CLASS_11_MATHS_SOLUTIONS_NCERT

Class XI Chapter 10 – Straight Lines Maths
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Equation (3) is in the normal form.
On comparing equation (3) with the normal form of the equation of the line
xcos
ysin p , we obtain = 315 and p = 2 2.
Thus, the perpendicular distance of the line from the origin is 2 2, while the angle between the
perpendicular and the positive x-axis is 315 .
Question 4:
Find the distance of the points (-1, 1) from the line 12(x + 6) = 5(y – 2).
Solution 4:
The given equation of the line is 12(x + 6) = 5(y – 2).
12x + 72 = 5y – 10
12x – 5y + 82 = 0 …(1)
On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 12, B
= - 5, and C = 82.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point
1 1
x, y is given by d =
1 1
Ax
By
C
2 2
A B
The given point is x1,
y
1
= (-1, 1).
Therefore, the distance of point (-1, 1) from the given line
121 51
82 12 582 65
= units units units 5units
2 2
12 5
169 13
Question 5:
x y
Find the points on the x-axis whose distance from the line 1are 4 units.
3 4
Solution 5:
The given equation of line is
x y
1
3 4
Or,4x + 3y -12 = 0 ….(1)
On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain 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.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point
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