Chapter 10: Straight Lines Miscellaneous Exercise

Here,

9h + 6k – 7 = 3 (3h + 2k + 6) or 9h + 6k – 7 = – 3 (3h + 2k + 6)

9h + 6k – 7 = 3 (3h + 2k + 6) is not possible as

9h + 6k – 7 = 3 (3h + 2k + 6)

By further calculation,

– 7 = 18 (is not correct)

We know that

9h + 6k – 7 = -3 (3h + 2k + 6)

By multiplication,

9h + 6k – 7 = -9h – 6k – 18

We get

18h + 12k + 11 = 0

Hence, the required equation of the line is 18x + 12y + 11 = 0.

Consider the coordinates of point A as (a, 0)

Construct a line (AL) which is perpendicular to the x-axis

Here, the angle of incidence is equal to the angle of reflection

∠BAL = ∠CAL = Φ

∠CAX = θ

It can be written as

∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]

On further calculation,

= 180° – θ – 180° + 2θ

= θ

So, we get

∠BAX = 180° – θ

By cross multiplication,

3a – 3 = 10 – 2a

We get

a = 13/5

Hence, the coordinates of point A are (13/5, 0).

It is given that

We can write it as

bx cos θ + ay sin θ – ab = 0 ….. (1)

It is given that

2x – 3y + 4 = 0 …… (1)

3x + 4y – 5 = 0 ……. (2)

6x – 7y + 8 = 0 …… (3)

Here, the person is standing at the junction of the paths represented by lines (1) and (2).

By solving equations (1) and (2), we get

x = – 1/17 and y = 22/17

Hence, the person is standing at point (-1/17, 22/17).

We know that the person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point (-1/17, 22/17).

Here, the slope of the line (3) = 6/7

We get the slope of the line perpendicular to the line (3) = -1/ (6/7) = – 7/6

So, the equation of the line passing through (-1/17, 22/17) and having a slope of -7/6 is written as

By further calculation,

6 (17y – 22) = – 7 (17x + 1)

By multiplication,

102y – 132 = – 119x – 7

We get

1119x + 102y = 125

Therefore, the path that the person should follow is 119x + 102y = 125.