Get the important questions for class 11 Maths Chapter 10 -Straight Lines here. The important questions from this chapter are taken from the previous year papers and sample paper, which helps the students to achieve good marks in the annual examination. The questions are framed as per the syllabus of the CBSE board, and the questions are majorly asked from the NCERT textbook. It includes objective type questions, short answers, long answers, HOTS and VBA (Value-Based Questions). Practice all the important Maths questions from class 11 chapters here at BYJU’S.
Class 11 Maths Chapter 10 – Straight Lines covers the following important concepts such as:
- General equation of a line
- Various forms of the equation of a line
- The slope of a line
- The distance of a point from a line.
- Important 1 Mark Questions for CBSE Class 11 Maths
- Important 4 Marks Questions for CBSE Class 11 Maths
- Important 6 Marks Questions for CBSE Class 11 Maths
Class 11 Chapter 10 – Straight Lines Important Questions with Solutions
To score the good marks in the final examination, practice the problems provided here, which will help you to solve the problems in the annual examination.
Calculate the slope of a line, that passes through the origin, and the mid-point of the segment joining the points P (0, -4) and B (8, 0).
The coordinates of the mid-point of the line segment joining the points P (0, -4) and B (8, 0) are:[(0+8)/2 , (-4+0)/2] = (4, -2)
It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2,
y2) is given by the formula
m = (y2 -y1) / ( (x2 -x1), where (x2 is not equal to x1)
Therefore, the slope of the line passing through the points (0, 0,) and (4, -2) is
Hence, the required slope of the line is -1/2
Find the equation of the line which is at a perpendicular distance of 5 units from the
origin and the angle made by the perpendicular with the positive x-axis is 30°.
If p is the length of the normal from the origin to a line and ω is the angle made by the
normal with the positive direction of the x-axis
Then, the equation of the line for the given condition is written by
x cos ω + y sin ω = p.
Here, p = 5 units and ω = 30°
Thus, the required equation of the given line is
x cos 30°+ y sin 30° = 5
x(√3/2) + y(½) = 5
√ 3x +y = 10
Thus, the required equation of a line is √ 3x + y = 10
Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3
The equation of the line is given as x – 7y + 5 = 0.
The above equation can be written in the form y = mx+c
Thus, the above equation is written as:
y= (1/7)x + (5/7)
From the above equation, we can say that,
The slope of a line, m = 5/7
The slope of the line perpendicular to the line having a slope of 1/7 is
m = -1/(1/7) = -7
Hence, the equation of a line with slope -7 and intercept 3 is given as:
y = m (x – d)
⇒ y= -7(x-3)
⇒ y=-7x + 21
7x+ y = 21
Hence, the equation of a line which is perpendicular to the line x – 7y + 5 = 0 with x-intercept 3 is 7x+ y = 2.
The perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2). Find the values of m and c.
The given equation of the line is y = mx + c.
From the given condition, the perpendicular from the origin meets the given line at (-1, 2).
Hence, the line joining the points (0, 0) and (-1, 2) is perpendicular to the given line.
The slope of the line joining (0, 0) and (-1, 2) is
= 2/-1 = -2
m (– 2) = -1 (Since the two lines are perpendicular)
Since points (-1, 2) lies on the given line, it satisfies the equation y = mx + c.
Now, substitute the value of m, (x, y) coordinates in the equation:
2 = m(-1) + c
2 = ½(-1) + c
2 = -½ + c
C = 2 + (½)
C = 5/2
Therefore, the value of m and c are ½ and 5/2 respectively.
Find the points on the x-axis whose distance from the line equation (x/3) + (y/4) = 1 is given as 4units.
The equation of a line = (x/3) + (y/4) = 1
It can be written as:
4x + 3y -12 = 0 …(1)
Compare the equation (1) with general line equation Ax + By + C = 0,
we get the values A = 4, B = 3, and C = -12.
Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.
we know that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
D = |Ax1+By1 + C|/ √A2 + B2
Now, substitute the values in the above formula, we get:
4 = |4a+0 + -12|/ √42 + 32
⇒4 = |4a-12|/5
⇒|4a-12| = 20
⇒± (4a-12)= 20
⇒ (4a-12)= 20 or -(4a-12) =20
Therefore, it can be written as:
4a = 20+12
4a = 32
a = 8
-4a +12 =20
-4a = 20-12
⇒ a= 8 or -2
Hence, the required points on x axis are (-2, 0) and (8, 0).
Practice Problems for Class 11 Maths Chapter 10 Straight Lines
Go through the concepts covered in the class 11 Maths Chapter 10 straight lines and solve the problems given below:
- If the slopes of the two lines are ½ and 3, then find the angle between the two lines.
- Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).
- Find the equations to the line parallel to the axes and passing through the point (-3, 5).
- Find the equation of the line which is passing through the point (-2, -4) and perpendicular to the line 3x – y + 5 = 0.
- Find the equation of the line which passes through the point(3,4) and the sum of whose intercepts on the axes is 14.
- Find the distance of the point (-1,1) from the line 12(x+6) = 5(y-2)
- Find the point of intersection of the ines 5a
+ 7b = 3 and 2a – 3b =7.
- At what point is the origin be shifted, if the coordinates of the point (,5)become (3,7)?
- Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – c= 0.
- Find the intercepts cut off by the line 2a – b + 16 = 0
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