Class 9 Advanced Maths Chapter 4 Coordinate Geometry Ex 4.6 Solutions - #NCSOLVE 📚

Accurate and updated NCERT Class 9 Advanced Maths Solutions Chapter 4 Coordinate Geometry Ex 4.6 help students build a strong mathematical foundation.

Ex 4.6 Class 9 Advanced Maths Solutions

Advanced Maths Class 9 Exercise 4.6 Solutions

Exercise 4.6 Class 9 Advanced Maths 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 sq units, find the values of its x-intercept a and y-intercept b.
Solution:
Given, the line passes through the point (2, 3) and forms a right angled triangle with positive X and Y-axes.
∵ The intercepts are on the positive axes.
∴ a > 0 and h > 0
The area of a triangle formed by a line with the axes is given by 12.

∴ Area = \(\frac{1}{2}\) × Base × Height
⇒ \(\frac{1}{2}\) × a × b = 12
⇒ ab = 24
⇒ b = \(\frac{24}{a}\) …………… (i)
The equation of a line in intercept form is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
∵ The line passes through the point (2, 3).
∴ \(\frac{2}{a}\) + \(\frac{3}{b}\) = 1 ………… (ii)
On substituting the value ofb from Eq. (i) in Eq. (ii), we get
\(\frac{2}{a}+\frac{3}{\left(\frac{24}{a}\right)}\) = 1
\(\frac{2}{a}\) + \(\frac{3a}{24}\) = 1 ⇒ \(\frac{2}{a}\) + \(\frac{a}{8}\) = 1
\(\frac{16+a^2}{8 a}\) = 1
⇒ a² + 16 = 8a
⇒ a² – 8a + 16 = 0 ⇒ (a – 4)² = 0
∴ a = 4
On substituting a = 4 into Eq. (i), we get
b = \(\frac{24}{4}\)
∴ b = 6
Hence, the value of the x-intercept is 4 and the y-intercept is 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 the product of its intercepts.
Solution:
Given, the sum of the x-intercepts and y-intercept is 9.
∴ a + b = 9
⇒ b = 9 – a ……….. (i)
Let the equation of the given line is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
∵ The line passes through the point (2, 2)

⇒ 9a – a² = 18
⇒ a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a(a – 6) -3 (a – 6) = 0
⇒ (a – 3) (a – 6) = 0
∴ a = 3 or a = 6
Case I When a = 3, b = 9 – 3 = 6 [From Eq. (i)]
Case II When a = 6, b = 9 – 6 = 3 [from Eq.(i)]
∴ The product of intercepts = 3 × 6 = 18 or 6 × 3 = 18.
Hence, the value of the product of its 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:
Given, the straight line passes through the point (3, 5).
Let the x-intercept be a and the y-intercept be b.
∴ a + b = 0 [given]
Case I Intercepts are zero.
If a = 0 and b = 0, the line passes through the origin (0, 0).
The equation of a line passing through (0, 0) and (x₁, y₁) is
y = \(\frac{y_1}{x_1}\)x. [∵ slope = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{y_1-0}{x_1-0}\) = \(\frac{y_1}{x_1}\)]
∵ The line passes through (3, 5).
∴ y = \(\frac{5}{3}\)x
⇒ 5x – 3y = 0
Case II Intercepts are non-zero.
The equation of the line in intercept form is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
∴ Sum of intercepts, a + b = 0
⇒ b = -a
On substituting b = -a in the intercept form, we get
\(\frac{x}{a}\) + \(\frac{y}{-a}\) = 1
⇒ \(\frac{x-y}{a}\) = 1
⇒ x – y = a ……… (i)
∵ The line passes through the point (3, 5).
∴ 3 – 5 = a
⇒ a = -2
On substituting a = -2 in Eq. (i), we get
x – y = -2
⇒ x – y + 2 = 0
Hence, the equations of possible lines are 5x – 3y = 0 and x – y + 2 = 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 sq 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:
Given, the straight line forms a right angled triangle with the positive X and Y-axes.
Let the x-intercept be a and the y-intercept be b.
∵ The intercepts are on the positive axes.
∴ a > 0 and b> 0
Also, given the area of the triangle is 24 sq units.
∴ \(\frac{1}{2}\) × a × b = 24
⇒ ab = 48 …………… (i)

The hypotenuse is the line segment between (a, 0) and (0, b).
∵ Length of hypotenuse is 10 units.
∴ \(\sqrt{a^2+b^2}\) = 10
⇒ a² + b² = 100
We know that (a + b)² = a² + b² + 2ab
(a + b)² – 2ab = 100 ………. (ii)
⇒ (a + b)² – 2(48) = 100 [using Eq. (i)]
⇒ (a + b)² = 196
⇒ a + b = 14 [∵ a, b > 0]
⇒ b = 14 – a ………………. (iii)
On substituting b = 14 – a in Eq. (i), we get
⇒ a(14 – a) = 48
⇒ 14a – a² = 48
⇒ a² – 14a + 48 = 0
⇒ a² – 8a – 6a + 48 = 0
⇒ a(a – 8) – 6(a – 8) = 0
⇒ (a – 8) (a – 6) = 0
∴ a = 8 or a = 6
Case I When a = 8, b = 14 – 8 = 6 [from Eq. (iii)]
Case II When a = 6, b = 14 – 6 = 8 [from Eq. (iii)]
The intercept form of a line is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
For Case I, the equation is \(\frac{x}{8}\) + \(\frac{y}{6}\) = 1.
For Case II, the equation is \(\frac{x}{6}\) + \(\frac{y}{8}\) = 1.
Hence, the possible equations of the line in intercept form
are \(\frac{x}{8}\) + \(\frac{y}{6}\) = 1 and \(\frac{x}{6}\) + \(\frac{y}{8}\) = 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:
Given, the straight line passes through the point (3, 2).
The equation of the line in intercept form is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
Also, given the x-intercept a and y-intercept b are both positive numbers.
∴ a > 0 and b > 0
∵ The sum of its intercepts is 12.
∴ a + b = 12
⇒ b = 12 – a ……..(i)
∵ The line passes through the point (3, 2). …….. (ii)
∴ \(\frac{3}{a}\) + \(\frac{2}{b}\) = 1 ……. (ii)
On substituting the value of b from Eq. (i) in Eq. (ii), we get
\(\frac{3}{a}\) + \(\frac{2}{12 – a}\) = 1
⇒ \(\frac{3(12-a)+2 a}{a(12-a)}\) = 1 ⇒ \(\frac{36-3 a+2 a}{12 a-a^2}\) = 1
⇒ 36 – a = 12a – a²
⇒ a² – 13a + 36 = 0
⇒ a² – 9a – 4a + 36 = 0
⇒ a(a – 9) – 4 (a – 9) = 0
⇒ (a – 4)(a – 9) = 0
∴ a = 4 or a = 9
Case I When a = 4, b = 12 – 4 = 8 [from Eq. (i)]
∴ The equation of required line is \(\frac{x}{4}\) + \(\frac{y}{8}\) = 1
i.e. 2x + y = 8.
Case II When a = 9 ⇒ b = 12 – 9 = 3 [fromEq. (i)]
∴ The equation of required line is \(\frac{x}{9}\) + \(\frac{y}{3}\) = 1 = 1 i.e. x + 3y = 9.
Hence, the possible equations for the line are 2x + y = 8 and x + 3y = 9.

The post Class 9 Advanced Maths Chapter 4 Coordinate Geometry Ex 4.6 Solutions appeared first on Learn CBSE.

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