between
Question 4: Find a point on the x-axis, which is equidistant from the points
Answer:
Point is on the x-axis, therefore, y coordinate is 0
Let's assume the point is (x, 0)
Now, it is given that the given point (x, 0) is equidistance from point (7, 6) and (3, 4)
We know that
Distance between two points is given by
Now,
and
Now, according to the given condition
Squaring both the sides
Therefore, the point is
Answer:
Mid-point of the line joining the points
It is given that line also passes through origin which means passes through the point (0, 0)
Now, we have two points on the line so we can now find the slope of a line by using formula
Therefore, the slope of the line is
Question 6: Without using the Pythagoras theorem, show that the points
Answer:
It is given that point A(4,4) , B(3,5) and C(-1,-1) are the vertices of a right-angled triangle
Now,
We know that the distance between two points is given by
Length of AB
Length of BC
Length of AC
Now, we know that Pythagoras theorem is
Is clear that
Hence proved
Answer:
It is given that the line makes an angle of
Now, we know that
line makes an angle of
Therefore, the angle made by line with the positive x-axis is =
Now,
Therefore, the slope of the line is
Question 8: Without using the distance formula, show that points
Answer:
Given points are
We know the pair of the opposite side are parallel to each other in a parallelogram
Which means their slopes are also equal
The slope of AB =
The slope of BC =
The slope of CD =
The slope of AD
=
We can clearly see that
The slope of AB = Slope of CD (which means they are parallel)
and
The slope of BC = Slope of AD (which means they are parallel)
Hence pair of opposite sides are parallel to each other
Therefore, we can say that points
Question 9: Find the angle between the x-axis and the line joining the points
Answer:
We know that
So, we need to find the slope of line joining points (3,-1) and (4,-2)
Now,
Therefore, angle made by line with positive x-axis when measure in anti-clockwise direction is
Answer:
Let
Then, we know that
It is given that
Now,
Now,
=
Now,
According to which value of
Therefore,
Question 11: A line passes through
Answer:
Given that A line passes through
Now,
Hence proved
Also read
1) Slope of a Line:
If θ is the inclination of a line L, then tan θ is called the slope of the line L. The slope of a line whose inclination is 90° is not defined. It may be observed that the slope of the x-axis is zero, and the slope of the y-axis is not defined. The slope of a line is denoted by m.
2) Slope of a line when the coordinates of any two points on the line are given:
Let P(x1, y1) and Q(x2, y2), then the slope of the line which is passing through these points will be m:
3)Conditions for parallelism and perpendicularity of lines in terms of their slope:
If the line L1 ( slope m1) is parallel to the line L2 (slope m2), then their slopes will always be equal; therefore, m1 = m2.
The lines L1 ( slope m1) and the line L2 (slope m2) are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
4)Angle between two lines :
Let L1 and L2 be two non-vertical lines with slopes m1 and m2, and theta is the angle between them, then theta will be
where
Also Read
Students can refer subject wise NCERT solutions. The links to solutions are given below
Students can access the NCERT exemplar solutions to enhance their deep understanding of the topic. These solutions are aligned with the CBSE syllabus and also help in competitive exams.
The angle made by line l with the positive direction of x-axis and measured anticlockwise is called the inclination of the line 'l'.
The tan θ is called the slope or gradient of line 'l' where θ is the inclination of line 'l'.
Slope of line = tan 45 = 1
Condition for collinearity of points A, B, and C.
Slope of AB = Slope of BC
The slope of the x-axis is zero.
The slope of the y-axis is not defined.
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