Class 9 Advanced Maths Chapter 4 Coordinate Geometry Ex 4.4 Solutions

Accurate and updated NCERT Class 9 Advanced Maths Solutions Chapter 4 Coordinate Geometry Ex 4.4 help students build a strong mathematical foundation.

Ex 4.4 Class 9 Advanced Maths Solutions

Advanced Maths Class 9 Exercise 4.4 Solutions

Exercise 4.4 Class 9 Advanced Maths 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 sq units, find all possible values for the slope of this line.
Solution:
Given, the equation of the line is kx + 3y – 12 = 0.
To find the intercepts, we put y = 0 for the x-intercept and x = 0 for the y-intercept.
For x-intercept,
kx – 12 = 0 ⇒ x = \(\frac{12}{k}\)
∴ The coordinates of the vertex on the X-axis are A(\(\frac{12}{k}\), 0)
For y-intercept,
3y – 12 = 0 ⇒ y = 4
∴ The coordinates of the vertex on the Y-axis are B(0, 4).
The line forms a right angled triangle with the origin O(0, 0) and the axes.
The lengths of the base and height of this triangle are \(\frac{12}{k}\) and 4, respectively.
We know that the area of a triangle = \(\frac{1}{2}\) × Base × Height
Given, area of the triangle = 12 sq units
⇒ \(\frac{1}{2}\) × |\(\frac{12}{k}\)| × |4| = 12
⇒ \(\frac{24}{k}\) = 12
⇒ |k| = 2
⇒ k = 2 or k = -2
Now, the slope (m) of the line kx + 3y – 12 = 0 is given by m = \(-\frac{\text { Coefficient of } x}{\text { Coefficient of } y}\)
⇒ m = –\(\frac{k}{3}\)
Case I If k = 2, m = –\(\frac{2}{3}\)
Case II If k = -2, m = \(\frac{(-2)}{3}\) = \(\frac{2}{3}\)
Hence, the possible values for the slope of the line are \(\frac{2}{3}\) and – \(\frac{2}{3}\)

Question 2.
The straight line px + qy + r = 0 (where p, q, and r ≠ 0) forms an isosceles right angled triangle with the coordinate axes in the quadrant. What must be the algebraic relationship between the coefficients p and q?
Solution:
Given, the equation of the line is px + qy + r = 0, where p, q, r ≠ 0.
To find the intercepts made by the line on the coordinate axes, we put y = 0 for the x-intercept and x = 0 for the y-intercept.
For x-intercept
px + r = 0 ⇒ x = –\(\frac{r}{p}\)
∴ The vertex of the triangle on the X-axis is A (-\(\frac{r}{p}\), 0).
For y-intercept
qy + r = 0 ⇒ y = –\(\frac{r}{p}\),
∴ The vertex of the triangle on the X-axis is B (0, \(\frac{r}{q}\)).
The triangle is formed by these intercepts and the origin O(0, 0).
We know that a triangle formed by the coordinate axes is always right angled at the origin.
Also, given the triangle is an isosceles right angled triangle.
∴ Length of base OA = Length of height OB r 9

Hence, the algebraic relationship between the coefficients p and q must be |p| = |q|.

Question 3.
Line l₁ has the equation 3x – 5y + 10 = 0.
Line l₂ 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:
Given, the equation of line l₁, is 3x – 5y + 10 = 0
and the equation of line l₂ is 5x + 3y + K = 0.
We know that the slope m of a line Ax + By + C- 0 is
given by m = –\(\frac{A}{B}\).
Slope of line l₁, m₁ = –\(\frac{3}{-5}\) = \(\frac{3}{5}\)
and slope of line l₂, m₂ = –\(\frac{5}{3}\).
Now, m₁ × m₂ = (\(\frac{3}{5}\)) –\(\frac{5}{3}\) = -1
Since, the product of the slopes is -1, the lines l₁ and l₂ are perpendicular.
For x-intercept of l₁ we put y = 0 in its equation.
∴ 3x – 5(0) + 10 = 0 ⇒ 3x = -10 ⇒ x = –\(\frac{10}{3}\)
For y-intercept of l₂, we put x = 0 in its equation.
∴ 5(0) + 3y + K = 0
⇒ 3y = -K ⇒ y = –\(\frac{K}{3}\)
Also, given the x-intercept of l₁ is identical to the
y-intercept of l₂ i.e. –\(\frac{10}{3}\) = –\(\frac{K}{3}\) ⇒ 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 general form of the line is kx – y + C = 0.
For x-intercept, put y = 0 in given equation of line.
∴ kx – 0 + C = 0 ⇒ x = –\(\frac{C}{k}\)
For y-intercept, put x = 0 in given equation of line
∴ k(0) – y + C = 0
⇒ y = C
Now, slope of given line = –\(\frac{k}{(-1)}\) = k
Since, the given line passes through the coordinate point (3, 10).
∴ k(3) – 10 + C = 0
⇒ 3k + C = 10
⇒ C = 10 – 3k ……….. (i)
Also, the sum of x-intercept and y-intercept of given line is equal to its slope.
∴ –\(\frac{C}{k}\) + C = k
⇒ \(\frac{-C+C k}{k}\) = k ⇒ C(k – 1) = k²
On substituting the value of C from Eq. (i) into Eq.(ii), we get
(10 – 3k) (k – 1) = k²
⇒ 10k – 10 – 3k² + 3k = k²
⇒ -3k² + 13k – 10 = k²
⇒ 4k² – 13k + 10 = 0
⇒ 4k² – 8k – 5k + 10 = 0
⇒ 4k(k – 2) – 5(k – 2) = 0
⇒ (4k – 5)(k – 2) = 0
∴ k = 2 or k = \(\frac{5}{4}\)
Case I When k = 2, C = 10 – 3 (2) = 4 [from Eq. (i)]
∴ The equation of required line is 2x – y + 4 = 0.
∴ Case II When k = \(\frac{5}{4}\),
C = 10 – 3(\(\frac{5}{4}\)) = 10 – \(\frac{15}{4}\) = \(\frac{25}{4}\) [from Eq. (i)]
∴ The equations of required line is
\(\frac{5}{4}\) x – y + \(\frac{25}{4}\) = 0i.e. 5x – 4y + 25 = 0.

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