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VOOZH | about |
In Chapter 23 of RD Sharma's Class 11 Mathematics "The Straight Lines" students explore the fundamental concepts and equations related to the straight lines in coordinate geometry. This chapter covers the various forms of the equations of the line, slope-intercept form general form, point-slope form, and intercept form along with topics such as the parallel and perpendicular lines angle between the two lines and distance of the point from a line.
A straight line is a fundamental concept in geometry and algebra representing the shortest distance between the two points. In coordinate geometry, a line can be described using the various equations based on the different forms such as the slope-intercept form y=mx+c, where m is the slope and c is the y-intercept. Understanding these forms and properties of the straight lines is crucial for solving problems related to the angles, distances, and intersections in both the two and three dimensions.
Solution:
Given: (x1, y1) = A (1, 2) and θ = 60°
By using the formula, the equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ = r
When we substitute the values, we get:
(x - 1) / cos60° = (y - 1) / sin60° = r
(x - 1) / (1/2) = (y - 1) / (√3/2) = r
Where, r represents the distance of any point on the line from A (1, 2).
The coordinate of point P on this line are
(1 + r/2, 2 + √3r/2)
and, P lies on the line x + y = 6
1 + r/2 + 2 + √3r/2 = 6
r = 6 / (1 + √3) = 3(√3 - 1)
Therefore, the value of AP = 3(√3 – 1)
Solution:
Given: (x1, y1) = A (3, 4) and θ = π/6 = 30°
By using the formula, the equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ = r
When we substitute the values, we get:
(x - 3) / cos30° = (y - 4) / sin30° = r
(x - 3) / (√3/2) = (y - 4) / (1/2) = r
x – √3 y + 4√3 – 3 = 0
Let PQ = r
The coordinate of Q are:
(x - 3) / cos30° = (y - 4) / sin30° = r
x = 3 + √3r/2, y = 4 + r/2
Given that the Q lies on 12x + 5y + 10 = 0
So,
12(3 + √3r/2) + 5(4 + r/2) + 10 = 0
66 + (12√3 + 5)r / 2 = 0
r = - 132 / 12√3 + 5
PQ = |r| = 132 / 12√3 + 5
Therefore, the length of PQ = 132 / 12√3 + 5.
Solution:
Given: (x1, y1) = A (2, 1) and θ = π/4 = 45°
By using the formula, the equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ = r
When we substitute the values, we get:
(x - 2) / cos45° = (y - 1) / sin45° = r
(x - 2) / 1/√2 = (y - 1) / 1/√2 = r
x – y – 1 = 0
Let AB = r
The coordinate of B are:
(x - 2) / cos45° = (y - 1) / sin45° = r
x = 2 + r/√2, y = 1 + r/√2
Given that the B lies on x + 2y + 1 = 0
So,
(2 + r/√2) + 2(1 + r/√2) + 1 = 0
5 + (3r/√2)r = 0
r = 5√2/3
Therefore, the length of AB = 5√2/3
Solution:
Given: (x1, y1) = A (4, -1)
Also give equation of Line: 3x – 4y + 1 = 0
4y = 3x + 1
y = 3x/4 + 1/4
Slope tan θ = 3/4
Thus,
Sin θ = 3/5
Cos θ = 4/5
By using the formula, the equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ
When we substitute the values, we get:
(x - 4) / (4/5) = (y + 1) / (3/5)
3x – 4y = 16
Let, AP = r
and r = ±5
The coordinate of P are:
(x - 4) / (4/5) = (y + 1) / (3/5) = r
x = 4r/5 + 4 and y = 3r/5 + 4
x = 4(±5)/5 + 4 and y = 3(±5)/5 + 4
x = ±4 + 4 and y = ±3 –1
x = 8, 0 and y = 2, – 4
Therefore, the coordinates are (8, 2) and (0, −4) which are at the distance of 5 unit from the point A 4, -1).
Solution:
Given that the line passes through P(x1, y1) and makes an angle of θ with the x–axis.
The equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ
Let PQ = r
The coordinates of Q are:
Now, by using the formula, the equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ = r
x = x1 + rcosθ , y = y1 + rsinθ
Given that the Q lies on ax + by + c = 0
So,
a(x1 + rcosθ) + b(y1 + rsinθ) + c = 0
r = PQ = | (ax1 + by1 + c) / (acosθ + bsinθ ) |
Therefore, the value of PQ = | (ax1 + by1 + c) / (acosθ + bsinθ) |
Solution:
Given: (x1, y1) = (2, 3) and θ = 45°
By using the formula, the equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ = r
When we substitute the values, we get:
The coordinate are:
(x - 2) / cos45° = (y - 3) / sin45° = r
x = 2 + r/√2, y = 3 + r/√2
Given that the point lies on 2x - 3y + 9 = 0
So,
2(2 + r/√2) - 3(3 + r/√2) + 9 = 0
2[(2√2 + r)/√2] - 3[(3√2 + r)/√2] + 9 = 0
4√2 + 2r - 9√2 - 3r + 9√2 = 0
r = 4√2
Therefore, the distance is = 4√2 units.
Solution:
Given: (x1, y1) = (3, 5)
Also give equation of Line: 2x + 3y = 14
and Slope tan θ = 1/2
Thus,
Sin θ = 1/√5
Cos θ = 2/√5
By using the formula, the equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ = r
When we substitute the values, we get the coordinate of P as:
(x - 3) / (2/√5) = (y - 5) / (1/√5) = r
x = 2r/√5 + 3 and y = r/√5 + 5
Given that the point lies on 2x + 3y = 14
Therefore,
2(2r/√5 + 3) + 3(r/√5 + 5) = 14
4r/√5 + 6 + 3r/√5 +15 = 14
7r/√5 + 21 = 14
7r/√5 = 14 - 21
7r/√5 = -7
r = ±√5
Therefore, the distance is √5 units.
Solution:
Given: (x1, y1) = (2, 5)
Also give equation of Line: 3x + y + 4 = 0
and Slope tan θ = 3/4
Thus,
Sin θ = 3/5
Cos θ = 4/5
By using the formula, the equation of the line is given by:
(x - x1) / cosθ = (y - y1) / sinθ = r
When we substitute the values, we get the coordinate of P as:
(x - 2) / (4/5) = (y - 5) / (3/5) = r
x = 4r/5 + 2 and y = 3r/5 + 5
Given that the point lies on 3x + y + 4 = 0
Therefore,
3(4r/5 + 2) + 3r/5 + 5 + 4 = 0
12r/5 + 6 + 3r/5 +9 = 0
15r/5 + 15 = 0
15r/5 = -15
r = ±5
Therefore, the distance is 5 units.
