ML Aggarwal ICSE Solutions Class 10 Math 11th Chapter Section Formula
Class 10 Chapter 11 Section FormulaChapter 11 Section Formula
(1) Find the Coordinates 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)
Solution:
(i) Given two points are (2, -3) and (-6, 7)
Mid Point formula is given by,
(x, y) = (x1 + x2/2, y1 + y2/2)
(x, y) = (2 – 6)/2, (-3 + 7)/2 = (-2, 2)
Thus, the Co – ordinates of the mid Point of line segment joining the points (2, -3) & (-6, 7) is found to be (-2, 2)
(ii) Given two points are (5, – 11) and (4,3)
Mid – Point formula is given by
(x, y) = (x1 + x2/2, y1 + y2/2)
(x, y) = (5 + 4/2, -11 +3/2) = (9/2, -8/2) = (4.5, – 4)
Thus, the Co – ordinates of the mid – point of line segment joining the points (5, – 11) and (4,3) is found to be (4.5, – 4).
(iii) Given two points are (a + 3, 5b) and (2a – 1, 3b + 4)
Mid – point formula is given by,
(x, y) = (x1 + x2/2, y1 + y2/2)
= (a + 3 + 2a – 1/2, 5b+ 3b + 4/2) = (3a + 2/2, 8b + 4/2)
(x, y) = (3a + 2/2, 4b + 2)
Thus, the Co – ordinates of mid – point of line – segment joining the points (a + 3, 5b) and (2a – 1, 3b + 4) is found to be (3a + 2/2, 4b + 2)
(2) 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.
Solution:
Given that, point P divides the distance between A (-2, 1) and B (1,4) in the ratio of 2 :1.
Let us Consider the point P (x, y) divides AB in the ratio of m1 : m2 i.e. 2 : 1.
X = (m1 x2 + m2 x1/m1 + m2)
X = (2 × 1 + 1 × (-2)/ 2 + 1)
= (2 – 2/3) = 0/3 = 0 x = 0
And y = (m1 y1 + m2 y1/ m1 + m2) = (2 × 4 + 1 × 1/2 + 1) = (8 + 1/3) = 9/3 = 3 y = 3
Hence, the Co – ordinates of point P are found to be P = (0,3).
(3) (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 Co – ordinates of P and Q are (P, -2) and (5/3, 9) respectively, find the values P and q.
Solution:
(i) Given that,
The Points of line segment the points (3, -3) and (6, 9).
Let us consider the points P (x1, y1) and Q (x2, y2) are the points which trisect the line segment joining Points A (3, -3) and B (6, 9).
Here, point P (x1, y1) divides AB in the ratio 1:2.
Thus, the Co – ordinates of point P (x1, y1) are found to be P (4,1).
Let us consider, Q (x2, y2) divides the line segment AB in the ratio 2 :1.
Thus, the Co – ordinates of point Q (x2, y2) are found to be Q ( 5, 5).
(ii) Given that, the line segment joining the points (3, -4) and (1,2) is trisected at the points P and Q.
And the Co – ordinates of point P and Q are (P, -2) and (5/3, q) respectively.
Here, point P (p, -2) divides lines AB in the ratio 1:2 and point Q divides line AB in the ratio 2:1.
Hence, the co – ordinates of point Q are found to be Q (5/3, 0) and P (7/3, -2).
(4) 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.
Solution:
Let us consider the point P (x, y) divides the line segment joining the points A (3, 2) and B (5, 1) in the ratio 1: 2.
Given that point P is on the line 3x – 18y + k = 0 ….. (1)
Put x = 11/3 and y = 5/3 in equine (1)
= 3 (11/3) – 18 (5/3) + k = 0
= 11 – 30 + k = 0
= – 19 + k = 0
K = 19
(5) Find the co – ordinates of the point which is three – fourth of the way from A (3, 1) to B (-2, 5).
Solution:
Given that, the points having co – ordinates A (3, 1) and B (-2, 5) are the end points of the line segment.
Let us consider, P (x, y) be the point which is three – fourth of the way from A to B.
= AP/AB = 3/4
= AP/(AP + PB) = 3/4
4 AP = 3 AP + 3 PB
4 AP – 3 AP = 3PB
= AP = 3 PB
Hence, AP/PB = 3/1 = m1 = 3, m2 = 1
Now, to find the co – ordinates of point P (x, y):
Thus, the co – ordinates of point P will be (-3/4, 4).
(6) The line segment joining points 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.
Solution:
Given that, the line segment joining points A (-3, 1) and B (5, -4) is a diameter of a circle whose centre is C.
i.e. C is the midpoint of AB.
Let us consider the point C (x, y).
X = (-3 + 5)/2 = (1 – 4)/2 = 2/2 = 1
Y = (1 – 4)/2 = -3/2
Thus, the co – ordinates of Point c are found to be (1, -3/2).
(7) 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.
Solution:
Let us consider, the point P (1, 2m – 1) is the mid – point of the line segment joining the points (3m, 6) and (-4, 3n).
1 = (x1 + x2)/2
1 = (3m – 4)/2
2 = 3m – 4
m = 6/3
m = 2
and 2m – 1 = (6 + 3n) /2
4m – 2 = 6 + 3n
4 × 2 – 2 = 6 + 3n
8 – 2 = 6 + 3n
3n = 8 – 2 – 6
n = 0
Thus, Here the values of m and n are found to be m = 2 and n = 0
(8) 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.
Solution:
Let us consider the co – ordinates of point Q be (x, y). Given that, the co – ordinates of the mid – point of the line segment PQ are (1, -2) and P (-3, 2).
1 = (-3 + x)/2
-3 + x = 2
X = 2 + 3 = 5
and -2 = (2 + y) /2
-4 = 2 + y
Y = -4 – 2
Y = -6
Thus, the co – ordinates of point Q are found to be Q (5, -6).
(9) AB is a diameter of a circle with centre C (-2, 5). If point A is (3, -7).
Find (i) the length of radius AC.
(ii) the co – ordinates of B.
Solution:
Here, given that
AB is a diameter of a circle with centre c (-2, 5).
and point A having co – ordinates A (3, – 7).
(i) Thus, By distance formula
AC = √(3 + 2)2 + (-7 – 5)2 = √52 + 122 = √25 + 144
= √169 = 13
Thus, the length of radius AC is found to be 13 units.
(ii) Let us consider, co – ordinates of points B are (x, y).
∴ (3 + x)/2 = -2
3 + x = -4
X = -7
and (y – 7)/2 = 5
y = 10 + 7
y = 17
Thus, the co – ordinates of point B are found to be B (-7, 17).
(10) Find the reflection (image) of the point (5, – 3) in the point (-1, 3).
Solution:
Let us consider, the co – ordinates of the image of the point A (5, -3) be A1 (x, y) in the point (-1, 3).
= Point (-1, 3) is the midpoint of AA.
= -1 = (5 + x) /2
5 + x = -2
X = -7
and 3 = (-3 + Y) /2
-3 + y = 6
Y = 9
Hence, the co – ordinates of the image A will be A (-7, 9).
(11) 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.
Solution:
Given that, the line segment joining A (-1, 5/3) the points B (a, 6) is divided in the ratio 1:3 at P.
(i) Thus,
(ii) Given that, the line segment AB intersects Y axis at P.
∴ x = 0
= (a – 3)/4 = 0
a – 3 = 0
a = 3
And hence the co – ordinates of point P are found to be P (0, 5/2).
(12) The point P (-4, 1) divides the line segment joining the points A (-2, 2) and B in the ratio 3: 5. Find the point B.
Solution:
Given that,
The point P (-4, 1) divides the line segment joining the points A (2, -2) and B in the ratio 3: 5.
Let us consider, B be any point whose co – ordinates are found to be B (x, y).
= -4 = [3x + 5 (2)/ 3 + 5] = (3x + 10)/8
Also, 3x + 10 = -32
3x = -32 – 10
3x = -42
∴ x = -14
and 1 = [3y + 5 (-2)/ 3 + 5]
1 = (3y – 10) / 8
8 = 3y – 10
3y = 8 + 10
Y = 18/3
Y = 6
Thus, the co – ordinates of point B are found to be B (-14, 6).
(13) (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.
Solution:
(i) Given that, the point (5, 4) divide the line segment joining the points (2, 1) and (7, 6).
Let us consider the points be A (2, 1) & B (7, 6) and P (5, 4).
Let us consider point P divides AB in the ratio m1 : m2.
Then, 5 = (m1 x 7 + m2 x 2/m1+m2)
5m1 + 5m2 = 7m1 + 2m2
5m2 – 2m2 = 7m1 – 5m1
3m2 = 2m1
m1/m2 = 3/2
m1 : m2 = 3 : 2
Thus, point P (5, 4) divides the line segment A (2,1) & B (7, 6) joining points in the ratio 3 : 2.
(ii) Given that, point (-4, b) divide the line segment joining the points P (2, – 2), Q (-14, 6).
Let us consider point A (-4, b) divide the line segment joining points P & Q in the ratio m1 : m2.
Then, – 4 = (m1 x (-14) + m2 x 2/m1+m2)
-4m1 – 4m2 = -14m1 + 2m2
-4m1 + 14 m1 = 2m2 + 4m2
10m1 = 6m2
m1/ m2 = 6/10 = 3/5
m1/ m2 = 3: 5
Also, b =
Thus, point A divide the line segment joining points P and Q in the ratio 3: 5.
And the value of b is found to be b = 1.
(14) 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 point K. Hence, find the ratio in which K divides AB Also, find the co – ordinates of the point K.
Solution:
Given that,
The line segment joining A (2, 3) and B (6, -5) is intercepted by the x – axis at the point K.
Let us consider the co – ordinates of point K be K (x, o) Which intersects x – axis.
Let us consider point K divides AB in the ratio m1 : m2.
Thus, co – ordinates of point K are found to be K (7/2, 0).
(16) 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 co – ordinates of the point of division?
Solution:
Given that,
The line x – y – 2 = 0 divide the line segment the points (3, -1) and (8, 9).
Let us consider it divide the segment in the ratio m1 : m2 at point P (x, y).
8m1 + 3m2 – 9m1 + m2 – 2m1 – 2m2 = 0
-3m1 + 2m2 = 0
3m1 = 2m2
∴ m1: m2 = 2/3
Thus, the co – ordinates of point P are found to be P (5, 3)
(17) 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
Solution:
Given that, a line segment AB joining the points A (-4, 6) & B (8, -3).
(i) Let us consider AB is divided by the y – axis in the ratio m : 1.
O = (m × 8 – 4 × 1/ m + 1)
8m – 4 = 0
8m = 4
m = 1/2
Thus, AB is divided by the y – axis in the ratio 1/2 : 1 or 1 : 2
(ii) Here, y = (1 × (-3) + 2 × 6)/ (1 + 2) = 9/3 = 3
Thus, co – ordinates of the point of intersection are found to be (0, 3).
(iii) Here, By distance formula,
AB = √(8 – 4)2 + (-3 – 6)2
= √144 + 81
AB = √225 = 15 units
Thus, the length of line segment AB is found to be 15 units.
(18) 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).
Solution:
Given that, In triangle ABC,
The vertices are having co – ordinates A (7, -3), B (5, 3) and C (3, -1)
Let us consider the Point D (x, y) be the midpoint of BC.
Hence, co – ordinates of point D are calculated by,
X = (5 + 3)/ 2 = 8/2 = 4
and y = (3 – 1)/2 = 2/2 = 1
Hence, co – ordinates of point D are found to be D (4, 1).
Thus, length of DA = √(7 – 4)2 + (-3 -1)2
= √(3)2 + (-4)2
= √9 + 16
= √25
l (DA) = 5 units
Thus, the length of median is found to be 5 units.
(19) Three consecutive vertices of a parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0). Find the fourth vertex D.
Solution:
Given that, the three consecutive vertices of a Parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0).
Let us consider Point ‘P’ is the mid- point of diagonal AC.
Thus, co – ordinates of Point P = (1 + 4/2, 2 + 0/2) = (5/2, 1)
Let us consider the co – ordinates of point D (x, y).
Thus, 5/2 = (1 + x)/2
10 = 2 + 2x
2x = 10 – 2 = 8
X = 4
And 1 = (0 + y)/2
Y = 2
Thus, co – ordinates of point D are found to be D = (4, 2).
(20) If the Points A (-2, -1), B (1, 0), C (P, 3) and D (1, q) form a Parallelogram ABCD, Find the values of p and q.
Solution:
Given that, the points A (-2, -1), B (1, 0), C (P, 3) and D (1, q) forms a parallelogram ABCD.
- Let us consider, the diagonals of parallelogram ABCD which are AC and BD bisects each other at point P.
- Thus, By the properties of a parallelogram P is the mid – point of AC as well as BD.
- Let us consider point P has co – ordinates P(x, y).
Then, x = (P – 2)/2
and y = (3 – 1)/2 = 2/2 = 1
x = (P – 2)/ 2,
Y = 1
[Since, Point P is the mid – point of diagonal AC.]
- Now, x = (1 + 1)/ 2 = 2/2 = 1
(P – 2)/2 = 1
P – 2 = 2
P = 2 + 2 = 4
P = 4
and q/2 =1
q = 2
Thus, the values of P and q are found to be P = 4 and q = 2.
(21) If two vertices of a Parallelogram are (3, 2), (-1, 0) and its diagonals meet at (2,-5). Find the other two vertices of the Parallelogram.
Solution:
Given that, the two vertices of a Parallelogram are (3, 2), (-1, 0) and its diagonals meet at (2, -5).
Let us consider, A (3, 2), B (-1, 0) & P (2, – 5).
Then, P is the mid – point of diagonals Ac & BD.
Let us take, the co – ordinates P of point C be (x, y).
2 = (x + 3)/2
X + 3 = 4
X = 1
And, -5 = (y + 2)/2
Y = -10 – 2
Y = -12
Hence, co ordinates of point C are found to be C (1, -12).
Let us consider, co – ordinates of point ‘D’ be D (x, y).
2 = (x – 1)/2
X – 1 = 4
X = 5
and -5 = (y + 0)/2
y = -10
Thus, the co – ordinates of point D are found to be D (5, -10).
(22) Find the third vertex of a triangle if its two vertices are (-1, 4) and (5, 2) and mid – point of one side is (0, 3).
Solution:
Given that,
The third vertex of a triangle if its two vertices are A (-1, 4) and B (5, 2).
And also, the mid – point of one side is P (0, 3).
Thus, By mid – point formula,
0 = (x – 1)/2
X = 1
and 3 = (y + 4)/ 2
y + 4 = 6
y = 2
Hence, co – ordinates of third vertex be C (1, 2).
But, if we considered point P (0,3) is the mid – point of side BC then, By mid – point formula
O = (5 + x)/2
X + 5 = 0
X = – 5
and 3 = (2 + y)/2
2 + y = 6
Y = 4
Hence, co – ordinates of third vertex C be C (-5, 4).
Finally, the co – ordinates of point C found to be (1, 2) or (-5, 4).
(24) Show that by Section Formula the points (3, -2), (5, 2), and (8, 8) are collinear.
Solution:
Given points are (3, -2), (5, 2) and (8, 8).
To show the given points are collinear:
Let us consider (5, 2) divides the line joining points (3, -2) & (8, 8) in the ratio m: n.
Then, 5 = (m × 8 + n × 3/ m + n)
8m + 3n = 5m + 5n
8m – 5m = 5n – 3n
3m = 2n
m/n = 2/3 —– (1)
From (1) & (2), it is proved that point (5, 2) lies on the line joining the points (3, -2) & (8, 8).
Hence, the Points (3, -2), (5, 2) & (8, 8) are collinear. Hence proved.
(25) Find the value of P for which the points (-5, 1), (1, P) & (4, -2) are collinear.
Solution:
Given that, the Points (-5, 1), (1, P) & (4, -2) are collinear.
Let us take, A (-5, 1), B (1, P) & C (4, – 2).
And point A (-5, 1) divides BC in the ratio m: n.
From (1) & (2) = (P – 1)/3 = -2/3
= -3P + 3 = 6
-3P = 3
P = -3/3
P = 1 is the required value of P.
(26) The mid – point of the line segment AB shown in the adjoining diagram is (4, -3). Write down the co – ordinates of A and B.
Solution:
Given that, The mid – point of the line Segment AB shown in the adjoining diagram is (4, – 3).
A lies on the X – axis & B on the Y – axis as shown in fig.
Then, co – ordinates of A and B are found to be:
A (x, 0) and B (0, y).
As P (+4, -3) is the mid – point of AB, then by mid – point formula,
4 = (x + 0)/2
X = 8
and -3 = (0 + y)/ 2
y = -6
Thus, the co – ordinates of points A and B are found to be:
A (8, 0) and B (0, -6).
(27) Find the co – ordinates of the centroid of a triangle whose vertices are A (-1, 3), B (1, – 1) and C (5, 1).
Solution:
Given that, the three vertices of a triangle are A (-1, 3), B (1, -1) and C (5, 1).
Then, by Centroid Formula,
C (x, y) ≡ (-1 + 1 + 5/3, 3 – 1 + 1/3) ≡ (5/3, 3/3) ≡ (5/3, 1)
Thus, the co – ordinates of a centroid of a given triangle is found to be C (5/3, 1).
(28) Two vertices of a triangle are (3, -5) and (-7, 4). Find the third vertex given that the centroid is (2, -1).
Solution:
Given that, two vertices of a triangle are A (3, -5) & B (-7, 4).
And co – ordinates of a centroid of a triangle are C (2, 1).
Let us consider, the co – ordinates of third vertex of a triangle are D (x, y).
Then, By centroid formula,
2 = (3 – 7 + x/3)
(x – 4)/3 = 2
X – 4 = 6
X = 10
and -1 = (- 5 + 4 + y/3)
-3 = -1 + y
Y = -3 + 1
Y = -2
Thus, the co – ordinates of third vertex of a given triangle are found to be D (10, -2).
(29) 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 centriod of the triangle ABC is the point (1, -1).
Solution:
Given that, the vertices of a triangle are A (-5, 3), B (P, – 1), and C (6, q).
And centroid of a triangle is found to be D (1, -1).
Then, By Centroid formula,
(x, y) ≡ (-5 + P + 6/3, 3 – 1 + q/3) ≡ (1 + P/3, 2 + q/3)
= 1 + P/3 = 1
1 + P = 3
P = 2
and (2 + q/3) = -1
2 + q = – 3
q = -5
Thus, the values of p and q are found to be
P = 2 and q = – 5
