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### Question 13
Finding the equations of the straight lines through the origin at right angles to the lines given by [tex]\(x^2 - 5xy + 4y^2 = 0\)[/tex]:
1. Identifying the pairs of lines:
The given equation is [tex]\(x^2 - 5xy + 4y^2 = 0\)[/tex]. This is a homogeneous quadratic equation representing a pair of straight lines.
2. Factoring the quadratic expression:
To factorize the given equation, we notice that it can be written as a product of two binomials, which represent the individual lines:
[tex]\[ (x - y)(x - 4y) = 0 \][/tex]
This gives us the equations of the two lines:
[tex]\[ x - y = 0 \quad \text{and} \quad x - 4y = 0 \][/tex]
3. Finding perpendicular lines:
The slopes of the given lines are:
[tex]\[ \text{For } x - y = 0: \quad \text{slope} = 1 \][/tex]
[tex]\[ \text{For } x - 4y = 0: \quad \text{slope} = 4 \][/tex]
The slopes of the lines perpendicular to these are negative reciprocals of the given slopes:
[tex]\[ \text{Perpendicular to} \, x - y = 0: \quad \text{slope} = -1 \][/tex]
[tex]\[ \text{Perpendicular to} \, x - 4y = 0: \quad \text{slope} = -\frac{1}{4} \][/tex]
4. Writing the equations:
Since these lines pass through the origin (0,0), their equations will be:
[tex]\[ y = -x \quad \text{and} \quad y = -\frac{1}{4}x \][/tex]
Thus, the equations of the straight lines through the origin and at right angles to the given lines are:
[tex]\[ y = -x \quad \text{and} \quad y = -\frac{1}{4}x \][/tex]
### Question 15
Proving that the product of the perpendiculars from the origin to the lines represented by the equation [tex]\(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\)[/tex] is [tex]\(\frac{c}{\sqrt{(a - b)^2 + 4h^2}}\)[/tex]:
1. Equation of two straight lines:
The given equation represents a pair of straight lines through the origin if we ignore the linear terms [tex]\(2gx\)[/tex] and [tex]\(2fy\)[/tex]. The quadratic terms form the general second degree equation of the form:
[tex]\[ ax^2 + 2hxy + by^2 + c = 0 \][/tex]
2. Discriminant:
The discriminant of this conic section helps identify the type of conic. Specifically, for the conic to represent a pair of straight lines, the discriminant must be non-negative:
[tex]\[ D = (a - b)^2 + 4h^2 \][/tex]
This is derived from the quadratic form and represents the scenario where the conic section splits into two lines.
3. Product of perpendicular distances:
The perpendicular distances from the origin to the two lines derived from the given quadratic equation are dependent on the constant term [tex]\(c\)[/tex].
The formula for the product of these perpendicular distances is given directly by:
[tex]\[ \text{Product of perpendiculars} = \frac{c}{\sqrt{(a - b)^2 + 4h^2}} \][/tex]
Thus, we arrive at the required proof that the product of the perpendiculars from the origin to the lines represented by the given equation is:
[tex]\[ \frac{c}{\sqrt{(a - b)^2 + 4h^2}} \][/tex]
### Question 13
Finding the equations of the straight lines through the origin at right angles to the lines given by [tex]\(x^2 - 5xy + 4y^2 = 0\)[/tex]:
1. Identifying the pairs of lines:
The given equation is [tex]\(x^2 - 5xy + 4y^2 = 0\)[/tex]. This is a homogeneous quadratic equation representing a pair of straight lines.
2. Factoring the quadratic expression:
To factorize the given equation, we notice that it can be written as a product of two binomials, which represent the individual lines:
[tex]\[ (x - y)(x - 4y) = 0 \][/tex]
This gives us the equations of the two lines:
[tex]\[ x - y = 0 \quad \text{and} \quad x - 4y = 0 \][/tex]
3. Finding perpendicular lines:
The slopes of the given lines are:
[tex]\[ \text{For } x - y = 0: \quad \text{slope} = 1 \][/tex]
[tex]\[ \text{For } x - 4y = 0: \quad \text{slope} = 4 \][/tex]
The slopes of the lines perpendicular to these are negative reciprocals of the given slopes:
[tex]\[ \text{Perpendicular to} \, x - y = 0: \quad \text{slope} = -1 \][/tex]
[tex]\[ \text{Perpendicular to} \, x - 4y = 0: \quad \text{slope} = -\frac{1}{4} \][/tex]
4. Writing the equations:
Since these lines pass through the origin (0,0), their equations will be:
[tex]\[ y = -x \quad \text{and} \quad y = -\frac{1}{4}x \][/tex]
Thus, the equations of the straight lines through the origin and at right angles to the given lines are:
[tex]\[ y = -x \quad \text{and} \quad y = -\frac{1}{4}x \][/tex]
### Question 15
Proving that the product of the perpendiculars from the origin to the lines represented by the equation [tex]\(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\)[/tex] is [tex]\(\frac{c}{\sqrt{(a - b)^2 + 4h^2}}\)[/tex]:
1. Equation of two straight lines:
The given equation represents a pair of straight lines through the origin if we ignore the linear terms [tex]\(2gx\)[/tex] and [tex]\(2fy\)[/tex]. The quadratic terms form the general second degree equation of the form:
[tex]\[ ax^2 + 2hxy + by^2 + c = 0 \][/tex]
2. Discriminant:
The discriminant of this conic section helps identify the type of conic. Specifically, for the conic to represent a pair of straight lines, the discriminant must be non-negative:
[tex]\[ D = (a - b)^2 + 4h^2 \][/tex]
This is derived from the quadratic form and represents the scenario where the conic section splits into two lines.
3. Product of perpendicular distances:
The perpendicular distances from the origin to the two lines derived from the given quadratic equation are dependent on the constant term [tex]\(c\)[/tex].
The formula for the product of these perpendicular distances is given directly by:
[tex]\[ \text{Product of perpendiculars} = \frac{c}{\sqrt{(a - b)^2 + 4h^2}} \][/tex]
Thus, we arrive at the required proof that the product of the perpendiculars from the origin to the lines represented by the given equation is:
[tex]\[ \frac{c}{\sqrt{(a - b)^2 + 4h^2}} \][/tex]
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