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 xaxis 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 xaxis
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 midpoint of the segment joining the points P (0, 4) and B (8, 0).
Ans: Given that,
The coordinates of the midpoint 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 nonvertical line passing through the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by the formula
m = (y_{2} y_{1}) / ( (x_{2} x_{1}), where (x_{2} is not equal to x_{1})
Therefore, the slope of the line passing through the points (0, 0,) and (4, 2) is
m = (20)/(40)
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 xintercept 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(x3)
⇒ y=7x + 21
7x+ y = 21
Hence, the equation of a line that is perpendicular to the line x – 7y + 5 = 0 with xintercept 3 is 7x+ y = 21.
Q5: Find the points on the xaxis 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 xaxis 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 (x_{1}, y_{1}) is given by
D = Ax_{1 }+ By_{1} + C/ √A^{2} + B^{2}
Now, substitute the values in the above formula, we get:
4 = 4a+0 + 12/ √4^{2} + 3^{2}
⇒4 = 4a12/5
⇒4a12 = 20
⇒± (4a12)= 20
⇒ (4a12)= 20 or (4a12) =20
Therefore, it can be written as:
(4a12)= 20
4a = 20+12
4a = 32
a = 8
(or)
(4a12) =20
4a +12 =20
4a = 2012
4a= 8
a= 2
⇒ a= 8 or 2
Hence, the required points on the xaxis are (2, 0) and (8, 0).
75 videos238 docs91 tests


Explore Courses for Commerce exam
