Класс 11 RD Sharma Solutions – Глава 23 Прямые линии – Упражнение 23.12 | Набор 2

Вопрос 11. Найдите уравнение правой биссектрисы отрезка, соединяющего точки (a, b) и (a ₁ , b ₁ ).

Решение:

Let us considered P(a, b) and Q(a₁, b₁) are the given points and O be the mid-point of line PQ.

So the coordinate O = ((a + a₁)/, (b + b₁)/2)

Slope of line PQ = 

And the slope of right bisector of AB(m’) = 

The equation of required line is 

y – y₁ = m"(x – x₁)

2x(a₁ – a) + 2y(b₁ – b) + a² + b² = a₁² + b₁² 

2x(a₁ – a) + 2y(b₁ – b) + (a² + b²) – (a₁² + b₁²) = 0

Hence, the equation of line is 2x(a₁ – a) + 2y(b₁ – b) + (a² + b²) – (a₁² + b₁²) = 0

Вопрос 12. Найдите образ точки (2, 1) относительно зеркальной линии x + y – 5 = 0.

Решение:

Let us considered P(2, 1) be the image of Q(a, b) are the given points and O be the mid-point of line PQ.

So the coordinate O = ((2 + a)/2, (1 + b)/2)

And O point lie on the line x + y – 5 = 0

(2 + a)/2 + (1 + b)/2 – 5 = 0

a + b = 7                 ………(1)

Here, x + y – 5 = 0 is perpendicular to PQ

So, (Slope of PQ) x (slope of AB) = -1

b – 1 = a – 2

b – a = -1           ………(2)

On solving eq(1) and (2), we get

a = 5 and b = 2

Hence, the image of (1, 2) in x + y – 5 = 0 is (4, 3)

Вопрос 13. Если образ точки (2, 1) относительно прямого зеркала равен (5, 2), найдите уравнение зеркала.

Решение:

Let us considered A(5, 2) be image of B(2, -1) and O be the mid point of AB

So the coordinate O = (7/2, 3/2)

Let us considered PQ be the mirror and line AB perpendicular to PQ

So, (Slope of PQ) x (slope of AB) = -1

(2 – 1/5 – 2) x (slope of AB) = -1

(slope of AB) = -3

So the equation of the mirror is 

y – 3/2 = -3(x – 7/2)

2y – 3 = -6x + 21

Hence, the equation of the mirror is 3x + y – 12 = 0

Вопрос 14. Найдите уравнение прямой, параллельной 3x – 4y + 6 = 0 и проходящей через среднюю точку соединения точек (2, 3) и (4, -1).

Решение:

It is given that A(2, 3) and B(4, -1) and O be the mid point of AB

So the coordinate O = (3, 1)

It is given that the equation to the straight line parallel to 3x – 4y + 6 = 0

So, 

y = 3x/4 + 3/2 

On comparing y = mx + c, we get

m = 3/4

Now put the value of m and (x₁, y₁) is eq(1), we get

The required equation of line is 

y – y₁ = m(x – x₁)    

y – 1 = 3/4(x – 3)

4y – 4 = 3x – 9

3x – 4y = 5

Hence, the equation of line is 3x – 4y = 5

Вопрос 15. Докажите, что прямые 2x – 3y + 1 = 0, x + y = 3, 2x – 3y = 2 и x + y = 4 образуют параллелограмм.

Решение:

AS we know that in a parallelogram opposite sides are parallel and parallel sides have equal slope.

So, the slope of line 2x – 3y + 1 = 0

m₁ = 2/3             …….(1)

The slope of line x + y = 3 

m₂ = -1              …….(2)

The slope of line 2x – 3y – 2 = 0

m₃ = 2/3                 …….(3)

The slope of line x + y = 4

m₄ = -1                     …….(4)

From (1), (2), (3) and (4), we get 

We conclude that the opposite sides of ABCD have same slope 

Hence, the given quadrilateral is parallelogram. 

Вопрос 16. Найдите уравнение прямой, проведенной перпендикулярно прямой x/4 + y/6 = 1 через точку ее пересечения с осью y.

Решение:

Let us considered the required equation of line is 

y – y₁ = m(x – x₁)      ……….(1)

The required line is perpendicular to the given line x/4 + y/6 = 1

When x = 0

y/6 = 1

y = 6

So, the point (x₁, y₁) is (0, 6)

It is given that the required equation of line is perpendicular to the line x/4 + y/6 = 1

So,

(slope of required line) x (slope of given line) = -1

m1 = -1/(-6/4) = 4/6 = 2/3

Now put the value of m’ and (x₁, y₁) is eq(1), we get

(y – 6) = 2/3(x – 0)

2x – 3y = -18

2x – 3y + 18 = 0

Hence, the equation of line is 2x – 3y + 18 = 0

Вопрос 17. Перпендикуляр из начала координат к прямой y = mx + c пересекает ее в точке (-1, 2). Найдите значения m и c.

Решение:

Let us considered point O (0, 0) and P (-1, 2) and OP is perpendicular to the given line y = mx + c

So, (slope of OP) x (slope of line)=-1

And point P lie on the line  so,

2 = (1/2)(-1) + c

c = 2 + 1/2 = 5/2

Hence, the value of c = 5/2 and m = 1/2

Вопрос 18. Найдите уравнение правой биссектрисы отрезка, соединяющего точки (3, 4) и (-1, 2).

Решение:

Let us considered P(3, 4) and Q(-1, 2) are the given point and O be the mid point of AB

So, the coordinate O = (1, 3)

And the slope of line PQ is 

The right bisector of PQ is -2

The equation of the required line is 

y – 3 = (-2)(x – 1)

y – 3 = -2x + 2

2x + y – 5 = 0

Hence, the required equation of line is 2x + y – 5 = 0

Вопрос 19. Прямая, проходящая через (h, 3) и (4, 1), пересекает прямую 7x – 9y – 19 = 0 под прямым углом. Найдите значение h.

Решение:

Let us considered P (h, 3) and Q(4, 1) are the given points

Now the slope the line 7x – 9y – 19 = 0 is 7/9

It is given that line PQ is perpendicular to 7x – 9y – 19 = 0

so 

7/9 x (1 – 3)/(4 – h) = -1

9h = 22

h = 22/9

Hence, the value of h is 22/9

Вопрос 20. Найдите образ точки (3, 8) относительно прямой x + 3y = 7, считая эту прямую плоским зеркалом.

Решение:

Let us considered A(5, 2) be image of B(a, b) and O be the mid point of AB

So, the coordinate of O = ((3 + a)/2, (8 + b)/2)

O point lies on line x + 3y = 7

So, (3 + a)/2 + 3 x ((8 + b)/2) = 7

a + 13b + 13 = 0   ……(1)

It is given that line AB is perpendicular to mirror PQ

So, slope of AB x slope of PQ = -1

(b – 8/a – 3) x -1/3 = -1

3a – b – 1 = 0   ……(2)

On solving eq(1) and (2), we get

a = -1, b = -4

Hence, the image is (-1, -4)