ML Aggarwal Solutions for Chapter 11 Section Formula Class 10 Maths ICSE
Exercise 11
1. Find the co-ordinates of the mid-point of the line segments joining the following pairs of points :
(i) 2, - 3, -6, 7
(ii) 5, - 11, 4, 3
(iii) a + 3, 5b, 2a – 1, 3b + 4
Answer
(i) Co-ordinates of the mid-point of (2, -3), (- 6, 7)
{(x2 + x2)/2, (y1 + y2)/2} or
{(2 – 6)/2, (-3 + 7)/2} or
(-4/2, 4/2) or (-2, 2)
(ii) Mid-point of (5, - 11) and (4, 3)
= {(x1 + x2)/2, (y1 + y2)/2} or
{(5+4)/2, (-11+3)/2}
Or (9/2, -8/2) or (9/2, - 4)
(iii) Mid-point of (a + 3, 5b) and (2a – 1, 3b + 4)
= (x1 + x2)/2, (y1 + y2)/2
Or (a + 3 + 2a – 1)/2, (5b + 3b + 4)/2
Or (3a + 2)/2 , (8b + 4)/2
Or {(3a + 2)/2, (4b + 2)}
2. The co-ordinates of two points A and B are (- 3, 3) and (12, -7) respectively. P is a point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of P.
Answer
Points are A (-3, 3), B (12, - 7)
Let P (x1, y1) be the points which divides AB in the ratio of m1 : m2 i.e. 2 : 3 then co-ordinates of P will be
x = (m1x2 + m2x1)/(m1 + m2)
= {2×12 + 3×(-3)}/(2×3)
= (24 – 9)/5
= 15/5
= 3
y = (m1y2 + m2y1)/(m1 + m2)
= {2×(-7) + 3×3}/(2+3)
= (-14 + 9)/5
= -5/5
= -1
∴ Co-ordinates of P are (3, -1)
3. P divides the distance between A (-2, 1) and B (1, 4) in the ratio of 2 : 1. Calculate the co-ordinates of the point P.
Answer
Points are A (- 2, 1) and B (1, 4) and
Let P (x, y) divides AB in the ratio of m1 : m2 i.e., 2 : 1
Co-ordinates of P will be
x = (m1x2 + m2x1)/(m1 + m2)
= {(2×1 + 1×(-2)}/(2+1)
= (2 – 2)/3
= 0/3
= 0
y = (m1y2 + m2y1)/(m1 + m2)
= (2×4 + 1×1)/(2+1)
= (8 + 1)/3
= 9/3
= 3
∴ Co-ordinates of point P are (0, 3).
4. (i) Find the co-ordinates of the points of trisection of the line segment joining the point (3, -3) and (6, 9).
(ii) The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, - 2) and (5/3, q) respectively, find the values of p and q.
Answer
(i) Let P (x1, y1) and Q (x2, y2) be the points which intersect the line segment joining the points A (3, -3) and B (6, 9)
∵ P(x1, y1) divides AB in the ratio of 1 : 2
∴ x1 = (m1x2 + m2x1)/(m1 + m2)
= (1×6 + 2×3)/(1+2)
= (6 + 6)/3
= 12/3
= 4
y1 = (m1y2 + m2y1)/(m1 + m2)
= {(1×9 + 2×(-3)}/(1+2)
= (9 – 6)/3
= 3/3
= 1
∴ Co-ordinates of Pare (4, 1)
Again
∵ Q (x2, y2) divides the line segment
AB in the ratio of 2 : 1
∴ x2 = (m1x2 + m2x1)/(m1 + m2)
= (2×6 + 1×3)/(2+1)
= (12 + 3)/3
= 15/3
= 5
y2 = (m1y2 + m2y1)/(m1 + m2)
= {2×9 + 1(-3)}/(2+1)
= (18 – 3)/3
= 15/3
= 5
∴ Co-ordinates of Q are (5, 5)
(ii) Points P and Q trisect the line AB.
In other words, P divides it in the ratio 1 : 2 and Q divides it in the ratio 2 : 1
∴ p = (mx2 + nx1)/(m + n)
= (1×1 + 2×3)/(1+2)
= (1 + 6)/3
= 7/3
q = (my2 + ny1)/(m + n)
= {2×2 + 1×(-4)}/(2+1)
= (4 – 4)/2
= 0
∴ p = 7/3, q = 0
5. (i) The line segment joining the points A (3, 2) and B (5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k.
(ii) A point P divides the line segment joining the points A (3, -5) and B (- 4, 8) such that AP/PB = k/1. If P lies on the line x + y = 0, then find the value of k.
Answer
(i) The point P (x, y) divides the line segment joining the points A (3, 2) and B (5, 1) in the ratio 1 : 2
∴ x = (mx2 + nx1)/(m + n)
= (1 × 5 + 2 × 3)/(1+2)
= (5 + 6)/2
= 11/3
y = (my2 + ny1)/(m + n)
= (1 × 1 + 2 × 2)/(1+2)
= (1 + 4)/3
= 5/3
∵ P lies on the line 3x – 18y + k = 0
∴ It will satisfy it.
3(11/3) – 18(5/3) + k = 0
⇒ 11 – 30 + k = 0
⇒ - 19 + k = 0
⇒ k = 19
(ii) A point P divides the line segment joining the points A (3, -5), B (-4, 8) such that AP/BP = k/1
∴ Ratio = AP : PB = k : 1
Let co-ordinates of P be (x, y), then
x = (mx2 + nx1)/(m + n)
= {k×(-4) + 1×3}/(k+1)
x = (- 4k + 3)/(k+1)
and y = (8k – 5)/(k+1) {∵ (my2 + my1)/(m + n)}
= (8k – 5)/(k+1)
∵ This point lies on the line x + y = 0
∴ {(- 4k + 3)/(k + 1)} + {(8k – 5)/(k + 1)} = 0
⇒ - 4k + 3 + 8k – 5 = 0
⇒ 4k – 2 = 0
⇒ 4k = 2
⇒ k = 2/4 = 1/2
6. Find the coordinates of the point which is three-fourth of the way from A (3, 1) to B (- 2, 5).
Answer
Let P be the required point, then
AP/AB = 3/4
∴ AP/AB = 3/4
= AP/(AP + PB) = 3/4
⇒ 4AP = 3AP + 3PB
⇒ 4AP – 3AP = 3PB
AP = 3PB
AP/PB = 3/1
∴ m1 = 3, m2 = 1
Let co-ordinates of P be (x, y)
∴ x = (m1x2 + m2x1)/(m1 + m2)
= {3 × (-2) + 1 × (3)}/(3 + 1)
= (-6 + 3)/ 4
= -3/4
y = (m1y2 + m2y1)/(m1 + m2)
= (3 × 5 + 1 × 1)/(3 + 1)
= (15 + 1)/4
= 16/4
= 4
∴ Co-ordinates of P will be (-3/4, 4)
7. Point P (3 -5) is reflected to P’ in the x-axis. Also P on reflection in the y-axis is mapped as P’’.
(i) Find the co-ordinates of P’ and P’’.
(ii) Compute the distance P’ P’’.
(iii) Find the middle point of the line segment P’ P’’.
(iv) On which co-ordinate axis does the middle point of the line segment P P’’ lie ?
Answer
(i) Co-ordinates of P’, the image of P (3, -5) when reflected in x-axis will be (3, 5) and co-ordinates of P’’, the image of P (3, -5) when reflected in y-axis will be (- 3, -5) when reflected in y-axis will be (-3, - 5)
(ii)
(iii) Let co-ordinates of middle point M be (x, y)
∴ x = (x1 + x2)/2
= (3 – 3)/2
= 0/2
= 0
y = (y1 + y2)/2
= (-5 + 5)/2
= 0/2
= 0
∴ middle point is (0, 0)
(iv) Middle point of P P’’ be N (x1, y1)
∴ x1 = (3 – 3)/2 = 0/2 = 0
x2 = (- 5 – 5)/2
= -10/2
= -5
∴ Co-ordinates of middle point of PP" are (0, -5)
As x = 0, this point lies on y-axis.
8. Use graph paper for this question. Take 1 cm = 1 unit on both axes. Plot the points A (3, 0) and B (0, 4).
(i) Write down the co-ordinates of A1, the reflection of A in the y-axis.
(ii) Write down the co-ordinates of B1, the reflection of B in the x-axis.
(iii) Assign the special name to the quadrilateral ABA1B1.
(iv) If C is the mid point is AB. Write down the co-ordinates of the point C1, the reflection of C in the origin.
(iv) Assign the special name to quadrilateral ABC1B1.
Answer
Two points A (3, 0) and B (0, 4) have been plotted on the graph.
(i) ∵ A1 is the reflection of A (3, 0) in the y-axis. Its co-ordinates will be (-3, 0).
(ii) ∵ B1 is the reflection of B(0, 4) in the x-axis co-ordinates of B, will be (0, - 4).
(iii) The so formed figure ABA1B1 is a rhombus.
(iv) C is the mid point of AB co-ordinates of C’’ will be (AP/AB) = 3/4 ,
∵ C, is the reflection of C in the origin
Co-ordinates of C, will be (-3/2, -2)
(v) The name of quadrilateral ABC1B1 is a trapezium because AB is parallel to B1C1.
9. The line segment joining A (-3, 1) and B (5, -4) is a diameter of a circle whose centre is C. Find the co-ordinates of the point C.
Answer
∵ C is the centre of the circle and AB is the diameter
C is the midpoint of AB.
Let co-ordinates of C (x, y)
∴ x = (-3 + 5)/2, x = (1 – 4)/2
⇒ x = 2/2, y = -3/2
⇒ x = 1, y = -3/2
∴ Co-ordinates of C are (1, -3/2)
10.The mid-point of the line segment joining the points (3m, 6) and (-4, 3n) is (1, 2m – 1). Find the values of m and n.
Answer
Let the mid-point of the line segment joining two points A (3m, 6) and (-4, 3n) is P (1, 2m – 1)
∴ 1 = (x1 + x2)/2
= (3m – 4)/2
⇒ 3m - 4 = 2
⇒ 3m = 2 + 4 = 6
⇒ m = 6/3 = 2
And 2m – 1 = (6 + 3n)/2
⇒ 4m – 2 = 6 + 3n
⇒ 4 × 2 – 2 = 6 + 3n = 8 – 2 = 6 + 3n
⇒ 3n = 8 – 2 – 6 = 0
⇒ n = 0
Hence, m = 2, n = 0
11. The co-ordinates of the mid-point of the line segment PQ are (1, -2). The co-ordinates of P are (-3, 2). Find the co-ordinates of Q.
Answer
Let the co-ordinates of Q be (x, y) co-ordinates of P are (-3, 2) and mid-point of PQ are (1, -2) then
1 = (-3 + x)/2
⇒ - 3 + x = 2
⇒ x = 2 + 3 = 5
And -2 = (2 + y)/2
⇒ 2 + y = -4
⇒ y = - 4 – 2 = - 6
∴ x = 5, y = - 6
Hence, co-ordinates of Q are (5, -6)
12. AB is a diameter of a circle centre C (-2, 5). If point A is (3, -7). Find :
(i) the length of radius AC.
(ii) the coodinates of B.
Answer
Let co-ordinate of B are (x, y)
∴ (3 + x)/2 = -2 and (y – 7)/2 = 5
⇒ 3 + x = - 4 and y – 7 = 10
⇒ x = - 4 – 3 and y = 10 + 7
⇒ x = - 7 and y = 17
∴ B is (-7, 17)
13. Find the reflection (image) of the point (5, -3) in the point (-1, 3)
Answer
Let the co-ordinates of the images of the point A (5, -3) be
A1 (x, y) in the point (-1, 3) then the point (-1, 3) will be midpoint of AA1.
∴ - 1 = (5 + x)/2
⇒ 5 + x = - 2
x = - 2 – 5 = - 7
and 3 = (-3 + y)/2
⇒ - 3 + y = 6
⇒ y = 6 + 3 = 9
∴ Co-ordinates of the image A, will be (-7, 9).
14. The line segment joining A (- 1, 5/3) the points B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Calculate:
(i) the value of a
(ii) the co-ordinates of P.
Answer
Let P (x, y) divides the line segment joining the points (-1, 5/3), B (a, 5) in the ratio 1 : 3
∴ x = {1×a + 3×(-1)}/(1+3)
= (a – 3)/4
y = {1×a + 3×(-1)}/(1+3)
= (a – 3)/4
= (5 + 5)/4
= 10/4
= 5/2
(i) ∵ AB intersects y-axis at P
∴ x = 0
⇒ (a – 3)/4 = 0
⇒ a – 3 = 0
∴ a = 3
(ii) ∴ Co-ordinates of P are (0, 5/2)
15. The point P (- 4, 1) divides the line segment joining the points A (2, -2) and B in the ratio of 3 : 5. Find the point B.
Answer
Let the co-ordinates of B be (x, y)
Co-ordinates of A (2, -2) and point P (-4, 1) divides AB in the ratio of 3 : 5
∴ -4 = {3×x + 5×(2)}/(3+5)
= (3x +10)/8
And 3x + 10 = - 32
⇒ 3x = - 32 – 10
= - 42
∴ x = - 42/3 = - 14
l = {3×y + 5×(-2)}/(3+5)
⇒ l = (3y – 10)/8
⇒ 3y – 10 = 8
⇒ 3y = 8 + 10
= 18
∴ y = 18/3 = 6
∴ Co-ordinates of B = (- 14, 6)
16. (i) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7, 6) ?
(ii) In what ratio does the point (- 4, b) divide the line segment joining the points P (2, -2), Q (-14, 6) ? Hence find the value of b.
Answer
(i) Let the ratio be m1 : m2 that the point (5, 4) divides the line segment joining the points (2, 1), (7, 6)
5 = (m1×7 + m2×2)/(m1 + m2)
⇒ 5m1 + 5m2 = 7m1 + 2m2
⇒ 5m2 – 2m2 = 7m1 – 5m1
⇒ 3m2 = 2m1
⇒ m1/m2 = 3/2
⇒ m1 : m2 = 3 : 2
(ii) The point (- 4, b) divides the line segment joining the points P (2, -2) and Q (-14, 6) in the ratio m1 : m2.
∴ -4 = {m1(-14) + m2×2}/(m1+m2)
⇒ - 4 m1 – 4m2 = - 14 m1 + 2 m2
⇒ - 4m1 + 14 m1 = 2m2 + 4m2
⇒ 10m1 = 6 m2
⇒ m1/m2 = 6/10 = 3/5
⇒ m1 : m2 = 3 : 5
Again b = {m1×6 + m2×(-2)}/(m1+m2)
= (6m1 – 2m2)/(m1+m2)
⇒ b = (6×3 – 2×5)/(3+5)
= (18 – 10)/8
= 8/8
= 1
∴ b = 1
17. The line segment joining A (2, 3) and B (6, -5) is intercepted by the x-axis at the point K. Write the ordinate of the point k. Hence, find the ratio in which K divides AB. Also, find the coordinates of the point K.
Answer
Let the co-ordinates of K be (x, 0) as it intersects x-axis.
Let point K divides the line segment joining the points A (2, 3) and B (6, -5) in the ratio m1 : m2.
∴ 0 = (m1y2 + m2y1)/(m1+m2)
⇒ 0 = {m1×(-5) + m2×3}/(m1+m2)
⇒ -5m1 + 3m2 = 0
⇒ -5m1 = -3m2
⇒ m1/m2 = 3/5
⇒ m1 : m2 = 3 : 5
Now,
x = (m1x2 + m2x1)/(m1 + m2)
= (3×6 + 5×2)/(3 + 5)
= (18 + 10)/8
= 28/8
= 7/2
Co-ordinate of K are (7/2, 0)
18. If A (-4, 3) and B (8, -6),
(i) find the length of AB.
(ii) In what ratio is the line joining AB, divided by the x-axis.
Answer
Given A (- 4, 3), B (8, -6)
Let O divides AB in the ratio m1 : m2
∴ x = (m1x2 + m2x1)/(m1+m2)
⇒ 0 = {m1×8 + m2(-4)}/(m1 + m2)
⇒ 8m1 - 4m2 = 0
⇒ 8m1 = 4m2
⇒ m1/m2 = 4/8 = 1/2
∴ m1 : m2 = 1 : 2
∴ O, divides AB in the ratio 1 : 2
19. (i) Calculate the ratio in which the line segment joining (3, 4) and (- 2, 1) is divided by the x-axis.
(ii) In what ratio does the line x – y – 2 = 0 divide the line segment joining the points (3, - 1) and (8, 9) ? Also, find the coordinates of the point of division.
Answer
(i) Let the point P divided the line segment joining the points A (3, 4) and B (-2, 3) in the ratio of m1 : m2 and
Let the co-ordinates of P be (0, y) as it intersects the y-axis
∴ 0 = (m1x2 + m2x1)/(m1 + m2)
⇒ 0 = {m1(-2) + m2×3}/(m1 + m2)
⇒ 0 = - 2m1 + 3m2
⇒ 2m1 = 3m2
⇒ m1/m2 = 3/2
⇒ m1 : m2 = 3 : 2
(ii) Let the points be A (3, -1) and B (8, 9) and let line x – y – 2 = 0 divides the line segment joining the points A and B in the ratio m1 : m2 at point P (x, y) then
x = (m1x2 + m2x1)/(m1+m2)
= (m1×8 + m2×3)/(m1+m2)
and y = (m1y2 + m2y1)/(m1+m2)
= {m1×9 + m2(-1)}/(m1+m2)
= (9m1 – m2)/(m1+m1)
∵ The point P (x, y) lies on the line x - y – 2 = 0
∴ (8m1+3m2)/(m1+m2) – (9m1 – m2)/(m1+m2) – 2 = 0
⇒ 8m1 + 3m2 – 9m1 + m2 – 2m1 – 2m2 = 0
⇒ - 3m1 + 2m2 = 0
⇒ 3m1= 2m2
⇒ m1/m2 = 2/3
(i) ∵ Ratio = m1 : m2 = 2 : 3
∴ x = (2×8 + 3×3)/(2+3)
= (16 + 9)/5
= 25/5
= 5
And y = {2×9 + 3×(-1)}/(2+3)
= (18 – 3)/5
= 15/5
= 3
(ii) ∴ Co-ordinates of point P are (5, 3)
20. Given a line segment AB joining the points A (- 4, 6) and B (8, - 3). Find :
(i) the ratio in which AB is divided by the y-axis.
(ii) find the co-ordinates of the point of intersection.
(iii) the length of AB.
Answer
(i) Let the y-axis divide AB in the ratio m : 1. So,
0 = (m×8 – 4×1)/(m+1)
⇒ 8m – 4 = 0
⇒ m = 4/8
⇒ m = 1/2
So, required ratio = 1/2 : 1 or 1 : 2
(ii) Also, y = {1×(-3) + 2×6}/(1+2)
= 9/3
= 3
So, coordinates of the point of intersection are (0, 3)
(iii)
21. (i) Write down the co-ordinates of the point P that divides the line joining A (- 4, 1) and B (17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP where O is the origin.
(iii) In what ratio does the y-axis divide the line AB ?
Answer
(i) Let co-ordinate of P be (x, y) which divides the line segment joining the points
A (- 4, 1) and B (17, 10) in the ratio of 1 : 2.
∴ x = (m1x2 + m2x1)/(m1+m2)
= (1×17 + 2×(-4)}/(1+2)
= (17–8)/3
= 9/3
= 3
y = (m1y2 + m2y1)/(m1 + m2)
= (1×10 + 2×1)/(1+2)
= (10 + 2)/3
= 12/3
= 4
∴ Co-ordinates of P are (3, 4)
(ii) Distance of OP where O is the origin i.e., co-ordinates are(0, 0)
(iii) Let y-axis divides AB in the ratio of m1 : m2 at P and let co-ordinates of P be (0, y)
0 = (m1x2 + m2x1)/(m1+m2)
⇒ 0 = {m1×17 + m2×(-4)}/(m1+m2)
⇒ 17m1 – 4m2 = 0
⇒ 17m1 = 4m2
⇒ m1/m2 = 4/17
⇒ m1 : m2 = 4 : 17
22. Calculate the length of the median through the vertex A of the triangle ABC with vertices A (7, -3), B (5, 3) and C (3, -1)
Answer
Let D (x, y) be the median of ∆ABC through A to BC.
∴ D will be the midpoint of BC
∴ Co-ordinates of D will be,
x = (5 + 3)/2 = 8/2 = 4
and y = (3 – 1)/2 = 2/2 = 1
Co-ordinates of D are (4, 1)
23. Three consecutive vertices of a parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0). Find the fourth vertex D.
Answer
Let O in the mid-point of AC the diagonal of ABCD
∴ Co-ordinates of O will be {(1 + 4)/2, (2 + 0)/2} or (5/2, 1)
∵ OA also find the mid point of second diagonal BD and let co-ordinates of D be (x, y)
∴ 5/2 = (1 + x)/2
⇒ 10 = 2 + 2x
⇒ 2x = 10 – 2 = 8
∴ x = 8/2 = 4
And l = (0 + y)/2
⇒ y = 2
∴ Co-ordinates of D are (4, 2)
24. If the points A (-2, -1), B(1, 0), C (p, 3) and D (1, q) from a parallelogram ABCD, find the values of p and q.
Answer
A (- 2, -1), B (1, 0), C (p, 3) and D (1, q) are the vertices of a parallelogram ABCD
∴ Diagonal AC and BD bisect each other at O
O is the midpoint of AC as well as BD
Let co-ordinates of O be (x, y)
When O is mid-point of AC, then
Again when O is the mid-point of BD
Then x = (1 + 1)/2 = 2/2 = 1
And y = (0 + q)/2 = q/2
Now comparing, we get
(p – 2)/2 = 1
⇒ p – 2 = 2
⇒ p = 2 + 2 = 4
∴ p = 4 and q/2 = 1
⇒ q = 2
Hence p = 4, q = 2
25. If two vertices of a parallelogram are (2, 3) (-1, 0) and its diagonals meet at (2, -5), find the other two vertices of the parallelogram.
Answer
Two vertices of a || gm ABCD are A (3, 2), B (-1, 0) and point of intersection of its diagonals is P (2, - 5)
P is mid-point of AC and BD.
Let co-ordinates of C be (x, y), then
2 = (x + 3)/2
⇒ x + 3 = 4
⇒ x = 4 – 3 = 1
And – 5 = (y + 2)/2
⇒ y + 2 = - 10
⇒ y = - 10 – 2 = - 12
∴ Co-ordinates of C are (1, - 12)
Similarly we shall find the co-ordinates of D also
2 = (x – 1)/2
⇒ x – 1 = 4
⇒ x = 4 + 1 = 5
- 5 = (y + 0)/2
⇒ - 10 = y
∴ Co-ordinates of D are (5, - 10)
26. Prove that the points A (- 5, 4), B (- 1, - 2) and C (5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square.
Answer:
Points A (-5, 4), B (- 1, -2) and C (5, 2) are given.
If these are vertices of an isosceles ABC then
AB = BC
∵ AB = BC
∴ ∆ABC is an isosceles triangle.
27. Find the third vertex of a triangle if its two vertices are (- 1, 4) and (5, 2) and mid point of one sides is (0, 3).
Answer
Let A (- 1, 4) and B (5, 2) be the two points and let D (0, 3) be it’s the midpoint of AC and co-ordinates of C be (x, y).
∴ 0 = (x – 1)/2
⇒ x – 1 = 0
⇒ x = 1
3 = (y + 4)/2
⇒ y + 4 = 6
⇒ y = 6 – 4 = 2
∴ Co-ordinates of will be (1, 2)
If we take mid-point D (0, 3) of BC, then
0 = (5 + x)/2
⇒ x + 5 = 0
⇒ x = - 5
and 3 = (2 + y)/2
⇒ 2 + y = 6
⇒ y = 6 – 2 = 4
∴ Co-ordinates of C will be (-5, 4)
Hence co-ordinates of C, third vertex will be (1, 2) or (5, -4)
28. Find the coordinates of the vertices of the triangle the middle points of whose sides are (0, 1/2), (1/2, 1/2) and (1/2, 0)
Answer
Let ABC be a ∆ in which D (0, 1/2), E = (1/2, 1/2) and F = (1/2, 0), the mid-points of sides AB, BC and CA respectively.
Let co-ordinates of A be (x1, y1), B (x2, y2), C (x3, y3)
(x1 + x2)/2
⇒ x1 + x2 = 0 …(i)
1/2 = (y1 + y2)/2
⇒ y1 + y2 = 1 …(ii)
Again 1/2 = (x2 + x3)/2
⇒x2 + x3 = 1 …(iii)
And 1/2 = (y2 + y3)/2
⇒y2 + y3 = 1 …(iv)
And 1/2 = (x3 + x1)/2
⇒ x3 + x1 = 1 …(v)
0 = (y3 + y1)/2
⇒ y3 + y1 = 0 …(vi)
Adding (i), (iii) and (v)
2(x1 + x2 + x3) = 0 + 1 + 1 = 2
∴ x1 + x2 + x3 = 1
Now substituting (iii), (v) and (i) respectively, we get
x1 = 0, x2 = 0, x3 = 1
Again Adding (ii), (iv) and (vi)
2 (y1 + y2 + y3) = 1 + 1 + 0 = 2
∴ y1 + y2 + y3 = 1
Now subtracting (iv), (vi) and (ii) respectively we get,
y1 = 0, y2 = 1, y3 = 0
∴ Co-ordinates of A, B and C will be (0, 0), (0, 1) and (1, 0)
29. Show by section formula that the points (3, -2), (5, 2) and (8, 8) are collinear.
Answer
Let the point (5, 2) divides the line joining the points (3, -2) and (8, 8) in the ratio of m1 : m2.
∴ 5 = (m1 × 8 + m2 × 3)/(m1 + m2)
⇒ 8m1 + 3m2 = 5m1 + 5m2
⇒ 8m1 – 5m1
⇒ 5m2 – 3m2
⇒ 3m1 = 2m2
⇒ m1/m2 = 2/3 ….(i)
Again 2 = (8m1 – 2m2)/(m1+m2)
⇒ 8m1 – 2m2 = 2m1 + 2m2
⇒ 8m1 – 2m1 = 2m2 + 2m2
⇒ 6m1 = 4m2
⇒ m1/m2 = 4/6 = 2/3 …(ii)
From (i) and (ii) it is clear that point (5, 2) lies on the line joining the points (3, - 2) and (8, 8).
Hence, proved.
30. Find the value of p for which the points (-5, 1), (1, p) and (4, -2) are collinear.
Answer
Let points A (- 5, 1), B (1, p) and C (4, -2) are collinear and let point A (-5, 1) divides BC in the the ratio in m1 : m2
∴ x = (m1x2 + m2x1)/(m1 + m2)
⇒ -5 = (m1×4 + m1×1)/(m1 + m2) = (4m1 + m2)/(m1 + m2)
⇒ - 5m1 – 5m2 = 4m1 + m2
⇒ - 5m1 – 4m1 = m2 + 5m2
⇒ - 9m1 = 6m2
⇒ m1/m2 = 6/-9 = 2/-3 …(i)
And {m1×(-2) + m2×p}/(m1+m2)
= (-2m1 + m2p)/(m1+m2)
⇒ m1 + m2 = -2m1 + m2p
⇒ m1 + 2m1 = m2p – m2
⇒ 3m1 – m2 (p – 1)
⇒ m1/m2 = (p – 1)/3 …(ii)
From (i) and (ii)
(p – 1)/3 = 2/(-3)
⇒ - 3p + 3 = 6
⇒ - 3p = 6 – 3
⇒ - 3p = 3
⇒ p = 3/-3
= - 1
31. A (10, 5), B (5, -3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM = 1/2 BC.
Answer
Co-ordinates of L will be
{(10 + 6)/2, (5 – 3)/2} or (16/2, 2/2) or (8, 1)
Co-ordinates of M will be ={(10 + 2)/2, (5 + 1)/2} or (12/2, 6/2) or (6, 3)
From (i) and (ii)
LM = 1/2 BC
32. A (2, 5), B (- 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.
(i) Find the co-ordinates of P and Q
(ii) Show that PQ = 1/3 BC.
Answer
A (2, 5), B (-1, 2) and C (5, 8) are the vertices of a ∆ABC,
P and Q are points on AB
And AC respectively such that AP/PB = AQ/QC = 1/2
Let co-ordinates of P be (x1, y1) and of Q be (x2, y2)
∵ P divides AB in the ratio 1 : 2
∴ x1 = (m1x2 + m2x1)/(m1 + m2)
= {1×(-1) + 2×2}/(1 + 2)
= (- 1 + 4)/3
= 3/3
= 1
y1 = (m1y2 + m2y1)/(m1 + m2)
= (1×2 + 2×5)/(1+2)
= (2 + 10)/3
= 12/3
= 4
∴ Co-ordinates of P will be (1, 4)
Similarly Q divides AC in the ratio 1 : 2
∴ x2 = (m1x2 + m2x1)/(m1 + m2)
= (1×5 + 2×2)/(1 + 2)
= (5 + 4)/3
= 9/3
= 3
And y2 = (m1y2 + m2y1)/(m1 + m2)
= (1×8 + 2×5)/(1+2)
= (8+10)/3
= 18/3
= 6
∴ Co-ordinates of Q will be (3, 6)
(ii)
= (6√2)/3
= BC/3
= 1/3 BC
33. The mid-point of the line segment AB shown in the adjoining diagram is (4, - 3). Write down die co-ordinates of A and B.
A lies on x-axis and B on the y-axis.
Let co-ordinates of A be (x, 0) and of B be (0, y)
P (4, -3) is the mid-point of AB
∴ 4 = (x + 0)/2
⇒ x = 8
And – 3 = (0 + y)/2
⇒ y = - 6
Co-ordinates of A will be (8, 0) and of B will be (0, - 6)
34. Find the co-ordinates of the centroid of a triangle whose vertices are A (- 1, 3), B (1, - 1) and C (5, 1)
Answer:
Co-ordinates of the centroid of a triangle, whose vertices are (x1, y1), (x2, y2) and (x3, y3) are {(x1 + x2 + x3)/3 , (y1 + y2 + y3)/3}
∴ Co-ordinates of the centroid of the given triangle are {(-1 + 1 + 5)/3, (3 – 1 + 1)/2} i.e., (5/3, 1)
35. Two vertices of a triangle are (3, -5) and (-7, 4). Find the third vertex given that the centroid is (2, -1)
Answer:
Let the co-ordinates of third vertices be (x, y)
And other two vertices are (3, -5) and (- 7, 4)
And centroid = (2, 1)
∴ 2 = (3 – 7 + x)/3
⇒ (x – 4)/3 = 2
x – 4 = 6
⇒ x = 6 + 4
⇒ x = 10
And ⇒ - 1 = (- 5 + 4 + y)/3
⇒ - 3 = - 1 + y
⇒ y = - 3 + 1
= 2
∴ Co-ordinates are (10, - 2)
36. The vertices of a triangle are A (- 5, 3), B (p – 1) and C (6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, - 1).
Answer:
The vertices of ∆ABC are A (- 5, 3), B (p, -1), C (6, q) and the centroid of ∆ABC is O (1, - 1)
Co-ordinates of the centroid of ∆ ABC will be [(-5 + p + 6)/3, (3–1+q)/3]
⇒ {(1+p)/3, (2+q)/3}
But centroid is given (1, -1)
∴ Comparing, we get (1 + p)/3 = 1
⇒ 1 + p = 1
⇒ 1 + p = 3
⇒ p = 3 – 1 = 2
And (2 + q)/3 = - 1
⇒ 2 + q = - 3
⇒ q = - 3 – 2
⇒ q = - 5
Hence, p = 2, q = - 5
Multiple Choice Questions
Choose the correct answer from the given four options (1 to 12):
1. The points A (9, 0), B (9, 6), C(- 9, 6) and D (- 9, 0) are the vertices of a
(a) rectangle
(b) square
(c) rhombus
(d) trapezium
Answer
(a) rectangle
A(9, 0), B(9, 6), C(-9, 6), D(-9, 0)
AB2 = (x2 – x1)2 + (y2 – y1)2
= (9 – 9)2 + (6 – 0)2
= 02 + 62
= 02 + 36
= 36
CD2 = (-9 + 9)2 + (6 – 0)2
= 02 + 62
= 0 + 36
= 36
BC2 = (9 + 9)2 + (6 – 6)2
= 182 + 02
= 324 + 0
= 324
AD2 = (9 + 9)2 + (0)2
= 182 + 02
= 324 + 0
= 324
AB = CD and BC = AD
But these are opposite sides of a rectangle
ABCD is a rectangle.
2. The mid-point of the line segment joining the points A (- 2, 8) and B (- 6, - 4) is
(a) (-4, -6)
(b) (2, 6)
(c) (-4, 2)
(d) (4, 2)
Answer
(b) (2, 6)
Mid-point of the line segment joining the points A (- 2, 8), B (- 6, - 4)
= {(-2+6)/2, (8+4)/2}
= (4/2, 12/2)
= (2, 6)
3. If P (a/3, 4) segment joining the points Q (-6, 5) and R (-2, 3), the value of a is
(a) –4
(b) –6
(c) 12
(d) –12
Answer
(d) -12
P (a/3, 4) is mid-point of the line segment joining the points Q (-6, 5) and R (-2, 3)
∴ a/3 = (-6-2)/2
= -8/2 = -4
a = -4 × 3
⇒ a = -12
4. If the end points of a diameter of a circle are A (- 2, 3) and B (4, - 5), then the coordinates of its centre are
(a) (2, -2)
(b) (1, -1)
(c) (-1, 1)
(d) (-2, 2)
Answer
(b) (1, -1)
End points of a diameter of a circle are (- 2, 3) and B (4, - 5) then co-ordinates of the centre of the circle = {(- 2 + 4)/2, (3 – 5)/2} or (2/2, -2/2)
= (1, -1)
5. If one end of a diameter of a circle is (2, 3) and the centre is (- 2, 5), then the other end is
(a) (-6, 7)
(b) (6, -7)
(c) (0, 8)
(d) (0, 4)
Answer
(a) (-6, 7)
One end of a diameter of a circle is (2, 3) and center is (-2, 5)
Let (x, y) be the other end of the diameter (2 + x)/2 = - 2
⇒ 2 + x = - 4
⇒ x = -4-2 = - 6
And (3+y)/2 = 5
⇒ 3 + y = 10
⇒ y = 10 – 3
= 7
∴ Co-ordinates of other end are (-6, 7)
6. If the mid-point of the line segment joining the points P (a, b–2) and Q (-2, 4) is R (2, -3), then the values of a and b are
(a) a = 4, b = -5
(b) a = 6, b = 8
(c) a = 6, b = -8
(d) a = -6, b = 8
Answer
(c) a = 6, b = -8
The mid-point of the line segment joining the points P (a, b – 2) and Q (- 2, 4) is R (2, -3)
2 = (a – 2)/2
⇒ a – 2 = 4
⇒ a = 4 + 2 = 6
- 3 = (b – 2 + 4)/2 = (b + 2)/2
⇒ b + 2 = - 6
⇒ b = - 6 – 2 = - 8
∴ a = 6, b = - 8
7. The point which lies on the perpendicular bisector of the line segment joining the points A (- 2, - 5) and B (2, 5) is
(a) (0, 0)
(b) (0, 2)
(c) (2, 0)
(d) (- 2, 0)
Answer
(a) (0, 0)
The line segment joining the points A (- 2, - 5) and B (2, - 5), has mid-point = {(-2 + 2)/2, (-5 + 5)/2} = (0, 0)
(0, 0) lies on the perpendicular bisector of AB.
8. The coordinates of the point which is equidistant from the three vertices of ∆AOB (shown in the given figure) are
(a)
(b) (y, x)
(c) (x/2, y/2)
(d) (y/2, x/2)
(a) (x, y)
In the given figure, vertices of a ∆OAB are (0, 0), (0, 2y) and (2x, 0)
The point which is equidistant from O, A and B is the mid-point of AB.
∴ Coordinates are {(0 + 2x)/2, (2y + 0)/2} or (x, y)
9. The fourth vertex D of a parallelogram ABCD whose three vertices are A (- 2, 3), B (6, 7) and C (8, 3) is
(a) (0, 1)
(b) (0, - 1)
(c) (- 1, 0)
(d) (1, 0)
Answer
(b) (0, -1)
ABCD is a || gm whose vertices A (- 2, 3), B (6, 7) and C (8, 3)
The fourth vertex D will be the point on which diagonals AC and BD bisect each other at O.
∴ Co-ordinates of O are (-2+8)/2, (3+3)/2 or (6/2, 6/2) or (3, 3)
Let co-ordinates of D be (x, y), then
3 = (x + 6)/2 = 6
= x + 6
⇒ x = 6 – 6 = 0
And 3 = (y + 7)/2
⇒ y + 7 = 6
⇒ y = 6 – 7 = - 1
∴ Co-ordinates of D are (0, - 1)
10. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, - 5) is the mid-point of PQ, then the coordinates of P and Q respectively.
(a) (0, - 5) and (2, 0)
(b) (0, 10) and (- 4, 0)
(c) (0, 4) and (- 10, 0)
(d) (0, - 10) and (4, 0)
Answer
(d) (0, - 10)
A line intersects y-axis at P and x-axis a Q.
R (2, - 5) is the mid-point
Let co-ordinates of P be (0, y) and of Q be (x, 0), then
2 = (0 + x)/2
⇒ x = 4
And – 5 = (y + 0)/2
⇒ y = - 10
∴ Co-ordinates of P are (4, 0) and of Q are (0, - 10)
11. The points which divides the line segment joining the points (7, - 6) and (3, 4) in the ratio 1 : 2 internally lies in the
(a) Ist quadrant
(b) IInd quadrant
(c) IIIrd quadrant
(d) IVth quadrant
Answer
(d) IVth quadrant
A point divides line segment joining the points
A (7, - 6) and B (3, 4) in the ratio 1 : 2 internally.
Let (x, y) divides it in the ratio 1 : 2
∴ x = (mx2 + nx1)/(m + n)
= (1 × 3 + 2 × 7)/(1 + 2)
= (3 + 14)/3
= 17/3
y = (my2 + ny1)/(m + n)
= (1 × 4 + 2 × (-6)/(1 + 2)
= (4 – 12)/3
= -8/3
We see that x is positive and y is negative.
∴ It lies in the fourth quadrant.
12. The centroid of the triangle whose vertices are (3, - 7), (- 8, 6) and (5, 10) is
(a) (0, 9)
(b) (0, 3)
(c) (1, 3)
(d) (3, 3)
Answer
(b) (0,3)
Centroid of the triangle whose vertices are (3, -7), (-8, 6) and (5, 10) is {(3–8+5)/3, (-7+6+10)/3}
or (0, 9/3)
or (0, 3)
Chapter Test
1. The base BC of an equilateral triangle ABC lies on y-axis. The co-ordinates of the point C are (0, - 3). If origin is the mid-point of the base BC, find the coordinates of the point A and B.
Answer
Base BC of an equilateral ∆ABC lies on y-axis co-ordinates of point C are (0, - 3) origin (0, 0) is the mid-point of BC.
Let co-ordinates of B be (x, y)
∴ 0 = (x + 0)/2
⇒ x/2 = 0
⇒ x = 0
(y – 3)/2 = 0
⇒ y – 3 = 0
⇒ y = 3
∴ Co-ordinates of B are (0, 3)
Again let co-ordinates of A be (x, 0) as it lies on x-axis.
∵ AB = AC = BC = 6 units
x2 + (- 3)2 = 62
x2 + 9 = 36
⇒ x2 = 36 – 9 = 27
⇒ x = ± 3√3
∴ Co-ordinates of A will be (±3√3, 0)
2. A and B have co-ordinates (4, 3) and (0, 1), find
(i) the image A’ of A under reflection in the y-axis.
(ii) the image of B’ of B under reflection in the line AA’.
(iii) the length of A’B’.
Answer
(ii) Co-ordinates of B’ the image of B (0, 1) reflected in the line AA’ will be (0, 5)
(iii)
3. Find the co-ordinates of the point that divides the line segment joining the points P (5, - 2) and Q (9, 6) internally in the ratio of 3 : 1.
Answer
Let R be the point whose co-ordinates are (x, y) which divides PQ in the ratio of 3 : 1.
∴ x = (m1x2 + m2x1)/(m1 + m2)
= (3×9 + 1×5)/(3 + 1)
= (27+5)/4
= 32/4
= 8
y = (m1y2 + m2y1)/(m1 + m2)
= {3×6 + 1×(-2)}/(3+1)
= (18–2)/4
= 16/4
= 4
∴ Co-ordinates of R will be (8, 4)
4. Find the coordinates of the point P which is three-fourth of the way from A (3, 1) to B (- 2, 5).
Answer
Co-ordinates of A (3, 1) and B (- 2, 5)
P lies on AB such that
AP = 3/4 AB = 3/4 (AP + PB)
⇒ AP = 3PB
⇒ AP : PB = 3 : 1
Let co-ordinates of P be (x, y)
∴ x = (mx2 + nx1)/(m + n) = {(3 × (-2) + 1 × 3}/(3 + 1)
= (-6 + 3)/4
= -3/4
y = (my2 + ny1)/(m + n)
= (3×5 + 1×1)/(3+1)
= (15+1)/4
= 16/4
= 4
∴ Co-ordinates of Pare (-3/4, 4)
5. P and Q are the points on the line segment joining the points A (3, -1) and B (- 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.
Answer
Given
AP = PQ = QB
∴ P divides AB in the ratio of 1 : 2 and Q divides it in 2 : 1.
Let co-ordinates of P will be (x1, y1) and of Q will (x2, y2)
∴ x1 = (m1x2 + m2x1)/(m1 + m2)
= {1×(-6) + 2×3}/(1+2)
= (- 6 + 6)/3
= 0/3
= 0
y1 = (m1y2 + m2y1)/(m1 + m2)
= {1×5 + 2(-1)}/(1+2)
= (5-2)/3
= 3/3
= 1
∴ Co-ordinates of P will be (0, 1)
Again
x2 = (m1x2 + m2x1)/(m1 + m2)
= {2×(-6) + 1×3}/(2+1)
= (-12+3)/3
= -9/3
= - 3
y2 = (m1y2 + m2y1)/(m1 + m2)
= {2×5 + 1×(-1)}/(2+1)
= (10–1)/3
= 9/3
= 3
∴ Co-ordinates of Q will be (-3, 3).
∴ Co-ordinates of Q will be (-3, 3).
6. The center of a circle is (a + 2, a – 5). Find the value of a given that the circle passes through the points (2, - 2) and (8, -2).
Answer
Let A (2, -2), B (8, - 2) and center of the circle be O (𝝰 + 2, 𝝰 – 5)
∵ OA = OB = radii of the same will
Squaring both sides,
𝝰2 + (𝝰 + 1)2 = (6 – 𝝰)2 = (1 + a)2
⇒ a2 = (6 – 𝝰)2 [dividing by (𝝰 + 1)2]
⇒ 𝝰2 = 36 - 12𝝰 + 𝝰2
= 𝝰2 – 𝝰2 + 12𝝰
= 36
⇒ 12𝝰 = 36
∴ = 36/12 = 3
7. The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5) . Calculate the numerical values of p and q.
Answer
Given,
(3, 5) is the mid-point of A (2, p) and B (q, 4)
∴ 3 = (2 + q)/2
⇒ 2 + q = 6
⇒ q = 6 – 2 = 4
∴ q = 4
And 5 = (p + 4)/2
⇒ p + 4 = 10
⇒ p = 10 – 4 = 6
∴ p = 6
Hence, p = 6, q = 4
8. The ends of a diameter of a circle have the co-ordinates (3, 0) and (- 5, 6). PQ is another diameter where Q has the coordinates (- 1, - 2). Find the co-ordinates of P and the radius of the circle.
Answer
Let AB be the diameter where co-ordinates of A are (3, 0) and of B are (- 5, 6).
Co-ordinates of its origin O will be {(3 – 5)/2, (0 + 6)/2} or (-2/2, 6/2) or (- 1, 3)
Now PQ is another diameter in which co-ordinates of Q are (-1, -2).
Let co-ordinates of P be (x, y)
Then co-ordinates of center O will be {(- 1 + x)/2, (-2 + y)/2}
∴ (-1+x)/2 = - 1
⇒ - 1+x = - 2
⇒ x = -2 + 1 = - 1
And (- 2 + y)/2 = 3
⇒ - 2 + y = 6
⇒ y = 6 + 2 = 8
∴ Co-ordinates of P will be (-1, 8)
9. In what ratio does the point (- 4, 6) divide the line segment joining the points A (- 6, 10) and B (3, - 8) ?
Answer
Let the point (- 4, 6) divides the line segment joining the points
A (- 6, 10) and B (3, - 8), in the ratio m : n
∴ - 4 = (mx2 + nx1)/(m + n) = {m × 3 + n(-6)}/(m + n)
- 4 = (3m – 6n)/(m + n)
⇒ - 4m – 4n = 3m – 6n
⇒ - 4n + 6n = 3m + 4m
⇒ 7m = 2n
⇒ m/n = 2/7
∴ Ratio = 2 : 7
10. Find the ratio in which the point P (- 3, p) divides the line segment joining the points (- 5, - 4) and (-2, 3). Hence find the value of p.
Answer
Let P (- 3, p) divides AB in the ratio of m1 : m2 coordinates of A (- 5, - 4) and B (- 2, 3)
∴ - 3 = (m1x2 + m2x1)/(m1 + m2)
⇒ - 3 {m1(-2) + m2(-5)}/(m1 + m2)
⇒ - 3 = (- 2m1 – 5m2)/(m1 + m2)
⇒ - 3m1 – 3m2 = - 2m1 – 5m2
⇒ - 3m1 + 2m1 = - 5m2 + 3m2
⇒ - m1 = - 2m2
⇒ 2m2 = m1
⇒ m1/m2 = 2/1
⇒ m1: m2 = 2 : 1
Again,
p = (m1y2 + m2y1)/(m1 + m2)
= {(2×3 + 1×(-4)}/(2 + 1)
= (6 – 4)/3
= 2/3
Hence, p = 2/3
11. In what ratio is the line joining the points (4, 2) and (3, - 5) divided by the x-axis? Also find the co-ordinates of the point of division.
Answer
Let the point P which is on x-axis, divides the line segment
Joining the points A (4, 2) and B (3, - 5) in the ratio of m1 : m2.
and let co-ordinates of P be (x, 0)
∴ 0 = (m1y2 + m2y1)/(m1 + m2)
= {m1(-5) + m2(2)}/(m1 + m2)
⇒ (- 5m1 + 2m2)/(m1 + m2) = 0
⇒ - 5m1 + 2m2 = 0
⇒ - 5m1 = - 2m2
⇒ 5m1 = 2m2
⇒ m1/m2 = 2/5
⇒ m1 : m2 = 2 : 5
Again,
x = (m1x2 + m2x1)/(m1 + m2)
= {2×3 + 5×4}/(2+5)
= (6 + 20)/7
= 26/7
∴ Co-ordinates of P will be (26/7, 0)
12. If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points (- 4, - 3) and (6, 3). Hence, find the co-ordinates of P.
Answer
Let co-ordinates of A be (- 4, 3) and of B (6, 3) and of B (6, 3) and P be (2, y)
Let the ratio in which the P divides AB be m1 : m2
⇒ 2 = (m1×6 + m2×(-4)}/(m1+m2)
⇒ 2 = (6m1 – 4m2)/(m1 + m2)
⇒ 2m1 + 2m2 = 6m1 – 4m2
⇒ 6m1 – 2m1 = 2m2 + 4m2
⇒ 4m1 = 6m2
⇒ m1/m2 = 6/4 = 3/2
∴ m1 : m2 = 3 : 2
∴ y = (m1y2 + m2y1)/(m1+m2)
= (3×3 + 2×3)/(3+2)
= (9 + 6)/5
= 15/5
= 3
∴ Co-ordinates of P will be (-2, 3)
13. Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, - 2) and B (3, 7). Also find the co-ordinates of the point of division.
Answer
Points are given A (2, - 2), B (3, 7)
And let the line 2x + y – 4 = 0 divides AB in the ratio m1 : m2
At P and let co-ordinates of
x = (m1x2 + m2x1)/(m1 + m2)
= (m1×3 + m2×2)/(m1 + m2)
= (3m1 + 2m2)/(m1 + m2)
And y = {m1×7 + m2×(-2)}/(m1+m2)
= {7m1 – 2m2}/(m1+m2)
∴ P lies on the line 2x + y – 4 = 0, then
2(3m1 + 2m2)/(m1+m2) + (7m1 – 2m2)/(m1+m2) – 4 = 0
⇒ 6m1 + 4m2 + 7m1 – 2m2 – 4m1 – 4m2 = 0
⇒ 9m1 – 2m2 = 0
⇒ 9m1 = 2m2
⇒ m1/m2 = 2/9
or m1 : m2 = 2 : 9
∴ x = (2×3 + 2×9)/(2 + 9)
= (6 + 18)/11
= 24/11
And y = (2 × 7 – 2 × 9)/(2 + 9)
= (14 – 18)/11
= -4/11
∴ Co-ordinates of P will be (24/11, -4/11)
14. The point A (2, -3) is reflected in the y-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the point A’’.
(i) Write the coordinates of A’ and A’’.
(ii) Find the ratio in which the line segment AA’’ is divided by the x-axis. Also find the coordinates of the point of division.
Answer
A’ is the reflection of A (2, -3) in the x-axis
(i) ∴ Co-ordinates of A’ will be (2, 3)
Draw a line x = 4 which is parallel to y-axis
A’’ is the reflection of A’ (2, 3)
∴ Co-ordinates OA’’ will be (6, 3)
(ii) Join AA’’ which intersect x-axis at P whose co-ordinates are (4, 0)
Let P divide AA’’ in the ratio in m1 : m2
⇒ 0 = {m1×3 + m2×(-3)}/(m1 + m2)
⇒ 3m1 – 3m2 = 0
⇒ 3m1 = 3m2
⇒ m1/m2 = 3/3 = 1/1
∴ m1 : m2 = 1 : 1
Hence P (4, 0) divides AA’’ in the ratio 1 : 1
15. ABCD is a parallelogram. If the coordinates of A, B and D are (10, - 6), (2, - 6) and (4, - 2) respectively, find the co-ordinates of C.
Answer
Let the co-ordinates of C be (x, y) and other three vertices of the given parallelogram are A (10, - 6), B (2, - 6) and D (4, - 2)
∴ ABCD is a parallelogram
Its diagonals bisect each other.
Let AC and BD intersect each other at O.
∴ O is mid-point of BD
∴ Co-ordinates of O will be
{(2+4)/2, (-6–2)/2) or (6/2, -8/2) or (3, -4)
Again, O is the mid-point of AC then
3 = (10 + x)/2
⇒ 10 + x = 6
⇒ x = 6 – 10 = - 4
And – 4 = (-6 + y)/2
⇒ 6 + y = - 8
⇒ - 8 + 6
∴ y = - 2
Hence Co-ordinates of C will be (- 4, - 2).
16. ABCD is a parallelogram whose vertices A and B have co-ordinates (2, - 3) and (- 1, - 1) respectively. If the diagonals of the parallelogram meet at the point M (1, - 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. Find the perimeter of the parallelogram.
Answer
ABCD is a ||gm, m which co-ordinates of A are (2, - 3) and B (- 1, -1)
Its diagonals AC and BD bisect each other at M (1, - 4)
∴ M is the midpoint of AC and BD
Let co-ordinates of C be (x1, y1) and D be (x2, y2)
When M is the midpoint of AC then
∴ l = (2 + x1)/2 and – 4 = (-3 + y1)/2
⇒ 2 + x1 = 2
⇒ x1 = 2 – 2 = 0
And –8 = -3 + y1
⇒ y1 = -8 + 3 = -5
∴ Co-ordinates of C are (0, -5)
Again M is mid-point of BD, then
1 = (-1 + x2)/2, -4 = (-1 + y2)/2
⇒ -1 + x2 = 2
⇒ x2 = 2 + 1 = 3
And –1 + y2 = -8
⇒ y2 = -8 + 1 = -7
∴ Co-ordinates of D are (3, -7)
17. In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.
A lies on x-axis and
B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
And P (3, 1) divides it in the ratio of 2 : 3.
∴ 3 = (m1x2 + m2x1)/(m1 + m2)
= (2×0 + 3×x)/(2+3)
= (0 + 3x)/5
⇒ 3x = 15
⇒ x = 15/3 = 5
Again l = (m1y2 + m2y1)/(m1 + m2)
= (2×y + 3×0)/(2 + 3)
= (2y + 0)/5
= 2y/5
⇒ 2y = 5
⇒ y = 5/2
∴ Co-ordinates of A will be (5, 0) and of B will be (0, 5/2)
18. Given, O (0, 0), P (1, 2), S (- 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram. Find :
(i) the co-ordinates of Q.
(ii) the co-ordinates of R.
(iii) the ratio in which RQ is divided by y-axis.
(i) Let co-ordinates of Q be (x’, y’) and of R (x’’, y’’)
Point P (1, 2) divides OQ in the ratio of 2 : 3
∴ l = (m1x2 + m2x1)/(m1 + m2)
= (2x’ + 3×0)/(2+3)
⇒ (2x’+0)/5 = 1
⇒ 2x’ = 5
⇒ x’ = 5/2
And 2 = (m1y2 + m2y1)/(m1+m2)
= (2y’ + 3×0)/(2+3)
⇒ 2y’/5 = 2
⇒ 2y’ = 10
⇒ y’ = 5
∴ Co-ordinates of Q will be (5/2, 5)
∵ the diagonals of a parallelogram bisect each other
∴ In ||gm OPRS, diagonals OR and PS bisect each other at M.
∵ M is the mid-point of PS
∴ Co-ordinates of M will be = {(-2+1)/2, (0+2)/2}
or (-2/2, 2/2)
or (-1, 1)
(ii) ∵ M is the mid-point of OR also
∴ - 1 = (0 + x’’)/2
⇒ x’’ = - 2
And 1 = (0 + y’’)/2
⇒ y’’ = - 2
∴ Co-ordinates of R will be (- 2, 2)
(iii) RQ is dividing by y-axis in N
Let the ratio in which N divides RQ in m1 : m2
∵ N lies on y-axis
∴ its abscissa (x) = 0
0 = (m2x’ + m2x”)/(m1 + m2)
⇒ 0 = {m1×5/2 + m2×(-2)}/(m1+m2)
⇒ (5m1/2 – 2m2)/(m1+m2) = 0
⇒ 5/2 m1 – 2m2 = 0
⇒ 5/2 m1 = 2m2
⇒ m1/m2 = (2×2)/5 = 4/5
∴ m1 : m2 = 4 : 5
19. If A (5, -1), B (- 3, - 2) and C (- 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.
Answer
A (5, - 1), B (- 3, -2) and C (- 1, 8) are the vertices of ∆ABC D, E and F are the midpoints of sides BC, CA and AB respectively
And G is the centroid of the ∆ABC
∵ D is the midpoint of BC
∴ Co-ordinates of D will be {(x1 + x2)/2}, (y1 + y2)/2} or {(-3 –1)/2 , (-2+8)/2} or (-4/2, 6/2) or (-2, 3)
∵ G is the centroid
∴ Co-ordinates of G will be {(x1 + x2 + x3)/3 , (y1 + y2 + y3)/3}
Or {(5 – 3 – 1)/3, (-1–2 + 8)/3} or (1/3, 5/3)
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