Решения NCERT для класса 11. Глава 10. Прямые линии. Упражнение 10.2 | Набор 2
Глава 10 Прямые линии – упражнение 10.2 | Набор 1
Вопрос 11. Прямая, перпендикулярная отрезку, соединяющему точки (1, 0) и (2, 3), делит его в отношении 1:n. Найдите уравнение прямой.
Решение:
Slope of line segment will be m1 = (y2 – y1)/(x2 – x1)
= (3 – 0)/(2 – 1)
= 3/1
= 3
If two line are perpendicular to each other than slope of line1 X slope of line2 = -1
So, the slope of line will be m2 = (-1/m1)
m2 = -1/3
As we know that the coordinates of a point (p, q) dividing the line
segment joining the points (x1, y1) and (x2, y2) internally
In the ratio m : n are:
(p, q) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))
(p, q) = ((1 × 2 + n × 1)/(1 + n), (1 × 3 + n × 0)/(1 + n))
p = (2 + n)/(1 + n)
q = 3/(1 + n)
We know that the point (p, q) lies on the line with slope m2,
Equation of line will be y – q = m2(x – p)
y – 3/(1 + n) = (–1/3)(x – (2 + n)/(1 + n))
3((1 + n) y – 3) = (–(1 + n) x + 2 + n)
3(1 + n) y – 9 = – (1 + n) x + 2 + n
(1 + n) x + 3(1 + n) y – n – 9 – 2 = 0
(1 + n) x + 3(1 + n) y – n – 11 = 0
So, the equation of the line is (1 + n) x + 3(1 + n) y – n – 11 = 0
Вопрос 12. Найдите уравнение прямой, отсекающей равные отрезки на осях координат и проходящей через точку (2, 3).
Решение:
Given that the line cuts off equal intercepts on the coordinate axes i.e. a = b.
So, the equation of the line intercepts a and b on x-and y-axis, respectively, which is
x/a + y/b = 1
So, x/a + y/a = 1
x + y = a -(1)
Given point is (2, 3)
2 + 3 = a
a = 5
On substituting the value of ‘a’ in eq(1), we get
x + y = 5
x + y – 5 = 0
So, the equation of the line is x + y – 5 = 0.
Вопрос 13. Найдите уравнение прямой, проходящей через точку (2, 2) и отсекающей отрезки на осях, сумма которых равна 9.
Решение:
Equation of the line making intercepts a and b on x-and y-axis, respectively, is
x/a + y/b = 1 -(1)
Given that sum of intercepts = 9
a + b = 9
b = 9 – a
On substituting the value of b in the eq(1), we get
x/a + y/(9 – a) = 1
Given that the line passes through the point (2, 2),
So, 2/a + 2/(9 – a) = 1
[2(9 – a) + 2a] / a(9 – a) = 1
[18 – 2a + 2a] / a(9 – a) = 1
18/a(9 – a) = 1
18 = a (9 – a)
18 = 9a – a2
a2 – 9a + 18 = 0
By using factorization method for quadratic equation solving, we get
a2 – 3a – 6a + 18 = 0
a (a – 3) – 6 (a – 3) = 0
(a – 3) (a – 6) = 0
a = 3 or a = 6
Let us substitute in eq(1),
Case 1 (a = 3):
Then b = 9 – 3 = 6
x/3 + y/6 = 1
2x + y = 6
2x + y – 6 = 0
Case 2 (a = 6):
Then b = 9 – 6 = 3
x/6 + y/3 = 1
x + 2y = 6
x + 2y – 6 = 0
So, the equation of the line is 2x + y – 6 = 0 or x + 2y – 6 = 0
Вопрос 14. Найдите уравнение прямой, проходящей через точку (0, 2) и образующую с положительной осью x угол 2π/3. Кроме того, найдите уравнение прямой, параллельной ей и пересекающей ось у на расстоянии 2 единиц ниже начала координат.
Решение:
Given that
Point (0, 2) and θ = 2π/3
slope (m)= tan θ
m = tan (2π/3) = -√3
Now, the equation of line passing through point(p,q) with slope m will be:
y – q = m (x – p)
y – 2 = -√3 (x – 0)
y – 2 = -√3 x
√3 x + y – 2 = 0
Now we need to find equation of line parallel to above obtained equation
crosses the y-axis at a distance of 2 units below the origin.
So, the point = (0, -2) and for parallel line slope will be same that is m = -√3
From point slope form equation,
y – (–2) = –√3 (x – 0)
y + 2 = -√3 x
√3 x + y + 2 = 0
So, the equation of line is √3 x + y – 2 = 0 and the line parallel to it is √3 x + y + 2 = 0.
Вопрос 15. Перпендикуляр из начала координат к прямой пересекает ее в точке (–2, 9), найдите уравнение прямой.
Решение:
Perpendicular line will pass through point (0,0) and (-2,9) as per given question,
Then slope of perpendicular line will be m1 = (y2 – y1)/(x2 – x1)
= (9 – 0)/(–2 – 0)
= –9/2
As we know that two non-vertical lines are perpendicular to each
other if their slopes are negative reciprocals of each other.
So, we need to find equation of line let say it AB.
Slope of line AB will be m = (–1/m1) = –1/(–9/2) = 2/9
As perpendicular line and line intersecting at (–2, 9) that means (–2, 9) lies on the line AB
By using Slope Point Form of line, we get
y – y1 = m (x – x1)
y – 9 = (2/9) (x – (–2))
9(y – 9) = 2(x + 2)
9y – 81 = 2x + 4
2x + 4 – 9y + 81 = 0
2x – 9y + 85 = 0
So, the equation of line is 2x – 9y + 85 = 0.
Вопрос 16. Длина медного стержня L (в сантиметрах) является линейной функцией его температуры по Цельсию C. В эксперименте, если L = 124,942 при C = 20 и L = 125,134 при C = 110, выразите L через С.
Решение:
Let us considered ‘L’ along X-axis and ‘C’ along Y-axis,
So, we have two points (124.942, 20) and (125.134, 110) in XY-plane.
As we know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
0.192(C-20) = 90(L – 124.942)
L can be express in Terms of C in following manner
L = (0.192 × (C – 20)/90) + 124.942
Вопрос 17. Владелец молочного магазина обнаружил, что он может продавать 980 литров молока каждую неделю по цене рупий. 14/литр и 1220 литров молока каждую неделю по цене рупий. 16/литр. Предполагая линейную зависимость между продажной ценой и спросом, сколько литров он мог бы продавать в неделю по цене рупий? 17/литр?
Решение:
Let us considered the relationship between selling price and demand is linear.
So, the selling price per liter along X-axis and demand along Y-axis,
We have two points (14, 980) and (16, 1220) in XY-plane
We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y – 980 = 120 (x – 14)
y = 120 (x – 14) + 980
When x = Rs 17/liter,
y = 120 (17 – 14) + 980
y = 120(3) + 980
y = 360 + 980 = 1340
So, the owner can sell 1340 liters weekly at Rs. 17/liter.
Вопрос 18. P (a, b) — середина отрезка между осями. Покажите, что уравнение прямой равно x/a + y/b = 2.
Решение:
Let us considered AB be a line segment whose midpoint is P (a, b).
So, the coordinates of A is (0, y) and B is (x, 0)
Mid-Point of AB will be P(a, b) = ((x + 0)/2, (y + 0)/2)
a = x/2
x = 2a
And
b = y/2
y = 2b
That means line AB is passing from (0, 2b) and (2a, 0)
We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
(y – 2b) = ((0 – 2b)/2a) × x
a (y – 2b) = –bx
ay – 2ab = –bx
bx + ay = 2ab
On dividing both the sides with ab, we get
x/a + y/b =2
Вопрос 19. Точка R(h,k) делит отрезок между осями в отношении 1:2. Найдите уравнение прямой.
Решение:
Let us consider, PQ be the line segment such that r (h, k) divides it in the ratio 1: 2.
So the coordinates of P and Q be (0, y) and (x, 0) respectively.
We know that the coordinates of a point dividing the line segment
joins the points (x1, y1) and (x2, y2) internally in the ratio m: n is
(h, k) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))
(h, k) = ((1 × 0 + 2 × x)/(1 + 2), (1 × y + 2 × 0)/(1 + 2))
h = 2x/3 and k = y/3
x = 3h/2 and y = 3k
So, P = (0, 3k) and Q = (3h/2, 0)
We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y – 3k = ((0 – 3k)/(3h/2))x
3h(y – 3k) = –6kx
3hy – 9hk = –6kx
6kx + 3hy = 9hk
On dividing both the sides by 9hk, we get,
2x/3h + y/3k = 1
So, the equation of the line is given by 2x/3h + y/3k = 1
Вопрос 20. Используя понятие уравнения прямой, докажите, что три точки (3, 0), (– 2, – 2) и (8, 2) лежат на одной прямой.
Решение:
Prove that the given three points (3, 0), (– 2, – 2) and (8, 2) are collinear,
So, we have to also prove that the line passing through the points
(3, 0) and (– 2, – 2) also passes through the point (8, 2).
The equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y – 0 = ((–2 – 0)/(–2 – 3)) × (x – 3)
–5y = –2 (x – 3)
–5y = –2x + 6
2x – 5y = 6
Checking whether (8,2) is on line 2x – 5y = 6 or not,
LHS = 2x – 5y = 2(8) – 5(2)
= 16 – 10
= 6
= RHS
So, the line passing through the points (3, 0) and (– 2, – 2)
also passes through the point (8, 2).
Hence, proved that the points (3, 0), (– 2, – 2) and (8, 2) are collinear,