Решения NCERT класса 11 - Глава 10 Прямые линии - Разное упражнение по главе 10 | Набор 1

Вопрос 1. Найдите значения k, при которых линия (k – 3)x – (4 – k ² )y + k ² – 7k + 6 = 0 является

(а) Параллельно оси x

(b) Параллельно оси Y

(c) Прохождение через начало координат.

Решение:

We are given the line, (k – 3)x – (4 – k²)y + k² – 7k + 6 = 0

=> (4 – k²)y = (k – 3)x + k² – 7k + 6

=> y =+

Now the equation of the line is of the form y = mx + c where m is the slope of the line and c is its y-intercept.

So, we get m =and c =

(а) Параллельно оси x

If the line is parallel to the x-axis, then we have,

Slope of the line = Slope of the x-axis

=> m = 0

=>= 0

=> k – 3 = 0

=> k = 3

Therefore, if the given line is parallel to the x-axis, then the value of k is 3.

(b) Параллельно оси Y

If the line is parallel to the y-axis, then we have,

Slope of the line = Slope of the y-axis

=> m = ∞ (undefined)

=>= ∞

=> k² − 4 = 0

=> k² = 4

=> k = ±2

Therefore, if the given line is parallel to the y-axis, then the value of k is ± 2.

(c) Прохождение через начало координат.

If the line is passing through the origin,

Y-intercept = 0

=> c = 0

=> k² – 7k + 6 = 0

=> (k – 6) (k – 1) = 0

=> k = 1 or k = 6

Therefore, if the given line is passing through the origin, then the value of k is either 1 or 6.

Вопрос 2. Найдите значения θ и p, если уравнение x cos θ + y sin θ = p является нормальной формой прямой √3x + y + 2 = 0.

Решение:

We are given the line, √3x + y + 2 = 0.

=> −√3x − y = 2

On converting this into normal form, we get

=>

=>

=>

By comparing this equation with the given normal form x cos θ + y sin θ = p, we get

=> cos θ =, sin θ =and p = 1

The values of sin θ and cos θ are negative. So, θ = π +=.

Therefore, the value of θ isand p is 1.

Вопрос 3. Найдите уравнения прямых, пересекающих отрезки на осях, сумма и произведение которых равны 1 и –6 соответственно.

Решение:

Let’s suppose the intercepts cut by the given lines on the axes are a and b. According to the question, we have,

=> a + b = 1 . . . . (1)

=> ab = – 6 . . . . (2)

By solving both the equations we get

a = 3 and b = –2 or a = –2 and b = 3

We know that the equation of the line whose intercepts are a and b axes is,

bx + ay – ab = 0

When a = 3 and b = –2

So the equation of the line is – 2x + 3y + 6 = 0, i.e. 2x – 3y = 6.

When a = –2 and b = 3

So the equation of the line is 3x – 2y + 6 = 0, i.e. –3x + 2y = 6.

Therefore, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6.

Вопрос 4. Какие точки на оси у находятся на расстоянии 4 единиц от прямой x/3 + y/4 = 1?

Решение:

Suppose (0, b) is the point on the y-axis whose distance from line x/3 + y/4 = 1 is 4 units.

The line can be written as 4x + 3y – 12 = 0

By comparing our equation to the general equation of line Ax + By + C = 0, we get

A = 4, B = 3 and C = –12

Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x₁, y₁) is given by,

d =

We are given d = 4. For the point (0, b), the value of d becomes,

=>= 4

=>= 4

=> |3b – 12| = 20

=> 3b – 12 = 20 or 3b – 12 = –20

=> b = 32/3 or b = –8/3

Therefore, (0, 32/3) and (0, –8/3) are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units.

Вопрос 5. Найдите расстояние по перпендикуляру от начала координат до прямой, соединяющей точки (cos θ, sin θ) и (cos Ø, sin Ø).

Решение:

The equation of the line joining the points (cos θ, sin θ) and (cos Ø, sin Ø) is given by,

=> y – sin θ =(x – cosθ)

=> y(cos Ø – cos θ) – sin θ(cos Ø – cos θ) = x(sin Ø – sin θ) – cos θ(sin Ø – sin θ)

=> x(sin Ø – sin θ) + y(cos Ø – cos θ) + cos θ sin Ø – sin θ cos θ – sin θ cos Ø + sin θ cos θ = 0

=> x(sin Ø – sin θ) + y(cos Ø – cos θ) + sin (Ø – θ) = 0

So, we get, A = sin Ø – sin θ, B = cos Ø – cos θ and C = sin (Ø – θ).

Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from the origin (0, 0) is given by,

d =

=

=

=

=

=

=

Therefore,is the perpendicular distance from the origin to the given line.

Вопрос 6. Найдите уравнение прямой, параллельной оси y и проведенной через точку пересечения прямых x – 7y + 5 = 0 и 3x + y = 0.

Решение:

Two given lines are

x – 7y + 5 = 0 . . . . (1)

3x + y = 0 . . . . (2)

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

x = −5/22 and y = 15/22

(−5/ 22, 15/22) is the point of intersection of lines (2) and (3)

Now the equation of any line parallel to the y-axis is of the form

x = a . . . . (1)

If the line x = a passes through point (−5/22, 15/22) we get a = −5/22.

Therefore, the required equation of the line is x = −5/22.

Вопрос 7. Найдите уравнение прямой, проведенной перпендикулярно прямой x/4 + y/6 = 1 через точку пересечения ее с осью y.

Решение:

Given line is, x/4 + y/6 = 1.

=> 3x + 2y – 12 = 0

=> y = −3/2 x + 6, which is of the form y = mx + c

Here the slope of the given line = −3/2

So the slope of line perpendicular to the given line = −1/(−3/2) = 2/3

Suppose the given line intersects the y-axis at (0, y). So, the equation of the given line becomes,

=> y/6 = 1

=> y = 6

Hence, the given line intersects the y-axis at (0, 6). We know that the equation of the line that has a slope of 2/3 and passes through point (0, 6) is given by,

=> (y – 6) = 2/3 (x – 0)

=> 3y – 18 = 2x

=> 2x – 3y + 18 = 0

Therefore, the required equation of the line is 2x – 3y + 18 = 0.

Вопрос 8. Найдите площадь треугольника, образованного прямыми y – x = 0, x + y = 0 и x – k = 0.

Решение:

It is given that

y – x = 0 . . . . (1)

x + y = 0 . . . . (2)

x – k = 0 . . . . (3)

Here the point of intersection of lines (1) and (2) is x = 0 and y = 0.

And the point of intersection of lines (2) and (3) is x = k and y = – k, lines (3) and (1) is x = k and y = k.

So the vertices of the triangle formed by the three given lines are (0, 0), (k, –k) and (k, k).

Here the area of triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

A =

=

=

=

= k² square units

Therefore, k² sq. units is the area of the triangle formed by the given lines.

Вопрос 9. Найдите такое значение p, чтобы три прямые 3x + y – 2 = 0, px + 2y – 3 = 0 и 2x – y – 3 = 0 пересекались в одной точке.

Решение:

It is given that

3x + y – 2 = 0 . . . . (1)

px + 2y – 3 = 0 . . . . (2)

2x – y – 3 = 0 . . . . (3)

By solving equations (1) and (3) we get

x = 1 and y = –1

It is given that the three lines intersect at one point and the point of intersection of lines (1) and (3) will also satisfy line (2).

=> p (1) + 2 (–1) – 3 = 0

=> p – 2 – 3 = 0

=> p = 5

Therefore, the value of p is 5.

Вопрос 10. Если параллельны три прямые, уравнения которых имеют вид y = m ₁ x + c ₁ , y = m ₂ x + c ₂ и y = m ₃ x + c ₃ , то покажите, что m ₁ (c ₂ – c ₃ ) + м ₂ (с ₃ – с ₁ ) + м ₃ (с ₁ – с ₂ ) = 0.

Решение:

It is given that

y = m₁x + c₁ . . . . (1)

y = m₂x + c₂ . . . . (2)

y = m₃x + c₃ . . . . (3)

By subtracting equation (1) from (2), we get,

=> 0 = (m₂ – m₁) x + (c₂ – c₁)

=> (m₁ – m₂) x = c₂ – c₁

=> x =

And y =+ c₁

=> y =

=> y =

Hence, (,) is the point of intersection of lines (1) and (2).

As the given three lines are concurrent, this point must satisfy the equation (3).

=>

=>

=>

=> m₁ (c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0

Hence proved.

Вопрос 11. Найдите уравнение прямых, проходящих через точку (3, 2), которая составляет угол 45° с прямой x – 2y = 3.

Решение:

Suppose a is the slope of the line which passes through the point (3, 2).

Given line is x – 2y = 3.

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

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

We know that the angle between the required line and line x – 2y = 3 is 45^o. The angle is given by,

tan θ =

=> tan 45⁰ =

=>= 1

=>

=> 2 + a = 1 – 2a or 2 + a = – 1 + 2a

=> a = –1/3 or a = 3

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

=> y – 2 = 3 (x – 3)

=> y – 2 = 3x – 9

=> 3x – y = 7

When a = –1/3, the equation of the line passing through (3, 2) and having a slope –1/3 is

=> y – 2 = –1/3 (x – 3)

=> 3y – 6 = – x + 3

=> x + 3y = 9

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

Вопрос 12. Найдите уравнение прямой, проходящей через точку пересечения прямых 4x + 7y – 3 = 0 и 2x – 3y + 1 = 0, имеющую равные точки пересечения на осях.

Решение:

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

=> x/a + y/a = 1

=> 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.

Putting in (1), we get,

=> a = 1/13 + 5/13

=> a = 6/13

Here the equation (1) becomes

=> x + y = 6/13

=> 13x + 13y = 6

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