Class 10 SELINA Solutions Maths Chapter 13 - Section and Mid-Point Formula
Section and Mid-Point Formula Exercise TEST YOURSELF
Solution 1(e)
Correct option: (ii) 3 : 2
Le the required ratio be k : 1.
Hence, the required ratio is 3 : 2.
Solution 1(d)
Correct option: (iv) (0, 8)
As A, B, C and D divided the join of O and P into five equal parts.
D divides the join of O and P in the ratio 4 : 1.
Let D = (x, y)
⇒ (0, 8) = (x, y)
Thus, D = (0, 8).
Solution 1(c)
Correct option: (ii) (–2, 10)
Let the coordinates of the third vertex be (x, y).
⇒ x = –2 and y = 10
Thus, coordinates of the third vertex are (–2, 10).
Solution 1(b)
Correct option: (iii) x = –4, y = –8
Mid-point of AB = (0, 0)
⇒ x = –4 and y = –8
Solution 1(a)
Correct option: (iv) A = (8, 0) and B = (0, 6)
As A lies on x-axis and P = (4, 3) is the mid-point of AB.
∴ A = (8, 0)
As B lies on y-axis and P = (4, 3) is the mid-point of AB.
∴ B = (0, 6)
Solution 2
Given, BP: PC = 3: 2
Using section formula, the co-ordinates of point P are
Using distance formula, we have:
Solution 3
Using section formula,
Given, AB = 6AQ
Using section formula,
Solution 4
Given that, point P lies on AB such that AP: PB = 3: 5.
The co-ordinates of point P are
Also, given that, point Q lies on AC such that AQ: QC = 3: 5.
The co-ordinates of point Q are
Using distance formula,
Hence, proved.
Solution 5
Since, the line segment AB intersects the y-axis at point P, let the co-ordinates of point P be (0, y).
P divides AB in the ratio 1: 3.
Thus, the value of a is 3 and the co-ordinates of point P are.
Solution 6
Let the line segment AB intersects the x-axis at point P (x, 0) in the ratio k: 1.
Thus, the required ratio in which P divides AB is 3: 1.
Also, we have:
Thus, the co-ordinates of point P are (3, 0).
Solution 7
Since, point A lies on x-axis, let the co-ordinates of point A be (x, 0).
Since, point B lies on y-axis, let the co-ordinates of point B be (0, y).
Given, mid-point of AB is C (4, -3).
Thus, the co-ordinates of point A are (8, 0) and the co-ordinates of point B are (0, -6).
Solution 8
Solution 9
It is given that the mid-point of the line-segment joining (4a, 2b - 3) and (-4, 3b) is (2, -2a).
Solution 10
(i) Co-ordinates of point P are
(ii) OP =
(iii) Let AB be divided by the point P (0, y) lying on y-axis in the ratio k: 1.
Thus, the ratio in which the y-axis divide the line AB is 4: 17.
Solution 11
We have:
AB = BC and
ABC is an isosceles right-angled triangle.
Let the coordinates of D be (x, y).
If ABCD is a square, then,
Mid-point of AC = Mid-point of BD
x = 1, y = 8
Thus, the co-ordinates of point D are (1, 8).
Solution 12
Given, M is the mid-point of the line segment joining the points A (-3, 7) and B (9, -1).
The co-ordinates of point M are
Also, given that, R (2, 2) divides the line segment joining M and the origin in the ratio p: q.
Thus, the ratio p: q is 1: 2.
Solution 13
Solution 14
Solution 15
Solution 16
Let A' = (x, y) be the image of the point A(5, -3), under reflection in the point P(-1, 3).
⇒ P(-1, 3) is the mid - point of the line segment AA'.
Therefore the image of the point A(5, -3), under reflection in the point P(-1, 3) is A'(-7, 9).
Solution 17
Solution 18
Centroid of a △ABC = …… (i)
G(3, 4) is a centroid of △ABC ….. given
Therefore, the coordinates of B and C are (5, 4) and (1, 7) respectively.
Length of the side BC
Section and Mid-Point Formula Exercise Ex. 13(A)
Solution 2
Let the point P (1, a) divides the line segment AB in the ratio k: 1.
Using section formula, we have:
Solution 3
Let the point P (a, 6) divides the line segment joining A (-4, 3) and B (2, 8) in the ratio k: 1.
Using section formula, we have:
Solution 4
Let the point P (x, 0) on x-axis divides the line segment joining A (4, 3) and B (2, -6) in the ratio k: 1.
Using section formula, we have:
Thus, the required ratio is 1: 2.
Also, we have:
Thus, the required co-ordinates of the point of intersection are .
Solution 5
Let S (0, y) be the point on y-axis which divides the line segment PQ in the ratio k: 1.
Using section formula, we have:
Solution 6
Point A divides PO in the ratio 1: 4.
Co-ordinates of point A are:
Point B divides PO in the ratio 2: 3.
Co-ordinates of point B are:
Point C divides PO in the ratio 3: 2.
Co-ordinates of point C are:
Point D divides PO in the ratio 4: 1.
Co-ordinates of point D are:
Solution 7
Let the co-ordinates of point P are (x, y).
Solution 8
5AP = 2BP
The co-ordinates of the point P are
Solution 9
The co-ordinates of every point on the line x = 2 will be of the type (2, y).
Using section formula, we have:
Thus, the required ratio is 5: 3.
Thus, the required co-ordinates of the point of intersection are (2, 4).
Solution 10
The co-ordinates of every point on the line y = 2 will be of the type (x, 2).
Using section formula, we have:
Thus, the required ratio is 3: 5.
Solution 11
Point A lies on x-axis. So, let the co-ordinates of A be (x, 0).
Point B lies on y-axis. So, let the co-ordinates of B be (0, y).
P divides AB in the ratio 2: 5.
We have:
Thus, the co-ordinates of point A are (7, 0).
Thus, the co-ordinates of point B are (0, -14).
Solution 12
Let P and Q be the points of trisection of the line segment joining the points A(-3, 0) and B(6, 6).
So, AP = PQ = QB
We have AP: PB = 1: 2
Co-ordinates of the point P are
We have AQ: QB = 2: 1
Co-ordinates of the point Q are
Solution 13
Let P and Q be the points of trisection of the line segment joining the points A (-5, 8) and B (10, -4).
So, AP = PQ = QB
We have AP: PB = 1: 2
Co-ordinates of the point P are
We have AQ: QB = 2: 1
Co-ordinates of the point Q are
So, point Q lies on the x-axis.
Hence, the line segment joining the given points A and B is trisected by the co-ordinate axes.
Solution 14
Let A and B be the point of trisection of the line segment joining the points P (2, 1) and Q (5, -8).
So, PA = AB = BQ
We have PA: AQ = 1: 2
Co-ordinates of the point A are
Hence, A (3, -2) is a point of trisection of PQ.
We have PB: BQ = 2: 1
Co-ordinates of the point B are
Solution 15
Since, point L lies on y-axis, its abscissa is 0.
Let the co-ordinates of point L be (0, y). Let L divides MN in the ratio k: 1.
Using section formula, we have:
Thus, the required ratio is 5: 3.
Solution 16
BP: PC = 2: 3
Co-ordinates of P are
Using distance formula, we have:
Solution 17
Since, point K lies on x-axis, its ordinate is 0.
Let the point K (x, 0) divides AB in the ratio k: 1.
Thus, K divides AB in the ratio 3: 5.
Also, we have:
Thus, the co-ordinates of the point K are .
Solution 18
Since, point K lies on y-axis, its abscissa is 0.
Let the point K (0, y) divides AB in the ratio k: 1.
Thus, K divides AB in the ratio 2: 3.
Also, we have:
Thus, the co-ordinates of the point K are .
Solution 19
(i) Let point R (0, y) divides PQ in the ratio k: 1.
We have:
Thus, PR: RQ = 4: 3
(ii) Also, we have:
Thus, the co-ordinates of point R are .
(iii) Area of quadrilateral PMNQ
= (PM + QN)
MN
= (5 + 2)
7
= 7
7
= 24.5 sq units
Solution 20
Given, A lies on x-axis and B lies on y-axis.
Let the co-ordinates of A and B be (x, 0) and (0, y) respectively.
Given, P is the point (-4, 2) and AP: PB = 1: 2.
Using section formula, we have:
Thus, the co-ordinates of points A and B are (-6, 0) and (0, 6) respectively.
Solution 21
(i)
(ii)
(iii)
Solution 22
Take (x1 , y1) = (-3, 3a + 1) ; (x2 , y2) = (5, 8a) and
(x, y) = (-b, 9a - 2)
Here m1 = 3 and m2 =1
Solution 1(a)
Correct option: (i)
Let the coordinates of point P are (x, y).
Thus,
Solution 1(b)
Correct option: (iv) 1 : 3
Let the ratio AP : PB be m1 : m2.
Thus, AP : PB = 1 : 3.
Solution 1(c)
Correct option: (ii) 5 : 3
Let the required ratio be k : 1 and the point on the line x = 7 is (7, y).
Thus, the line x = 7 divides the join of (2, 4) and (10, 12) is the ratio 5:3.
Solution 1(d)
Correct option: (i) 3 : 5
Let the required ratio be k : 1 and the point on the line y = 4 is (x, 4).
Solution 1(e)
Correct option: (ii) 2 : 5
Let the required ratio be k : 1 and the point on y-axis is (0, y).
Hence, the required ratio is 2 : 5.
Section and Mid-Point Formula Exercise Ex. 13(B)
Solution 2
Mid-point of AB = (2, 3)
Solution 3
(i) Let the co-ordinates of A be (x, y).
Hence, the co-ordinates of A are (7, 4).
(ii) Let the co-ordinates of B be (x, y).
Hence, the co-ordinates of B are (-5, 7).
Solution 4
Point A lies on y-axis, so let its co-ordinates be (0, y).
Point B lies on x-axis, so let its co-ordinates be (x, 0).
P (-3, 2) is the mid-point of line segment AB.
Thus, the co-ordinates of points A and B are (0, 4) and (-6, 0) respectively.
Solution 5
Let A (-5, 2), B (3, -6) and C (7, 4) be the vertices of the given triangle.
Let AD be the median through A, BE be the median through B and CF be the median through C.
We know that median of a triangle bisects the opposite side.
Co-ordinates of point F are
Co-ordinates of point D are
Co-ordinates of point E are
The median of the triangle through the vertex B(3, -6) is BE
Using distance formula,
Solution 6
Given, AB = BC = CD
So, B is the mid-point of AC. Let the co-ordinates of point A be (x, y).
Thus, the co-ordinates of point A are (-1, -2).
Also, C is the mid-point of BD. Let the co-ordinates of point D be (p, q).
Thus, the co-ordinates of point D are (2, 13).
Solution 7
Co-ordinates of the mid-point of AC are
Co-ordinates of the mid-point of BD are
Since, mid-point of AC = mid-point of BD
Hence, ABCD is a parallelogram.
Solution 8
Let the coordinates of R and S be (x,y) and (a,b) respectively.
Mid-point of PR is O.
O(-3,2) =
-6 = 4 + x, 4 = 2 + y
x = -10 , y = 2
Hence, R = (-10,2)
Similarly, the mid-point of SQ is O.
Thus, the coordinates of the point R and S are (-10, 2) and (-5, -1).
Solution 9
Let A(x1,y1), B and C
be the co-ordinates of the vertices of
ABC.
Midpoint of AB, i.e. D
Similarly,
Adding (1), (3) and (5), we get,
From (3)
From (5)
Adding (2), (4) and (6), we get,
From (4)
From (6)
Thus, the co-ordinates of the vertices of ABC are (3, 1), (1, -3) and (-5, 7).
Solution 10
Given, AB = BC, i.e., B is the mid-point of AC.
Solution 11
Given, PR = 2QR
Now, Q lies between P and R, so, PR = PQ + QR
PQ + QR = 2QR
PQ = QR
Q is the mid-point of PR.
Solution 12
Let G be the centroid of DPQR whose coordinates are (2, -5) and let (x,y) be the coordinates of vertex P.
Coordinates of G are,
6 = x + 5, -15 = y + 13
x = 1, y = -28
Coordinates of vertex P are (1, -28)
Solution 13
Given, centroid of triangle ABC is the origin.
Solution 1(a)
Correct option: (iii) (–1, 0)
Let P = (7, 8) and Q = (x, y) be the other end of the diameter.
As A is the centre of the circle.
⇒ 7 + x = 6, 8 + y = 8
⇒ x = –1, y = 0
Thus, Q = (–1, 0).
Solution 1(b)
Correct option: (i) (4, 0)
Coordinates of A are (x, 0)
Coordinates of B are (0, y)
As P bisects the line segment AB, we have
⇒ x = 4, y = –4
Thus, Q = (4, 0)
Solution 1(c)
Correct option: (ii) (–2, 0)
Let D = (x, y).
As ABCD is a parallelogram, we have
Mid-point of AC = Mid-point of BD
⇒ x = –2 and y = 0
Thus, D = (–2, 0).
Solution 1(d)
Correct option: (iv) (–2, 13)
When point P(2, –7) is reflected in the point (0, 3), the coordinates of its image are (–2, 13).
Solution 1(e)
Correct option: (iii) (2, –3)
Coordinates of the centroid of the triangle is given by: