Class 10 RD Sharma Solutions — Глава 14 Координатная геометрия — Упражнение 14.5 | Набор 2
Вопрос 12. Докажите, что точки (a, b), (a1, b1) и (a – a1, b – b1) лежат на одной прямой, если ab1 = a1b.
Решение:
As we know that 3 points are collinear is the area of the triangle formed by them is zero
So, let us assume ABC is a triangle whose vertices A(a, b), B(a1, b1) and C(a – a1, b – b1)
Area of triangle = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
= 1/2 [a(b1 – b + b1) + a1(b – b1 – b) + (a – a1)(b – b1)]
= 1/2 [2b1a – b – a1b1 + ab – ab1 – a1b + a1b1]
= 1/2 (ab1 – a1b)
The points are collinear
So, Area of ∆ABC = 0
1/2(ab1 – a1b) = 0
⇒ ab1 – a1b = 0
⇒ ab1 = a1b
Hence, Proved
Вопрос 13. Если треугольник имеет вершины (1, -3), (4, p) и (-9, 7), а его площадь равна 15 квадратных единиц, найдите значение (значения) p.
Решение:
Let us assume ABC is a triangle whose vertices are A(1, -3), B(4, p), and C(-9, 7)
It is given that, area of triangle = 15 sq. units
So, Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
15 = 1/2[1(p – 7) + 4(7 + 3) + (-9)(-3 – p)]
15 = 1/2[p – 7 + 40 + 27 + 9p]
15 = 1/2[10p + 60] = 5p + 30
5p + 30 = 15
5p = 15 – 30 = -15
p = -15/5 = -3
Hence, the value of p is -3
Вопрос 14. Если (x, y) лежит на прямой, соединяющей две точки (1, -3) и (-4, 2), докажите, что x + y + 2 = 0.
Решение:
Given that point (x, y) be on the line joining the two points (1, -3) and (-4, 2)
So, points (x, y), (1, -3) and (-4, 2) are collinear
Now, let us assume that the points (x, y) (1, -3) and (-4, 2) are the vertices of a triangle ABC,
So, Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
= 1/2[x(-3 – 2) + 1(2 – y) + (-4)(y + 3)]
= 1/2[-5x + 2 – y – 4y – 12]
= 1/2[-5x – 5y – 10]
= -5/2(x + y + 2)
As we know that the points are collinear
So, the area of ∆ABC = 0
⇒ =5/2 (x + y + 2) = 0
⇒ x + y + 2 = 0
Вопрос 15. Найдите значение k, если точки (k, 3), (6, -2) и (-3, 4) лежат на одной прямой.
Решение:
Let us assume ABC is a triangle whose vertices are A(k, 3), B(6, -2), and C(-3, 4)
So,
Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
= 1/2[k(-2 – 4) + 6(4 – 3) + (-3)(3 + 2)]
= 1/2[-6k + 6 * 1 + (-3 * 5)]
= 1/2[-6k + 6 – 15]
= 1/2[-6k – 9]
= -3/2 [2k + 3]
It is given that the points are collinear
So, the area of ∆ABC = 0
-3/2 [2k + 3] = 0
⇒2k = -3 ⇒ k = -3/2
Hence, the value of k is-3/2
Вопрос 16. Найдите значение k, если точки A (7, -2), B (5, 1) и C (3, 2k) лежат на одной прямой.
Решение:
Let us assume ABC is a triangle whose vertices are A (7, -2), B (5, 1), and C (3, 2k)
So,
Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
= 1/2 [7(1 – 2k) + 5(2k + 2) + 3(-2 – 1)]
= 1/2 [7 – 14k + 10k + 10 – 6 – 3]
= 1/2 [8 – 4k] = 4 – 2k
It is given that the points are collinear
So, the area of ∆ABC = 0
⇒ 4 – 2k = 0
⇒ 2k = 4
⇒ k = 2
Hence, the value of k is 2
Вопрос 17. Если точка P (m, 3) лежит на отрезке, соединяющем точки A (−2/5, 6) и B (2, 8), найдите значение m.
Решение:
It is given that points P(m, 3) lies on the line segment joining the points A (−2/5, 6) and B (2, 8)
So, points A, P, B are collinear
Now the area of area ∆APB = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
= 1/2[-2/5(3 – 8) + m(8 – 6) + 2(6 – 3)]
= 1/2[-2/5 * (-5) + 2m + 2 * 3]
= 1/2[2 + 2m + 6]
= 1/2 [2m + 8]
= m + 4
As we know that the points are collinear
So, the area of ∆APB = 0
m + 4 = 0
⇒m = -4
Hence, the value of m is -4
Вопрос 18. Если R (x, y) — точка на отрезке, соединяющем точки P (a, b) и Q (b, a), то докажите, что x + y = a + b.
Решение:
It is given that point R (x, y) lies on the line segment joining the points P (a, b) and Q (b, a)
So, points R, P, Q are collinear
Now area of ∆PRQ = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
= 1/2 [a(y – a) + x(a – b) + b(b – y)]
= 1/2 [ay – a2 + ax – bx + b2 – by]
= 1/2 [x(a – b) + y(a – b) – (a2 – b2)]
= 1/2 [x(a – b) + y(a – b) – (a + b)(a – b)]
= 1/2 (a – b)(x + y – a – b) = 0
⇒x + y – a – b = 0
⇒x + y = a + b
Hence, c + y = a + b
Вопрос 19. Найдите значение k, если точки A (8, 1), B (3, -4) и C (2, k) лежат на одной прямой.
Решение:
Let us assume ABC is a triangle whose vertices are A (8, 1), B (3, -4), and C (2, k)
So,
Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
= 1/2[8(-4 – k) + 3(k – 1) + 2(1 + 4)]
= 1/2[-32 – 8k + 3k – 3 + 10]
= 1/2 [-25 – 5k]
It is given that the points are collinear
So, the area of ∆ABC = 0
⇒ 1/2[-25 -5k] = 0
⇒ 1/2 × (-5)(5 + k) = 0
⇒ 5 + k = 0
⇒ k = -5
Hence, the value of m is -5
Вопрос 20. Найдите значение а, при котором площадь треугольника, образованного точками А (а, 2а), В (-2, 6) и С (3, 1), равна 10 квадратным единицам.
Решение:
It is given that, ABC is a triangle whose vertices are A(a, 2a), B(-2, 6), and C(3, 1)
So, Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
= 1/2[a(6 – 1) + (-2)(1 – 2a) + 3(2a – 6)]
= 1/2[5a – 2 + 4a + 6a – 18]
= 1/2[15a – 20]
It is given that the area of ∆ABC = 10 sq. units
So,
1/2[15a – 20] = 10
⇒ 15a – 20 = 20
⇒ 15a = 20 + 20 = 40
⇒a = 40/15 = 8/3
Hence, the value of a is 8/3
Вопрос 21. Если a ≠ b ≠ 0, докажите, что точки (a, a 2 ), (b, b 2 ), (0, 0) никогда не лежат на одной прямой.
Решение:
Let us assume that A(a, a2), B(b, b2), and C(0, 0) be the coordinates of the given points.
As we know that the area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is
= 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
So area of ∆ABC = 1/2[a(b2 – 0) + b(0 – a2) + 0(a2 – b2)]
= 1/2[ab2 – a2b]
= 1/2[ab(b – a)] ≠ 0
Therefore, a ≠ b ≠ 0
Hence, the area of the triangle formed by the points (a, a2), (b, b2), (0, 0) is not zero,
so the given points are not collinear
Вопрос 22. Площадь треугольника 5 кв. единиц. Две его вершины находятся в точках (2, 1) и (3, -2). Если третья вершина равна (7/2, y), найдите y.
Решение:
Let us assume ABC is a triangle whose vertices are A(2, 1), B(3, -2), and C (7/2, y)
Also, the area of ∆ABC = 5 sq.units
So,
Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]
⇒ ±5 = 1/2[2(-2 – y) + 3(y – 1) + 7/2(1 + 2)]
⇒ ±10 = [-4 – 2y + 3y – 3 + 21/2]
⇒ ±10 = [y – 7 + 21/2]
⇒ ±10 = (2y – 14 + 21)/2
⇒ 2y + 7 = ±20
⇒ 2y = 20 – 7 or 2y = -20 – 7
⇒ y = 13/2 or y = -27/2