Решения NCERT класса 11 - Глава 10 Прямые линии - Разное упражнение по главе 10 | Набор 2
Вопрос 13. Покажите, что уравнение прямой, проходящей через начало координат и образующей угол θ с прямой y = mx + c, имеет вид 
.
Решение:
Suppose y = m1x is the equation of the line passing through the origin. Hence, slope of this line is,
m1 =
Given line is y = mx + c whose slope is m. It makes an angle θ with the line y = m1x.
=> tan θ =
=>
= tan θ
=>
= tan θ
=>
or
=>
or
=>
or
=>
Hence proved.
Вопрос 14. В каком отношении прямая, соединяющая (–1, 1) и (5, 7), делится прямой x + y = 4?
Решение:
We know that the equation of the line joining (–1, 1) and (5, 7) is given by,
=> y – 1 =
(x + 1)
=> y – 1 = x + 1
=> x – y + 2 = 0 . . . . (1)
We are given the line, x + y = 4 . . . . (2)
Solving (1) and (2), we get the point of intersection of these lines,
x = 1 and y = 3
Suppose the point (1, 3) divides the line joining (–1, 1) and (5, 7) in the ratio 1 : k.
Using section formula, we get,
(1, 3) =
(1, 3) =
We get,
=> – k + 5 = 1 + k
=> 2k = 4
=> k = 2
Therefore, the line joining the points (–1, 1) and (5, 7) is divided by the line x + y = 4 in the ratio 1 : 2.
Вопрос 15. Найдите расстояние прямой 4x + 7y + 5 = 0 от точки (1, 2) по прямой 2x – y = 0.
Решение:
We are given,
2x – y = 0 . . . . (1)
4x + 7y + 5 = 0 . . . . (2)
Solving (1) and (2), we get the point of intersection of lines,
x = –5/18 and y = –5/9.
So, we get B (–5/18, –5/9) and we have A (1, 2). Now we have to find the distance AB.
Using the distance formula, we get,
AB =
=
=
=
=
=
units
Therefore, the required distance is
Вопрос 16. Найдите направление, в котором надо провести прямую через точку (–1, 2) так, чтобы точка ее пересечения с прямой x + y = 4 находилась на расстоянии 3 единиц от этой точки.
Решение:
Suppose y = mx + c is the line passing through the point (–1, 2). So we get
=> 2 = m (–1) + c
=> 2 = –m + c
=> c = m + 2
Putting the value of c in the equation, we get,
y = mx + m + 2 . . . . (1)
Now the given line is
x + y = 4 . . . . (2)
By solving both the equations we get the point of intersection of these lines,
x =
, y =
Now the point
is at a distance of 3 units from the point (–1, 2).
From distance formula, we get,
=>
=>
=>
=>
=>
=>
=> 1 + m2 = m2 + 1 + 2m
=> 2m = 0
=> m = 0
As the slope of the required line is zero, the line must be parallel to the x-axis.
Вопрос 17. Гипотенуза прямоугольного треугольника имеет концы в точках (1, 3) и (−4, 1). Найдите уравнение катетов (перпендикулярных сторон) треугольника.
Решение:
Suppose △ABC is right-angled triangle where ∠ C = 90o and m is the slope of AC.
So the slope of BC = –1/m. And the equation of AC is given by,
=> y – 3 = m (x – 1)
=> x – 1 = 1/m (y – 3)
We know the equation of BC is given by,
=> y – 1 = – 1/m (x + 4)
=> x + 4 = – m (y – 1)
If m = 0, we get the equations of perpendicular sides,
y – 3 = 0 and x + 4 = 0 or,
y = 3 and x = –4
If m = ∞, we get the equations,
x – 1 = 0 and y – 1 = 0 or,
x = 1 and y = 1
Вопрос 18. Найдите образ точки (3, 8) относительно прямой x + 3y = 7, считая прямую плоской.
Решение:
We are given that
x + 3y = 7 . . . . (1)
Suppose B (a, b) is the image of point A (3, 8). Hence line (1) is the perpendicular bisector of AB.
Slope of line (1) = −1/3
And slope of line AB =
We know, AB and line (1) are perpendicular, so we have,
=>
=>
=> 3a − b = 1 . . . . (2)
Now, mid-point of AB =
. As the mid-point is satisfying line (1), we get,
=>
=> a + 3 + 3b + 24 = 14
=> a + 3b = −13 . . . . (3)
Solving (2) and (3), we get, a = –1 and b = –4
Therefore, the image of the given point with respect to the given line is (–1, –4).
Вопрос 19. Если прямые y = 3x + 1 и 2y = x + 3 одинаково наклонены к прямой y = mx + 4, найти значение m.
Решение:
We are given,
y = 3x + 1 . . . . (1)
2y = x + 3 . . . . (2)
y = mx + 4 . . . . (3)
So the slope of line (1), a = 3,
line (2), b = 1/2
line (3), c = m
It is given that the lines (1) and (2) are equally inclined to line (3) which implies that the angle between lines (1) and (3) equals the angle between lines (2) and (3).
=>
=>
=>
=>
=>
or
=>
or
=> 5m2 + 5 = 0 (ignored) or 7m2 − 2m − 7 = 0
=> 7m2 − 2m − 7 = 0
=> m =
=> m =
Hence the required value of m is
.
Вопрос 20. Если сумма расстояний по перпендикуляру переменной точки P (x, y) от прямых x + y – 5 = 0 и 3x – 2y + 7 = 0 всегда равна 10. Докажите, что точка P должна двигаться по прямой.
Решение:
We are given lines,
x + y – 5 = 0 . . . . (1)
3x – 2y + 7 = 0 . . . . (2)
Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x, y) is given by,
d =
So, perpendicular distance(d1) of a line x + y – 5 = 0 from (x, y) is,
=
And the perpendicular distance(d2) of a line 3x – 2y + 7 = 0 from (x, y) is,
=
According to the question,
=> d1 + d2 = 10
=>
=>
=>
=>
=>
As the equation represents a line, P must move on a line.
Hence, proved.
Вопрос 21. Найдите уравнение прямой, равноудаленной от параллельных прямых 9x + 6y – 7 = 0 и 3x + 2y + 6 = 0.
Решение:
We are the given the lines,
9x + 6y – 7 = 0 . . . . (1)
3x + 2y + 6 = 0 . . . . (2)
Let (a, b) be the point lying on the line which is equidistant from (1) and (2).
Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x, y) is given by,
So, perpendicular distance(d1) of a line 9x + 6y – 7 = 0 from (a, b) is,
=
And the perpendicular distance(d2) of a line 3x + 2y + 6 = 0 from (a, b) is,
=
According to the question,
=> d1 = d2
=>
=> |9a + 6b − 7| = 3 |3a + 2b + 6|
=> 9h + 6k – 7 = 3 (3h + 2k + 6) or 9h + 6k – 7 = – 3 (3h + 2k + 6)
=> 9h + 6k – 7 = 9h + 6k + 18 or 9h + 6k – 7 = –9h – 6k – 18
=> – 7 = 18 (ignored) or 18h + 12k + 11 = 0
=> 18h + 12k + 11 = 0
Therefore, the required equation of the line is 18x + 12y + 11 = 0.
Вопрос 22. Луч света, проходящий через точку (1, 2), отражается от оси абсцисс в точке А, и отраженный луч проходит через точку (5, 3). Найдите координаты А.
Решение:
The perpendicular line from the axes divides the ∠ BAC = 90o in two equal parts, each having value of θ.
Therefore, θ = 45o. And hence lines AC and AB subtend equal angles (θ) at x-axis.
Slope of AC = tan θ =
. . . . (1)
Slope of AB = tan (180−θ) =
=> −tan θ =
=> tan θ =
. . . . (2)
From (1) and (2), we get,
=>
=> 3a – 3 = 10 – 2a
=> a = 13/5
Therefore, the coordinates of point A are (13/5, 0).
Вопрос 23. Докажите, что произведение длин перпендикуляров, проведенных из точек 
а также 
к линии 
это б 2 .
Решение:
We are given the line,
=> bx cos θ + ay sin θ – ab = 0 . . . . (1)
Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x, y) is given by,
So, perpendicular distance(d1) of (1) from point
=
And the perpendicular distance(d2) of (1) from point
=
Now, L.H.S. = d1d2
=
=
=
=
=
=
=
=
= b2
= R.H.S.
Hence, proved.
Вопрос 24. Человек, стоящий на стыке (пересечении) двух прямых путей, представленных уравнениями 2x – 3y + 4 = 0 и 3x + 4y – 5 = 0, хочет достичь пути, уравнение которого 6x – 7y + 8 = 0 в наименьшее время. Найдите уравнение пути, по которому он должен пройти.
Решение:
We are given that,
2x – 3y + 4 = 0 . . . . (1)
3x + 4y – 5 = 0 . . . . (2)
6x – 7y + 8 = 0 . . . . (3)
It is given that the person is standing at the intersection of the paths represented by lines (1) and (2).
By solving equations (1) and (2) we get
x = –1/17 and y = 22/17
As perpendicular distance is the shortest distance, 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).
Now, the slope of the line (3) = 6/7
Hence, the slope of the line perpendicular to line (3) = –1/(6/7) = –7/6
So the equation of line passing through (–1/17, 22/17) and having a slope of –7/6 is given by,
=>
=> 6 (17y – 22) = –7 (17x + 1)
=> 102y – 132 = –119x – 7
=> 1119x + 102y = 125
Therefore, the required path is 119x + 102y = 125.
