Class 9 Advanced Maths Chapter 4 Coordinate Geometry Ex 4.6 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 4 Coordinate Geometry Ex 4.6 to verify their answers.

Advanced Maths Class 9 Exercise 4.6 Solutions

Class 9 Advanced Maths Ex 4.6 Solutions

Question 1.
A straight line passes through the point (2, 3) and forms a right-angled triangle with the positive x and y axes. If the area of this triangle is 12 square units, find the values of its x-intercept (a) and y-intercept (b).
Solution:
Let the line cut the positive x-axis at (a, 0) and the positive y-axis at (0, b).
Then its intercept form is: \(\frac{x}{a}+\frac{y}{b}\) = 1
Since the line passes through (2, 3).
Then, \(\frac{2}{a}+\frac{3}{b}\) = 1
The triangle formed with the coordinate axes has area:
\(\frac{1}{2}\)ab = 12
ab = 24
So, b = \(\frac{24}{a}\)
Substitute into the first equation:
\(\frac{2}{a}+\frac{3}{\frac{24}{a}}\) = 1
\(\frac{2}{a}+\frac{3 a}{24}\) = 1
\(\frac{2}{a}+\frac{a}{8}\) = 1
Multiply by 8a:
16 + a² = 8a
a² – 8a + 16 = 0
(a – 4)² = 0
a = 4
Now, b = \(\frac{24}{4}\) = 6
Therefore, a = 4, b = 6.

Question 2.
A straight line passes through the point (2, 2). The sum of its x-intercept (a) and its y-intercept (b) is 9. Determine the value of the product of its intercepts.
Solution:
Let the x-intercept be a and the y-intercept be b.
The intercept form of the line is:
\(\frac{x}{a}+\frac{y}{b}\) = 1
Since the line passes through (2, 2)
Then, \(\frac{2}{a}+\frac{2}{b}\) = 1
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{2}\)
\(\frac{a+b}{a b}=\frac{1}{2}\)
Given, a + b = 9
Put the value in (i), we get
\(\frac{9}{a b}=\frac{1}{2}\)
ab = 18
Therefore, the product of the intercepts is 18.

Question 3.
A straight line passes through the point (3, 5). The sum of its x-intercept and its y-intercept is zero. Find the equation(s) of all possible lines that satisfy these conditions.
Solution:
Case I: Both intercepts are non-zero
Let the x-intercept be a and the y-intercept be b.
The intercept form of a line is:
\(\frac{x}{a}+\frac{y}{b}\) = 1
Given, a + b = 0
Then, b = -a
\(\frac{x}{a}+\frac{y}{-a}\) = 1
x – y = a
Since the line passes through (3, 5),
3 – 5 = a
a = -2
Thus, x – y = -2 or y = x + 2

Case II: The line passes through the origin
In this case, x-intercept = 0, y-intercept = 0, and therefore
0 + 0 = 0.
The line through (0, 0) and (3, 5) has slope:
m = \(\frac{5-0}{3-0}=\frac{5}{3}\)
Therefore, y = \(\frac{5 x}{3}\)
or 5x – 3y = 0.

Question 4.
A straight line forms a right-angled triangle with the positive x and y-axes. The total area of this triangle is 24 square units and the length of its hypotenuse (the line segment intercepted between the axes) is 10 units. Find all possible equations of this line in the intercept form.
Solution:
Let the x-intercept be a and the y-intercept be b.
Since, the line cuts the positive x-axis and y-axis, its equation in intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1
Area of triangle formed with the axes: \(\frac{1}{2}\) ab = 24
ab = 48 …(i)
The intercepted segment between (a, 0) and (0, b) is the hypotenuse.
Using distance formula:
\(\sqrt{a^2+b^2}\) = 10
a² + b² = 100 …(ii)
Using identity:
(a + b)² = a² + b² + 2ab
Substitute values (a + b)² = 100 + 96
(a + b) = 14
Now solve:
a + b = 14, ab = 48
Form quadratic equation,
t² – 14f + 48 = 0
(t – 6)(t – 8) = 0
t = 6, 8
So, a = 6, b = 8
Or, a = 8, b = 6
Then, required equations are:
\(\frac{x}{6}+\frac{y}{8}\) = 1 or \(\frac{x}{8}+\frac{y}{6}\) = 1

Question 5.
A straight line passes through the point (3, 2). The x-intercept (a) and y-intercept (b) of this line are both positive numbers. If the sum of its intercepts is 12, find all possible equations for this line.
Solution:
Let the x-intercept be a and the y-intercept be b, where a > 0 and b > 0.
The intercept form of the line is:
\(\frac{x}{a}+\frac{y}{b}\) = 1
Since the line passes through (3, 2), substitute x = 3 and y = 2:
\(\frac{3}{a}+\frac{2}{b}\) = 1 …(i)
We are also given, a + b = 12
So, b = 12 – a
Substitute into equation (i)
3(12 – a) + 2a = a(12 – a)
36 – 3a + 2a = 12a – a²
a² – 12a – a + 36 = 0
a² – 13a + 36 = 0
a² – 9a – 4a + 36 = 0
(a – 9)(a – 4) =0
a = 9, a = 4
Using a + b = 12
If a = 9, then b = 3
If a = 4, then b = 8
Then, first possible equation is :
\(\frac{x}{9}+\frac{y}{3}\) = 1 or x + 3y = 9
Second possible equation is:
\(\frac{x}{4}+\frac{y}{8}\) = 1 or 2x + y = 8