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• RD Sharma Class 11 Solutions for Maths

Class 11 RD Sharma Solutions- Chapter 23 The Straight Lines- Exercise 23.8

Question 1: A-line passes through a point A (1, 2) and makes an angle of 60° with the x-axis and intercepts the line x + y = 6 at the point P. Find AP.

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)

Question 2: If the straight line through the point P(3, 4) makes an angle π/6 with the x-axis and meets the line 12x + 5y + 10 = 0 at Q, find the length PQ.

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.

Question 3: A straight line drawn through the point A (2, 1) making an angle π/4 with positive x-axis intersects another line x + 2y + 1 = 0 in the point B. Find length AB.

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

Question 4: A line a drawn through A (4, – 1) parallel to the line 3x – 4y + 1 = 0. Find the coordinates of the two points on this line which are at a distance of 5 units from A.

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).

Question 5: The straight line through P(x1, y1) inclined at an angle θ with the x-axis meets the line ax + by + c = 0 in Q. Find the length of PQ.

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) |

Question 6: Find the distance of the point (2, 3) from the line 2x – 3y + 9 = 0 measure along a line making an angle of 45° with the x-axis.

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.

Question 7: Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to a line having slope 1/2.

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.

Question 8: Find the distance of the point (2, 5) from the line 3x + y + 4 = 0, measured parallel to a line having slope 3/4.

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.

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