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Sagot :
Let's solve the problem step-by-step.
### Part (a): Finding the equation of the perpendicular line passing through A(-9, -6)
1. Equation of the given line (line 4):
[tex]\[ 2x + 5y = 10 \][/tex]
2. Determine the slope of the given line:
For a line in the form [tex]\(Ax + By = C\)[/tex], the slope [tex]\(m\)[/tex] is given by [tex]\(-\frac{A}{B}\)[/tex].
[tex]\[ A = 2, \quad B = 5 \rightarrow \text{Slope of line 4} = -\frac{2}{5} \][/tex]
3. Find the slope of the perpendicular line (line 2):
The slope of a line perpendicular to another is the negative reciprocal of the given slope.
[tex]\[ \text{Slope of line 2} = -\left(-\frac{5}{2}\right) = \frac{5}{2} \][/tex]
4. Equation of the perpendicular line (line 2) passing through point [tex]\(A(-9, -6)\)[/tex]:
Using the point-slope form of a line equation: [tex]\(y - y_1 = m(x - x_1)\)[/tex], where [tex]\((x_1, y_1)\)[/tex] is the point [tex]\((-9, -6)\)[/tex] and [tex]\(m\)[/tex] is the slope [tex]\(\frac{5}{2}\)[/tex].
[tex]\[ y + 6 = \frac{5}{2}(x + 9) \][/tex]
5. Simplify the equation to standard form:
Distribute and rearrange to obtain the equation in the form [tex]\(Ax + By + C = 0\)[/tex].
[tex]\[ y + 6 = \frac{5}{2}x + \frac{45}{2} \][/tex]
Multiply through by 2 to clear the fraction:
[tex]\[ 2y + 12 = 5x + 45 \][/tex]
Rearrange terms to standard form:
[tex]\[ 5x - 2y + 33 = 0 \][/tex]
So, the equation of the perpendicular line (line 2) is:
[tex]\[ 5x - 2y + 33 = 0 \][/tex]
### Part (b): Finding the area of triangle [tex]\(ABO\)[/tex]
6. Find the intersection point [tex]\(B\)[/tex] of lines 4 and 2:
Solve the system of equations:
[tex]\[ \begin{cases} 2x + 5y = 10 \\ 5x - 2y + 33 = 0 \end{cases} \][/tex]
From the calculations, the intersection (point [tex]\(B\)[/tex]) is:
[tex]\[ B\left( -5, 4 \right) \][/tex]
7. Use the determinant method to find the area of triangle [tex]\(ABO\)[/tex] with vertices [tex]\(A(-9, -6)\)[/tex], [tex]\(B(-5, 4)\)[/tex], and [tex]\(O(0,0)\)[/tex]:
The area of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] is given by:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
Substituting the coordinates of points [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(O\)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \left| (-9)(4 - 0) + (-5)(0 - (-6)) + 0((-6) - 4) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -36 + 30 + 0 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -6 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} (6) \][/tex]
[tex]\[ \text{Area} = 33 \][/tex]
The area of triangle [tex]\(ABO\)[/tex] is:
[tex]\[ 33 \text{ square units} \][/tex]
### Part (a): Finding the equation of the perpendicular line passing through A(-9, -6)
1. Equation of the given line (line 4):
[tex]\[ 2x + 5y = 10 \][/tex]
2. Determine the slope of the given line:
For a line in the form [tex]\(Ax + By = C\)[/tex], the slope [tex]\(m\)[/tex] is given by [tex]\(-\frac{A}{B}\)[/tex].
[tex]\[ A = 2, \quad B = 5 \rightarrow \text{Slope of line 4} = -\frac{2}{5} \][/tex]
3. Find the slope of the perpendicular line (line 2):
The slope of a line perpendicular to another is the negative reciprocal of the given slope.
[tex]\[ \text{Slope of line 2} = -\left(-\frac{5}{2}\right) = \frac{5}{2} \][/tex]
4. Equation of the perpendicular line (line 2) passing through point [tex]\(A(-9, -6)\)[/tex]:
Using the point-slope form of a line equation: [tex]\(y - y_1 = m(x - x_1)\)[/tex], where [tex]\((x_1, y_1)\)[/tex] is the point [tex]\((-9, -6)\)[/tex] and [tex]\(m\)[/tex] is the slope [tex]\(\frac{5}{2}\)[/tex].
[tex]\[ y + 6 = \frac{5}{2}(x + 9) \][/tex]
5. Simplify the equation to standard form:
Distribute and rearrange to obtain the equation in the form [tex]\(Ax + By + C = 0\)[/tex].
[tex]\[ y + 6 = \frac{5}{2}x + \frac{45}{2} \][/tex]
Multiply through by 2 to clear the fraction:
[tex]\[ 2y + 12 = 5x + 45 \][/tex]
Rearrange terms to standard form:
[tex]\[ 5x - 2y + 33 = 0 \][/tex]
So, the equation of the perpendicular line (line 2) is:
[tex]\[ 5x - 2y + 33 = 0 \][/tex]
### Part (b): Finding the area of triangle [tex]\(ABO\)[/tex]
6. Find the intersection point [tex]\(B\)[/tex] of lines 4 and 2:
Solve the system of equations:
[tex]\[ \begin{cases} 2x + 5y = 10 \\ 5x - 2y + 33 = 0 \end{cases} \][/tex]
From the calculations, the intersection (point [tex]\(B\)[/tex]) is:
[tex]\[ B\left( -5, 4 \right) \][/tex]
7. Use the determinant method to find the area of triangle [tex]\(ABO\)[/tex] with vertices [tex]\(A(-9, -6)\)[/tex], [tex]\(B(-5, 4)\)[/tex], and [tex]\(O(0,0)\)[/tex]:
The area of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] is given by:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
Substituting the coordinates of points [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(O\)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \left| (-9)(4 - 0) + (-5)(0 - (-6)) + 0((-6) - 4) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -36 + 30 + 0 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -6 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} (6) \][/tex]
[tex]\[ \text{Area} = 33 \][/tex]
The area of triangle [tex]\(ABO\)[/tex] is:
[tex]\[ 33 \text{ square units} \][/tex]
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