Class 9 Advanced Maths Chapter 4 Coordinate Geometry Ex 4.2 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 4 Coordinate Geometry Ex 4.2 to verify their answers.

Advanced Maths Class 9 Exercise 4.2 Solutions

Class 9 Advanced Maths Ex 4.2 Solutions

Question 1.
Point P(a, b) lies in IV Quadrant. Its perpendicular distance from the y-axis is 3 units greater than its perpendicular distance from the x-axis. If the product of its coordinates is -28, find the value of the expression a² + b.
Solution:
Since point P(a, b) lies in the IV quadrant:
a > 0, b < 0
The perpendicular distance from the y-axis is |a| and from the x-axis is | b |.
Given that the distance from the y-axis is 3 units greater than the distance from the x-axis:
|a| = |b| + 3
In the IV quadrant: |a| = a, |b| = -b
So, a = – b + 3
Also, the product of the coordinates is -28:
ab = -28
Substitute a = 3 – b:
(3 – b)b = -28
3b – b² = -28
b² – 3b – 28 = 0
(b – 7)(b + 4) = 0
Since b < 0,
b = -4
Then, a = 3 – (-4) = 7
Now compute:
a² + b = 7² + (-4)
= 49 – 4 = 45

Question 2.
Point Q(k – 4, 2k + 7) lies strictly in II Quadrant. The perpendicular distance of Point Q from the y-axis is exactly twice its perpendicular distance from the x-axis. Find the value of k and evaluate the expression k³ + 10.
Solution:
Point Q(k – 4, 2k + 7) lies in the II quadrant.
So, x < 0 and y > 0
Thus, k – 4 < 0 ⇒ k < 4 And 2k + 7 > 0 ⇒ k > \(-\frac{7}{2}\)
The perpendicular distance from the y-axis is:
|k – 4|
The perpendicular distance from the x-axis is: |2k + 7|
Given: |k – 4| = 2|2k + 7|
Since Q is in II quadrant:
k – 4 < 0 ⇒ | k – 4| = 4 – k And 2k + 7 > 0
⇒ |2k + 7| = 2k + 7
Equation:
4 – k = 2(2k + 7)
4 – k = 4k + 14
-5k = 10 ⇒ k = -2
Now, k³ + 10 = (-2)² + 10 = -8 + 10 = 2

Question 3.
Let P(3, -4) be a point on the Cartesian plane. Point Q(a, b) is the image of P when reflected through y-axis. Point R(c, d) is the image of Q when reflected in the x-axis. Find the value of the \(\sqrt{a c-b d}\).
Solution:
Given: P = (3, -4)
Reflect P in the y-axis
Reflection in the y-axis changes the sign of the x-coordinate:
Q = (a, b) = (-3, -4)
So, a = -3, b = -4
Reflection in the x-axis changes the sign of the y-coordinate:
R = (c, d) = (-3, 4)
So, c = -3, d = 4
Now, ac = (-3) (-3) = 9
bd = (-4) (4) = -16
Therefore, ac – bd = 9 – (-16) = 25
\(\sqrt{a c-b d}\) = \(\sqrt{25}\) = 5