Important Questions: Straight Lines | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: 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°.
Ans: 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
It becomes
√ 3x +y = 10
Thus, the required equation of a line is √ 3x + y = 10

Q2: 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.
Ans: 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
Therefore,
m (– 2) = -1 (Since the two lines are perpendicular)
m= ½
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.

Q3: 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).
Ans: Given that,
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
m = (-2-0)/(4-0)
m = -2/4
m = -½
Hence, the required slope of the line is -1/2

Q4: Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3
Ans: 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 that is perpendicular to the line x – 7y + 5 = 0 with x-intercept 3 is 7x+ y = 21.

Q5: Find the points on the x-axis whose distance from the line equation (x/3) + (y/4) = 1 is given as 4 units.
Ans: Given that,
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-12)= 20
4a = 20+12
4a = 32
a = 8
(or)
-(4a-12) =20
-4a +12 =20
-4a = 20-12
-4a= 8
a= -2
⇒ a= 8 or -2
Hence, the required points on the x-axis are (-2, 0) and (8, 0).