Chapter 10: Straight Lines Miscellaneous Exercise

Consider m₁ as the slope of the required line

It can be written as

y = 1/2 x – 3/2 which is of the form y = mx + c

So, the slope of the given line m₂ = 1/2

We know that the angle between the required line and line x – 2y = 3 is 45^o

If θ is the acute angle between lines l₁ and l₂ with slopes m₁ and m_2,

It can be written as

2 + m₁ = 1 – 2m₁ or 2 + m₁ = -1 + 2m₁

m₁ = -1/3 or m₁ = 3

Case I – m₁ = 3

Here, the equation of the line passing through (3, 2) and having a slope of 3 is

y – 2 = 3 (x – 3)

By further calculation,

y – 2 = 3x – 9

So, we get

3x – y = 7

Case II – m₁ = -1/3

Here, the equation of the line passing through (3, 2) and having a slope of -1/3 is

y – 2 = -1/3 (x – 3)

By further calculation,

3y – 6 = – x + 3

So, we get

x + 3y = 9

Hence, the equations of the lines are 3x – y = 7, and x + 3y = 9.

Consider the equation of the line having equal intercepts on the axes as

x/a + y/a = 1

It can be written as

x + y = a ….. (1)

By solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we get

x = 1/13 and y = 5/13

(1/13, 5/13) is the point of intersection of two given lines.

We know that equation (1) passes through the point (1/13, 5/13)

1/13 + 5/13 = a

a = 6/13

So, equation (1) passes through (1/13, 5/13).

1/13 + 5/13 = a

We get

a = 6/13

Here, equation (1) becomes

x + y = 6/13

13x + 13y = 6

Hence, the required equation of the line is 13x + 13y = 6.

Consider y = m₁x as the equation of the line passing through the origin

By cross multiplication

– k + 5 = 1 + k

We get

2k = 4

k = 2

Hence, the line joining the points (-1, 1) and (5, 7) is divided by the line x + y = 4 in the ratio 1:2.

It is given that

2x – y = 0 ….. (1)

4x + 7y + 5 = 0 …… (2)

Here A (1, 2) is a point on line (1)

Consider B as the point of intersection of lines (1) and (2)

By solving equations (1) and (2), we get x = -5/18 and y = – 5/9

So, the coordinates of point B is (-5/18, -5/9)

From the distance formula, the distance between A and B

Hence, the required distance is
.

Consider y = mx + c as the line passing through the point (-1, 2)

So, we get

2 = m (-1) + c

By further calculation,

2 = -m + c

c = m + 2

Substituting the value of c,

y = mx + m + 2 …… (1)

So, the given line is

x + y = 4 ……. (2)

By solving both equations, we get

By cross multiplication

1 + m² = m² + 1 + 2m

So, we get

2m = 0

m = 0

Hence, the slope of the required line must be zero, i.e. the line must be parallel to the x-axis.

Consider ABC as the right angles triangle where ∠C = 90^o

Here, infinitely such lines are present.

m is the slope of AC

So, the slope of BC = -1/m

Equation of AC,

y – 3 = m (x – 1)

By cross multiplication,

x – 1 = 1/m (y – 3)

Equation of BC,

y – 1 = – 1/m (x + 4)

By cross multiplication,

x + 4 = -m (y – 1)

By considering the values of m, we get

If m = 0,

So, we get

y – 3 = 0, x + 4 = 0

If m = ∞,

So, we get

x – 1 = 0, y – 1 = 0 we get x = 1, y = 1

It is given that

x + 3y = 7 ….. (1)

Consider B (a, b) as the image of point A (3, 8)

So, line (1) is the perpendicular bisector of AB.

On further simplification,

a + 3b = – 13 ….. (3)

By solving equations (2) and (3), we get

a = -1 and b = -4

Hence, the image of the given point with respect to the given line is (-1, -4).

It is given that

y = 3x + 1 …… (1)

2y = x + 3 …… (2)

y = mx + 4 …… (3)

Here, the slopes of

Line (1), m₁ = 3

Line (2), m₂ = ½

Line (3), m₃ = m

We know that lines (1) and (2) are equally inclined to the line (3), which means that the angle between lines (1) and (3) equals the angle between lines (2) and (3).

On further calculation,

– m² + m + 6 = 1 + m – 6m²

So we get

5m² + 5 = 0

Dividing the equation by 5,

m² + 1 = 0

m = √-1, which is not real.

Therefore, this case is not possible.

If

In the same way, we can find the equation of the line for any signs of (x + y – 5) and (3x – 2y + 7).

Hence, point P must move on a line.