OP Malhotra Class 11 Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Ex 15(b)

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S Chand Class 11 ICSE Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Ex 15(b)

Question 1.
Find the mid-points of the lines joining
(i) (5, 8),(9, 11);
(ii) (0, 0),(8, -5);
(iii) (-7, 0),(0, 10);
(iv) (-4, 3),(6, -7)
Solution:
We know that, mid-point of the line joining the points (x₁, y₁) and (x₂, y₂) be
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
(i) ∴ Mid-point of the line joining the points (5, 8) and (9, 11) be
\(\left(\frac{5+9}{2}, \frac{8+11}{2}\right)\) i.e. \(\left(7, \frac{19}{2}\right) \text {. }\)
(ii) The mid-point of the line joining the points (0, 0) and (8, -5) be given by
\(\left(\frac{0+8}{2}, \frac{0-5}{2}\right)\) i.e. \(\left(4, \frac{-5}{2}\right)\)
(iii) The mid-point of the line joining the points (-7, 0) and (0, 10) be given by \(\left(\frac{-7+10}{2}, \frac{0+10}{2}\right) \text { i.e. }\left(\frac{-7}{2}, 5\right) \text {. }\)
(iv) The mid-point of the line joining the points (-4, 3) and (6, -7) be given by \(\left(\frac{-4+6}{2}, \frac{3-7}{2}\right) \text { i.e. }(1,-2) \text {. }\)

Question 2.
Find the mid-points of the sides of a triangle whose vertices are A(1, -1), B (4, – 1), C (4, 3).
Solution:
The mid-point of side BC of △ABC be given by \(\left(\frac{4+4}{2}, \frac{-1+3}{2}\right) \text { i.e. }(4,1)\)
The mid-point of side CA of △ABC be given by \(\left(\frac{1+4}{2}, \frac{-1+3}{2}\right) \text { i.e. }\left(\frac{5}{2}, 1\right) \text {. }\)

The mid-point of side AB of △ABC be given by \(\left(\frac{1+4}{2}, \frac{-1-1}{2}\right) \text { i.e. }\left(\frac{5}{2},-1\right) \text {. }\)

Question 3.
Find the centre of a circle if the end points of a diameter are A(-5, 7) and B (3, – 11).
Solution:
Clearly centre C of circle be the mid-point of diameter AB
using mid-point formula, we have coordinates of centre C are
\(\left(\frac{-5+3}{2}, \frac{7-11}{2}\right) \text { i.e. }(-1,-2) \text {. }\)

Question 4.
If M is the mid-point of AB, find the coordinates of :
(i) A if the co-ordinates of M and B are M(2, 8) and B(-4, 19) and
(ii) B if the co-ordinates of A and M are (-1, 2), M(-2, 4).
Solution:
(i) Given coordinates of M and B are M(2, 8) and B(-4, 19)
Let the coordinates of point A are (α, β) since M be the mid-point of AB
∴ (2,8) = \(\left(\frac{-4+\alpha}{2}, \frac{19+\beta}{2}\right)\)
i.e. 2 =\(\frac{-4+\alpha}{2}\) ⇒ 4 = – 4 + α
⇒ α = 8
and 8 = \(\frac{19+\beta}{2}\) ⇒ 16 = 19 + β
⇒ β = -3
Thus the coordinates of A are (8, -3).

(ii) Given coordinates of A and M are A(-1, 2) and M(-2, 4)
Let the coordinates of point B are (α, β).
∴ Mid-point of AB be \(M\left(\frac{-1+\alpha}{2}, \frac{2+\beta}{2}\right)\)
also coordinates of M are (-2, 4)
∴ -2 = \(\frac{-1+\alpha}{2}\) ⇒ – 1 + α = – 4
⇒ α = – 3
and 4 = \(\frac{2+\beta}{2}\) ⇒ 2 + β = 8 ⇒ β = 6
Thus the required coordinates of B are (-3, 6).

Question 5.
Find the distance between each of the following pairs of points :
(i) (7, 9),(4, 5);
(ii) (15, 11),(3, 6);
(iii) (4, – 5), (0, 0);
(iv) (2, – 11), (-4, – 3)
Solution:
(i) Distance between the points A (7, 9) and B (4, 5)
= |AB| = \(\sqrt{(4-7)^2+(5-9)^2}\)
= \(\sqrt{9+16}\) = 5

(ii) Distance between the points C (15, 11) and D (3, 6)
= |CD| = \(\sqrt{(3-15)^2+(6-11)^2}\)
= \(\sqrt{144+25}\) = \(\sqrt{169}\) = 13

(iii) Distance between the points L (4, -5) and M (0, 0)
= |LM| = \(\sqrt{(0-4)^2+(0+5)^2}\) = \(\sqrt{16+25}\) = \(\sqrt{41}\)

(iv) Distance between the points P(2, – 11) and Q(-4, – 3)
=|PQ| = \(\sqrt{(-4-2)^2+(-3+11)^2}\) = \(\sqrt{36+64}\) = \(\sqrt{100}\) = 10

Question 6.
Find the radius of the circle that has its centre at (0, -4) and passes through \((\sqrt{13}, 2)\)
Solution:
Clearly distance between the centre C and any point A on circle gives the required radius of circle

∴ radius of circle = |CA|
= \(\sqrt{(\sqrt{13}-0)^2+(2+4)^2}\) = \(\sqrt{13+36}\) = \(\sqrt{49} \) = 7

Question 7.
Find the lengths of the sides of the triangle whose vertices are A(3, 4), B(2, -1) and C(4, -6).
Solution:
|AB| = \(\sqrt{(2-3)^2+(-1-4)^2}\) = \(\sqrt{1+25}\) = \(\sqrt{26}\)
|BC| = \(\sqrt{(4-2)^2+(-6+1)^2}\) = \(\sqrt{4+25}\) = \(\sqrt{29}\)
|AC| = \(\sqrt{(4-3)^2+(-6-4)^2}\) = \(\sqrt{1+100}\) = \(\sqrt{101}\)

Question 8.
The vertices of △ABC are A(-1, 3), B(1, 1) and C(5, 1). Find the length of the median to (i) AB, (ii) AC, (iii) BC.
Solution:
(i) Let D be the mid-point of AB
∴ Coordinates of D are \(\left(\frac{-1+1}{2}, \frac{3+1}{2}\right)\)
i.e. D(0, 2)
Thus length of median to AB = |CD|
= \(\sqrt{(5-0)^2+(1-2)^2}\) = \(\sqrt{25+1}\) = \(\sqrt{26}\)

(ii) Let E be the mid-point of AC
∴ Coordinates of E are \(\left(\frac{-1+5}{2}, \frac{3+1}{2}\right)\) i.e. E(2, 2)
Thus, length of median to AC = |BE|
= \(\sqrt{(2-1)^2+(2-1)^2}\) = \(\sqrt{1+1}\) = \(\sqrt{2}\)

(iii) Let F be the mid-point of side BC of △ABC
∴ Coordinates of F are \(\left(\frac{1+5}{2}, \frac{1+1}{2}\right)\)
i.e. F(3, 1)
Thus, length of median to BC = |AF|
= \(\sqrt{(3+1)^2+(1-3)^2}\) = \(\sqrt{16+4}\) = \(\sqrt{20}\) = \(2 \sqrt{5}\)

Question 9.
A circle has its centre at the origin and a radius of \(\sqrt{12}\). State whether each of the following points is on, outside or inside the circle : (1, \(\sqrt{7}\)),(3, 5),(2, 2\(\sqrt{2}\)).
Solution:
Let the given points are
P(1, –\(\sqrt{7}\)), Q(3, 5) and R(2, 2\(\sqrt{2}\))

Here, |CP| = \(\sqrt{(1-0)^2+(-\sqrt{7}-0)^2}\) = \(\sqrt{1+7}\) = \(\sqrt{8}\) = \(2 \sqrt{2}\) < \(\sqrt{12}\) ∴ point P(1, – \(\sqrt{7}\)) lies inside the circle. |CQ| = \(\sqrt{(3-0)^2+(5-0)^2}\) = \(\sqrt{9+25}\) = \(\sqrt{34}\) > \(\sqrt{12}\) = r
Thus point Q(3, 5) lies outside the circle.
|CR| = \(\sqrt{(2-0)^2+(2 \sqrt{2}-0)^2}\) = \(\sqrt{4+8}\) = \(\sqrt{12}\) = r
Thus the point R(2, 2\(\sqrt{2}\)) lies on given circle.

Question 10.
A line is of length 10, and one end is at the point (-3, 2). If the ordinate of the other end be 10, prove that the abscissa will be 3 or -9 .
Solution:
Let x be the abscissa of other end of line.
∴ Coordinates of other end of line be Q(x, 10). Also coordinates of one end of line be P(-3, 2) s.t |PQ| = 10
⇒ \(\sqrt{(x+3)^2+(10-2)^2}\) = 10;
on squaring both sides; we have
(x + 3)² + 64 = 100 ⇒ (x + 3)² = 36
⇒ x + 3 = ± 6 ⇒ x = ± 6 – 3 = 3, – 9
Hence the abscissa of other end will be 3 or -9.

Question 11.
Find the coordinates of the point which divides internally the join of the points
(i) (8, 9) and (-7, 4) in the ratio 2 : 3;
(ii) (1, -2) and (4, 7) in the ratio 1 : 2.
Solution:
(i) Let the point P divides the line segment AB in the ratio 2 : 3 internally.

Then by section’s formula, we have coordinates of P are
\(\left(\frac{2 \times(-7)+3 \times 8}{2+3}, \frac{2 \times 4+3 \times 9}{2+3}\right)\)
i.e. \(\mathrm{P}\left(\frac{10}{5}, \frac{35}{5}\right) \text { i.e. } \mathrm{P}(2,7) \text {. }\)

(ii) Let the point Q divides the line segment AB in the ratio 1 : 2 internally.
Then by Section’s formula, we have coordinates of Q are

Aliter : Let the point P divides the line segment BA in the ratio 2 : 3 internally.

(iii) Let the point Q divides the line segment BA in the ratio 1 : 2 internally.
Then by Section’s formula, we have
∴ Coordinates of Q are

Question 12.
Find the coordinates of the point which divides externally the join of the points
(i) (-4, 4) and (1, 7) in the ratio 2 : 1;
(ii) (3, 4) and (-6, 2) in the ratio 3 : 2.
Solution:
(i) Let point P divides line segment AB in the ratio 2 : 1 externally i.e. 2 : – 1 internally

i.e. (6, 10)
Then by section formula,
coordinates of P are

(ii) Let point P divides line segment AB in the ratio 3 : 2 externally i.e. 3 : – 2 internally.
Then by Section formula,
coordinates of P are

Question 13.
Find the coordinates of the points of trisection of the line joining the points (2, 3) and (6, 5).
Solution:
Let P and Q are the points of trisection of line segment AB.

So P divides line segment AB in the ratio 1 : 2 internally. Then by section formula, we have, Coordinates of P are
\(\left(\frac{1 \times 6+2 \times 2}{1+2}, \frac{1 \times 5+2 \times 3}{1+2}\right)\)
\(\text { i.e. }\left(\frac{10}{3}, \frac{11}{3}\right)\)
Further point Q divides line segment AB in the ratio 2 : 1 internally.
Then by section formula, we have coordinates of Q are
\(\left(\frac{2 \times 6+1 \times 2}{1+2}, \frac{2 \times 5+1 \times 3}{2+1}\right) \text { i.e. }\left(\frac{14}{3}, \frac{13}{3}\right) \text {. }\)
Hence the points of trisection of given line segment are \(\left(\frac{10}{3}, \frac{11}{3}\right)\) and \(\left(\frac{14}{3}, \frac{13}{3}\right)\)

Question 14.
The line joining the points (3, 2) and (6, 8) is divided into four equal parts, find the coordinates of the points of section.
Solution:
Let the line segment joining the points A (3, 2) and B(6, 8) is divided into four equal parts at P, Q and R
s.t AP = PQ = QR = RB

Thus point P divides the line segment AB in the ratio 1 : 3 internally.
Then by Section formula, we have coordinates of point P are
\(\left(\frac{1 \times 6+3 \times 3}{1+3}, \frac{1 \times 8+3 \times 2}{1+3}\right)\)
\(\text { i.e. }\left(\frac{15}{4}, \frac{14}{4}\right) \text { i.e. }\left(\frac{15}{4}, \frac{7}{2}\right)\)
The point Q be the mid-point of line segment AB.
∴ Coordinates of Q are
\(\left(\frac{3+6}{2}, \frac{2+8}{2}\right) \text { i.e. } \mathrm{Q}\left(\frac{9}{2}, 5\right)\)
Further point R divides line segment AB in the ratio 3 : 1 internally.
Then by section formula, we have Coordinates of R are
\(\left(\frac{3 \times 6+1 \times 3}{3+1}, \frac{3 \times 8+1 \times 2}{3+1}\right)\)
\(\text { i.e. } \hat{\mathbf{R}}\left(\frac{21}{4}, \frac{13}{2}\right) \text {. }\)
Hence the points od section are
\(\left(\frac{15}{4}, \frac{7}{2}\right) ;\left(\frac{9}{2}, 5\right) ;\left(\frac{21}{4}, \frac{13}{2}\right) .\)

Question 15.
In what ratio does the point \(\left(1, \frac{-7}{2}\right)\) divide the join od (-2, -4) and \(\left(2, \frac{-10}{3}\right)\)?
Solution:
Let the point P divides the line segment AB in the ratio k : 1 internally.
Then by section formula, we have

Thus both eqn.’s yield the same value of k.
Hence the required ratio be k : 1 i.e. 3 : 1.

Question 16.
In what ratio is the line joining the points
(i) (2, -3) and (5, 6) divided by the x-axis ;
(ii) (3, -6) and (-6, 8) divided by the y-axis?
Solution:
(i) Let the point P divides the line segment AB in the ratio k : 1

Then by Section formula, we have
Coordinates of P are \(\left(\frac{5 k+2}{k+1}, \frac{6 k-3}{k+1}\right)\)
Since it is given that line segment AB is divided by x-axis.
∴ Ordinate of point P is 0 .
Thus \(\frac{6 k-3}{k+1}\) = 0 ⇒ 6k = 3 ⇒ k = \(\frac{1}{2}\)
Thus required ratio be k : 1 i.e. \(\frac{1}{2}\) : 1
i.e. 1 : 2

(ii) Let the point P divides the line segment AB in the ratio k : 1

Then by section formula, we have
coordinates of \(\left(\frac{-6 k+3}{k+1}, \frac{8 k-6}{k+1}\right)\)
Further it is given that line segment AB is divided by $y$-axis.
Therefore abscissa of point P is 0 .
Thus, \(\frac{-6 k+3}{k+1}\) = 0 ⇒ 6k = 3 ⇒ k = \(\frac{1}{2}\)
Therefore required ratio be k : 1 i.e. \(\frac{1}{2}\) : 1
i.e. 1 : 2.

Question 17.
Find the ratio in which the axes divide the line joining the points (2, 5) and (1, 9).
Solution:
Let the point P divides the line segment AB in the ratio k : 1 internally.

Then by Section formula, we have coordinates of \(\left(\frac{k+2}{k+1}, \frac{9 k+5}{k+1}\right)\)
It is given that line segment AB is divided by x-axis.
∴ ordinate of point P is 0 .
⇒ \(\frac{9 k+5}{k+1}\) = 0 ⇒ 9k = -5 ⇒ k = \(\frac{-5}{9}\)
Thus required ratio be k : 1 i.e. -5 : 9 i.e. 5 : 9 externally.
Further, it is given that line segment AB is divided by y-axis.
∴ abscissa of point P is 0 .
i.e. \(\frac{k+2}{k+1}=0\) ⇒ k + 2 = 0 ⇒ k = – 2
Thus required ratio be k : 1 i.e. – 2 : 1 internally i.e. 2 : 1 externally.

Question 18.
Show by using section formula that the point (3, -2 ),(5, 2) and (8, 8) are collinear.
Solution:
Let the given points are A(3, – 2), B(5, 2) and C(8, 8)
and let point B divide the line segment AC in the ratio k : 1.

Then by section formula, Coordinates of B are \(\left(\frac{8 k+3}{k+1}, \frac{8 k-2}{k+1}\right)\)
Also given coordinates of B are (5, 2).
\(\frac{8 k+3}{k+1}\) = 5 ⇒ 8k + 3 = 5k + 5
⇒ 3k = 2 ⇒ k = \(\frac{2}{3}\)
and \(\frac{8 k-2}{k+1}\) = 2 ⇒ 8k – 2 = 2k + 2
⇒ 6k = 4 ⇒ k = \(\frac{2}{3}\)
Hence both eqns. gives same value of k.
Thus the point B divide AC in the ratio k : 1
i.e. 2:3. Therefore A, B, C are lies on same line and hence collinear.

Question 19.
Find the centroid of the triangle whose angular points are (-4, 6),(2, – 2) and (2, 5) respectively.
Solution:
We know that the centroid of △ABC having vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be given by

Question 20.
If (x₁, y₁) = (2, 3); x₂ = 3 and y₃ = – 2 and G is (0, 0), find y₂ and x₃.
Solution:
Given (x₁, y₁) = (2, 3); x₂ = 3, y₃ = – 2 and centroid G be (0, 0)
We know that, centroid of △ABC are given by
G\(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\) i.e. G\(\left(\frac{2+3+x_3}{3}, \frac{3+y_2-2}{3}\right)\)
Also cooridnates of centroid G be G(0, 0).
i.e. 0 = \(\frac{2+3+x_3}{3}\) ⇒ x₃ = – 5
and 0 = \(\frac{3+y_2-2}{3}\) ⇒ y₂ = – 1

Question 21.
Find the coordinates of the in-centre of the triangle whose vertices are (-36, 7) and (20, 7) and (0, – 8 ).
Solution: