Class 9 Advanced Maths Chapter 4 Coordinate Geometry Ex 4.4 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 4 Coordinate Geometry Ex 4.4 to verify their answers.

Advanced Maths Class 9 Exercise 4.4 Solutions

Class 9 Advanced Maths Ex 4.4 Solutions

Question 1.
The line kx + 3y -12 = 0 forms a right-angled triangle with the x and y-coordinate axes. If the total area of this triangle is 12 square units, find all possible values for the slope of this line.
Solution:
Given, kx + 3y -12 = 0
Rewrite equation in slope – intercept form, we get
3y = – kx + 12
y = \(-\frac{k}{3} x\) + 4
So, the slope is: \(-\frac{k}{3}\)
For x-intercept
Put y = 0
kx – 12 = 0
x = \(\frac{12}{k}\)
For y-intercept
Put x = 0:
3y – 12 = 0
y = 4
The triangle formed with the coordinate axes has base: |\(\frac{12}{k}\)| and height: 4.
Area = \(\frac{1}{2}\)|\(\frac{12}{k}\)| = 12
\(\frac{24}{|k|}\) = 12
|k| = 2
Thus, k = 2 or k = -2
Since, slope, m = \(-\frac{k}{3}\)
For k = 2, m = \(-\frac{2}{3}\)
For k = -2, m = \(\frac{2}{3}\)
Therefore, the possible values of the slope are: \(\frac{2}{3}\) and \(-\frac{2}{3}\).

Question 2.
The straight line px + qy + r = 0 (where p, q, r ≠ 0) forms an isosceles right-angled triangle with the coordinate axes in the first quadrant. What must be the algebraic relationship between the coefficients p and q?
Solution:
For the line px + qy + r = 0, to form an isosceles right-angled triangle with the coordinate axes in the first quadrant, the x-intercept and y-intercept must be equal.
For x – intercept, put y = 0, in the equation
px + r = 0
x = \(-\frac{r}{p}\)
For y – intercept, put x = 0, in the equation
qy + r = 0
y = \(-\frac{r}{q}\)
For an isosceles right triangle:
x-intercept = y-intercept
\(-\frac{r}{p}=-\frac{r}{q}\)
Thus, p = q
This is the required relation.

Question 3.
Line l₁ has the equation 3x – 5y + 10 = 0. Line Z2 has the equation 5x + 3y + K = 0. Prove algebraically that l₁ and l₂ are perpendicular. Also, if the x-intercept of l₁, is identical to the y-intercept of l₂ then find the value of K.
Solution:
l₁: 3x – 5y + 10 = 0
Slope: m₁ = \(\frac{3}{5}\)
l₂: 5x + 3y + K = 0
Slope: m₂ = \(-\frac{5}{3}\)
m₁ × m₂ = \(\frac{3}{5} \times-\frac{5}{3}\) = -1
So, lines are perpendicular.
Now x-intercept of l₁
x = \(-\frac{10}{3}\)
y-intercept of l₂:
y = \(-\frac{K}{3}\)
Given equal:
\(-\frac{10}{3}=-\frac{K}{3}\) ⇒ K = 10

Question 4.
A line is written in the general form kx – y + C = 0. You are given two clues about this line:
Clue 1: The line passes through the coordinate point (3,10).
Clue 2: The sum of its x-intercept and y-intercept is exactly equal to its slope.
Find all possible equations for this line.
Solution:
Given the line: kx – y + C = 0
Rewrite in slope-intercept form: y = kx + C
So, the slope is: m = k
Since the line passes through (3,10),
10 = 3k + C
C = 10 – 3k
Thus the line becomes: y = kx + 10 – 3k
For y-intercept: Put x = 0:
y = 10 – 3k
So, y-intercept = 10 – 3k
For x-intercept: Put y = 0:
0 = kx + 10 – 3k
x = \(\frac{3 k-10}{k}\)
3k – 10 + k(10 – 3k) = k²
3k – 10 + 10k – 3k² = k²
13k – 10 = 4k²
4k² – 13k + 10 = 0
Factorising: (4k – 5)(k – 2) = 0
k = \(\frac{5}{4}\) or k = 2 4
When k = 2, then C = 10 – 3(2) = 10 – 6 = 4
Thus, the equation is 2x – y + 4 = 0
When k = \(\frac{5}{4}\), then C = 10 – 3\(\left(\frac{5}{4}\right)\) = 10 – \(\frac{15}{4}\) = \(\frac{25}{4}\)
So, the equation is:
\(\frac{5}{4}\)x – y + \(\frac{25}{4}\) = 0
OR 5x – 4y + 25 = 0