1st PUC Basic Maths Question Bank Chapter 15 Co-ordinate System in a plane

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Karnataka 1st PUC Basic Maths Question Bank Chapter 15 Co-ordinate System in a plane

Question 1.
Find the reflection of (-4, 3) through x-axis, y-axis and through the centre.
Reflection through x – axis → (-4, – 3)
Reflection through y – axis → (4,3)
Reflection through center → (4, -3)

Question 2.
Find ‘a’ if the distance between the points (a, 2) and (3, 4) is V8 units.
Let P = (a, 2) and Q = (3, 4)
By data |PQ| = √8
⇒ √8 = $$\sqrt{(3-a)^{2}+(4-2)^{2}}$$
⇒ 8 = (3 – a)2 + 22 = 8
⇒ 9 + a2 – 6a + 4
⇒ a2 – 6a + 5 = 0
⇒ (a – 5)(a – 1) = 0
⇒ a = 5 or a = 1

Question 3.
Show that the points (2, 2) (6, 3) and (4,11) form a right angled triangle.
Let A = (2, 2), B = (6, 3), C = (4, 1)
Consider

By pythagorus theorem AC2 = AB2 + BC2

⇒ 85 = 17 + 68 = 85
∴ the points form a right angled triangle.

Question 4.
The points (x, 2) is equidistant from (8, -2) and (2, -2). Find the value of x.
Let A = (2,2) B = (8,-2) C = (2,-2)
By data AB = AC = AB2 = AC2
⇒ (8 – x)2 + (-2 -2)2 = (x – 2)2 + (2 + 2)2

-16x + 4x = 4 -64 ⇒ – 12x = -60
∴ x = 5

Question 5.
If two vertices of an equilateral triangle are (3,4) and (-2,3). Find the co-ordinates of third vertex.
Let A = (3, 4), B = (-2, 3) and C = (x, y)
bydata AB = BC = CA
⇒ AB2 = BC2 = CA2
consider BC2 = CA2
⇒ (x + 2)2 + (y – 3)2 = (x – 3)2 + (y – 4)2

4x + 6x – 6y + 8y = 25 – 13
∴ 10x + 2y =12
5x + y = 6 …… (1)
Again AB = BC
⇒ AB2 = BC2
∴ 5(3 + 2)2 + (4 – 3)2 = (x + 2)2 + (y – 3)2 ………. (2)
from(1) y = 6 – 5x
∴ substitute (1) is (2)
(x + 2)2 + (6 – 5x – 3)2 = 26
x2 + 4 + 4x + 9 + 25x2 – 30x = 26
26x2 – 26x – 13 = 0
⇒ 2x2 – 2x – 1 = 0

Section Formula

Question 1.
Find the co-ordinates of the point which divide the line joining the points
(i) (1, -3) and (-3, 9) internally in the ratio 1 : 3 and the line joining the points
(ii) (2, -6) (4, 3) externally is the ratio 3 : 2.
(i) Let A = (1, -3) B = (-3 9), Ratio is 1 : 3

(ii) Let A = (2, -6) B = (4, 3) ratio 3 : 2

Question 2.
Find the ratio in which (2, 7) divides the line joining the point (8, 9) and (-7, 4).
Let
A = (x1 y1) = (8, 9)
B = (x2 y2) = (-7, 4)
p = (x1 y2) = (2, 7)
The ratio l : m in which p divides AB, is given by

⇒ The ratio is l : m = 2 : 3
∴ the points (2,7) divides internally

Question 3.
Determine the ratio in which the line 3x + y – 9 = 0 divide the segment joining the points (1, 3) and (2, 7).
Let the of division k : 1 as and point then the coordinates of the points are $$\left(\frac{2 k+1}{k+1}, \frac{7 k+3}{k+1}\right)$$
But the point ‘C’ lies on 3x + y – 9 = 0

⇒ 4x – 3 = 0
⇒ k = $$\frac { 3 }{ 4 }$$
∴ The required ratio is 3 : 4 internally

Question 4.
Find the lengths of the medians of a triangle whose vertices are (3, 5) (5, 3) and (7, 7).