Exercise 10.2 1. Find the values of the unknown variables in the given figure. DK 1100 y CA B 720
Answers
Answer:
The y-coordinate of every point on x-axis is 0.
∴ Equation of x-axis is y = 0.
The x-coordinate of every point on y-axis is 0.
∴ Equation of y-axis is y = 0.
2. Passing through the point (– 4, 3) with slope 1/2
Solution:
Given:
Point (-4, 3) and slope, m = 1/2
We know that the point (x, y) lies on the line with slope m through the fixed point (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0)
So, y – 3 = 1/2 (x – (-4))
y – 3 = 1/2 (x + 4)
2(y – 3) = x + 4
2y – 6 = x + 4
x + 4 – (2y – 6) = 0
x + 4 – 2y + 6 = 0
x – 2y + 10 = 0
∴ The equation of the line is x – 2y + 10 = 0.
3. Passing through (0, 0) with slope m.
Solution:
Given:
Point (0, 0) and slope, m = m
We know that the point (x, y) lies on the line with slope m through the fixed point (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0)
So, y – 0 = m (x – 0)
y = mx
y – mx = 0
∴ The equation of the line is y – mx = 0.
4. Passing through (2, 2√3) and inclined with the x-axis at an angle of 75o.
Solution:
Given: point (2, 2√3) and θ = 75°
Equation of line: (y – y1) = m (x – x1)
where, m = slope of line = tan θ and (x1, y1) are the points through which line passes
∴ m = tan 75°
75° = 45° + 30°
Applying the formula:
NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 14
We know that the point (x, y) lies on the line with slope m through the fixed point (x1, y1), if and only if, its coordinates satisfy the equation y – y1 = m (x – x1)
Then, y – 2√3 = (2 + √3) (x – 2)
y – 2√3 = 2 x – 4 + √3 x – 2 √3
y = 2 x – 4 + √3 x
(2 + √3) x – y – 4 = 0
∴ The equation of the line is (2 + √3) x – y – 4 = 0.
5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
Solution:
Given:
Slope, m = -2
We know that if a line L with slope m makes x-intercept d, then equation of L is
y = m(x − d).
If the distance is 3 units to the left of origin then d = -3
So, y = (-2) (x – (-3))
y = (-2) (x + 3)
y = -2x – 6
2x + y + 6 = 0
∴ The equation of the line is 2x + y + 6 = 0.
6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30o with positive direction of the x-axis.
Solution:
Given: θ = 30°
We know that slope, m = tan θ
m = tan30° = (1/√3)
We know that the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c.
If distance is 2 units above the origin, c = +2
So, y = (1/√3)x + 2
y = (x + 2√3) / √3
√3 y = x + 2√3
x – √3 y + 2√3 = 0
∴ The equation of the line is x – √3 y + 2√3 = 0.
7. Passing through the points (–1, 1) and (2, – 4).
Solution:
Given:
Points (-1, 1) and (2, -4)
We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 15
y – 1 = -5/3 (x + 1)
3 (y – 1) = (-5) (x + 1)
3y – 3 = -5x – 5
3y – 3 + 5x + 5 = 0
5x + 3y + 2 = 0
∴ The equation of the line is 5x + 3y + 2 = 0.
8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30o.
Solution:
Given: p = 5 and ω = 30°
We know that the equation of the line having normal distance p from the origin and angle ω which the normal makes with the positive direction of x-axis is given by x cos ω + y sin ω = p.
Substituting the values in the equation, we get
x cos30° + y sin30° = 5
x(√3 / 2) + y( 1/2 ) = 5
√3 x + y = 5(2) = 10
√3 x + y – 10 = 0
∴ The equation of the line is √3 x + y – 10 = 0.
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