A straight line is a line which is not curved or bent. In this lesson, all the basic and advanced concepts related to straight lines are covered here. This lesson can also be downloaded as PDF which will students to refer to the concepts in offline mode.
A line is simply an object in geometry that is characterized under zero width object that extends on both sides. A straight line is just a line with no curves. So, a line that extends to both sides till infinity and has no curves is called a straight line.
Equation of line with slope ‘m’ and which passes through (x1, y1) can be given as
y – y1 = m(x – x1 )
Slope Point form (Equation of a Line with 2 Points)
Equation of a line passing through two points (x1, y1) & (x2, y2) is given as
y−y1=(x2−x1y2−y1)(x−x1)
For example,
Example:Find the equation of the line that passes through the points (-2, 4) and (1, 2).
Solution:
We know general equation of a line passing through two points is:
y = mx + b, where m = x2–x1y2−y1
Here m = (2-4)/(1-(-2)) = -2/3
We can find the equation (by solving first for “b”) if we have a point and the slope. So we need to choose one of the points and use it to solve for b. Using the point (–2, 4), we get:
4 = (– 2/3)(–2) + b
4 = 4/3 + b
4 – 4/3 = b
b = 8/3
so, y = ( – 2/3 ) x + 8/3.
Which is the required equation of the line.
Straight Lines Video Lesson
Relation between two Lines
Let L1 and L2 be the two lines as
L1 : a1x + b1y + c1 = 0
L2 : a2x + b2y + c2 = 0
For Parallel lines
Two lines are said to be parallel if satify the below condition,
Equation of a Line with 2 Points (Slope Point Form)
(y – y1) = m(x – x1)
Angle Between Straight lines
θ=tan−1∣∣∣∣(1+m1+m2m2−m1)∣∣∣∣
Problems on Straight Lines
Question 1:
Find the equation to the straight line which passes through the point (-5, 4) and is such that the portion of it between the axes is divided by the given point in the ratio 1 : 2.
Solution:
Let the required straight line be ax+by=1.
Using the given conditions, P(2+12a+1.0,2+12.0+1.b) is the point which divides (a, 0) and (0, b) internally in the ratio 1 : 2.
But P is (-5, 4)
Hence -5 = 2a/3, 4 = b/3 a = -15/2, b = 12.
Hence the required equation is (−15/2)x+12y=1
Question 2:
Find the equation of the straight line which passes through the point (1, 2) and makes an angle θ with the positive direction of the x-axis where cosθ=−31.
Solution:
Here cosθ=−31. (a negative number) so that 2π<θ<π⇒tanθ=−8= slope of line.
We know that the equation of the straight line passing through the point (x1, y1) having slope m is
y – y1 = m(x – x1)
Therefore the equation of the required line is y−2=−8(x−1)⇒8x+y−8−2=0.
Question 3:
Find the equation of the line joining the points (-1, 3) and (4, -2).
Solution:
Equation of the line passing through the points (x1, y1) and (x2, y2) is y−y1=x1−x2y1−y2(x−x1)
Hence equation of the required line will be y−3=−1−43+2(x+1)⇒x+y−2=0
Question 4:
Which line is having greatest inclination with positive direction of x-axis?
(i) line joining points (1, 3) and (4, 7)
(ii) line 3x – 4y + 3 = 0
Solution:
(i) Slope of line joining points A(1, 3) and B(4, 7) is 4−17−3=34=tanα
(ii) Slope of line is −−43=43=tanβ
Now tan α > tan β
Question 5:
Angle of line positive direction of x-axis is θ. Line is rotated about some point on it in anticlockwise direction by angle 45° and its slope becomes 3. Find the angle θ.
Solution:
Originally slope of line is tan θ = m
Now slope of line after rotation is 3.
Angle between old position and new position of lines is 45°.
∴ we have tan45∘=1+3m3−m
1 + 3m = 3 – m
4m = 2
m = 1/2 = tan θ
θ = tan-1(1/2)
Question 6:
If line 3x−ay−1=0 is parallel to the line (a+2)x−y+3=0 then find the values of a.
Solution:
Slope of line 3x−ay−1=0 is a3
Slope of line (a+2)x−y+3=0is (a + 2)
Since lines are parallel then we have a+2=a3
or a2+2a−3=0
or (a−1)(a+3)=0
or a = 1 or a = 3.
Question 7:
Find the value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear.
Solution:
If points A(x, -1) B(2, 1), and C(4, 5) are collinear, then
⇒2−x1−(−1)=4−25−1⇒2−x2=2⇒x=1
Question 8:
The slope of a line is double of the slope of another line. It tangent of the angle between them is 31.Find the slopes of the lines.
Solution:
Let m1 and m be the slopes of the two given lines such that m1 = 2m
We know that if θ is the angle between the lines l1 and l2 with slopes m1 and m2, then tanθ=∣∣∣∣1+m1m2m2−m1∣∣∣∣
It is given that the tangent of the angle between the two lines is 31
If one of the lines of the pair ax2+2hxy+by2=0 bisects the angle between positive direction of the axes, then find relation for a, b and h.
Solution:
Bisector of the angle between the positive directions of the axes is y = x.
Since it is one of the lines of the given pair of lines ax2+2hxy+by2=0,
We have x2(a+2h+b)=0 or a+b=−2h.
Question 15:
If the angle between the two lines represented by 2x2+5xy+3y2+6x+7y+4=0 is tan-1(m), then find the value of m.
Solution:
The angle between the lines
2x2+5xy+3y2+6x+7y+4=0 is given by
tanθ=2+3±2425−6θ=tan−1(±51).
Question 16:
The pair of lines 3x2−4xy+3y2=0 are rotated about the origin by pi/6 in the anticlockwise sense. Find the equation of the pair in the new position.
Solution:
The given equation of pair of straight lines can be rewritten as (3x−y)(x−3y)=0. Their separate equations are y=3x and y=1/3xor y = tan 60° x and y = tan 30° x
After rotation, the separate equations are
y = tan 90° x and y = tan 60° x
or x = 0 and y=3x
the combined equation in the new position is x(3x−y)=0 or 3x2−xy=0