Answer :
Let's solve the problem step by step:
### a. Find the equation of line [tex]\( l_2 \)[/tex].
1. Identify the slope of line [tex]\( l_1 \)[/tex]:
The line [tex]\( l_1 \)[/tex] is given by the equation:
[tex]\[
2x + y - 10 = 0
\][/tex]
We can rearrange this equation into the slope-intercept form [tex]\( y = mx + c \)[/tex]:
[tex]\[
y = -2x + 10
\][/tex]
Here, the slope [tex]\( m \)[/tex] of line [tex]\( l_1 \)[/tex] is [tex]\(-2\)[/tex].
2. Determine the slope of line [tex]\( l_2 \)[/tex]:
Line [tex]\( l_2 \)[/tex] is perpendicular to line [tex]\( l_1 \)[/tex]. The slope of a line perpendicular to another line with slope [tex]\( m \)[/tex] is the negative reciprocal of [tex]\( m \)[/tex]. Therefore, the slope of [tex]\( l_2 \)[/tex] is:
[tex]\[
\text{slope of } l_2 = \frac{1}{2}
\][/tex]
3. Find the equation of line [tex]\( l_2 \)[/tex]:
Line [tex]\( l_2 \)[/tex] crosses the [tex]\( x \)[/tex]-axis at the point [tex]\( A(-2, 0) \)[/tex]. We'll use the point-slope form to find the equation:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Plug in [tex]\( m = \frac{1}{2} \)[/tex], [tex]\( x_1 = -2 \)[/tex], and [tex]\( y_1 = 0 \)[/tex]:
[tex]\[
y - 0 = \frac{1}{2}(x + 2)
\][/tex]
[tex]\[
y = \frac{1}{2}x + 1
\][/tex]
So, the equation of line [tex]\( l_2 \)[/tex] is [tex]\( y = \frac{1}{2}x + 1 \)[/tex].
### b. Find the coordinates of [tex]\( M \)[/tex], the point of intersection of [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex].
1. Set the equations of [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] equal to find the intersection:
[tex]\[
-2x + 10 = \frac{1}{2}x + 1
\][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[
-2x + 10 = \frac{1}{2}x + 1
\][/tex]
Multiply every term by 2 to eliminate the fraction:
[tex]\[
-4x + 20 = x + 2
\][/tex]
Combine like terms:
[tex]\[
-5x = -18
\][/tex]
Divide by -5:
[tex]\[
x = \frac{18}{5}
\][/tex]
[tex]\( x = 3.6 \)[/tex]
3. Substitute [tex]\( x = 3.6 \)[/tex] back into either equation to find [tex]\( y \)[/tex]:
Using equation [tex]\( l_1: y = -2x + 10 \)[/tex]:
[tex]\[
y = -2(3.6) + 10 = 2.8
\][/tex]
The coordinates of point [tex]\( M \)[/tex] are [tex]\( (3.6, 2.8) \)[/tex].
### c. Find the x-intercept of line [tex]\( l_1 \)[/tex], point [tex]\( B \)[/tex], and calculate the area of triangle [tex]\( \triangle AMB \)[/tex].
1. Find point [tex]\( B \)[/tex]:
To find the [tex]\( x \)[/tex]-intercept of [tex]\( l_1 \)[/tex], set [tex]\( y = 0 \)[/tex]:
[tex]\[
2x - 10 = 0
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
2x = 10 \Rightarrow x = 5
\][/tex]
So, point [tex]\( B \)[/tex] is [tex]\( (5, 0) \)[/tex].
2. Calculate the area of triangle [tex]\( \triangle AMB \)[/tex]:
The vertices of triangle [tex]\( \triangle AMB \)[/tex] are [tex]\( A(-2, 0) \)[/tex], [tex]\( M(3.6, 2.8) \)[/tex], and [tex]\( B(5, 0) \)[/tex].
Use the formula for the area of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], [tex]\((x_3, y_3)\)[/tex]:
[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]
Substitute the values:
[tex]\[
\text{Area} = \frac{1}{2} \left| (-2)(2.8 - 0) + 3.6(0 - 0) + 5(0 - 2.8) \right|
\][/tex]
[tex]\[
\text{Area} = \frac{1}{2} \left| -5.6 + 0 - 14 \right|
\][/tex]
[tex]\[
\text{Area} = \frac{1}{2} \times 19.6 = 9.8
\][/tex]
The exact area of triangle [tex]\( \triangle AMB \)[/tex] is [tex]\( 9.8 \)[/tex] square units.
### a. Find the equation of line [tex]\( l_2 \)[/tex].
1. Identify the slope of line [tex]\( l_1 \)[/tex]:
The line [tex]\( l_1 \)[/tex] is given by the equation:
[tex]\[
2x + y - 10 = 0
\][/tex]
We can rearrange this equation into the slope-intercept form [tex]\( y = mx + c \)[/tex]:
[tex]\[
y = -2x + 10
\][/tex]
Here, the slope [tex]\( m \)[/tex] of line [tex]\( l_1 \)[/tex] is [tex]\(-2\)[/tex].
2. Determine the slope of line [tex]\( l_2 \)[/tex]:
Line [tex]\( l_2 \)[/tex] is perpendicular to line [tex]\( l_1 \)[/tex]. The slope of a line perpendicular to another line with slope [tex]\( m \)[/tex] is the negative reciprocal of [tex]\( m \)[/tex]. Therefore, the slope of [tex]\( l_2 \)[/tex] is:
[tex]\[
\text{slope of } l_2 = \frac{1}{2}
\][/tex]
3. Find the equation of line [tex]\( l_2 \)[/tex]:
Line [tex]\( l_2 \)[/tex] crosses the [tex]\( x \)[/tex]-axis at the point [tex]\( A(-2, 0) \)[/tex]. We'll use the point-slope form to find the equation:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Plug in [tex]\( m = \frac{1}{2} \)[/tex], [tex]\( x_1 = -2 \)[/tex], and [tex]\( y_1 = 0 \)[/tex]:
[tex]\[
y - 0 = \frac{1}{2}(x + 2)
\][/tex]
[tex]\[
y = \frac{1}{2}x + 1
\][/tex]
So, the equation of line [tex]\( l_2 \)[/tex] is [tex]\( y = \frac{1}{2}x + 1 \)[/tex].
### b. Find the coordinates of [tex]\( M \)[/tex], the point of intersection of [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex].
1. Set the equations of [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] equal to find the intersection:
[tex]\[
-2x + 10 = \frac{1}{2}x + 1
\][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[
-2x + 10 = \frac{1}{2}x + 1
\][/tex]
Multiply every term by 2 to eliminate the fraction:
[tex]\[
-4x + 20 = x + 2
\][/tex]
Combine like terms:
[tex]\[
-5x = -18
\][/tex]
Divide by -5:
[tex]\[
x = \frac{18}{5}
\][/tex]
[tex]\( x = 3.6 \)[/tex]
3. Substitute [tex]\( x = 3.6 \)[/tex] back into either equation to find [tex]\( y \)[/tex]:
Using equation [tex]\( l_1: y = -2x + 10 \)[/tex]:
[tex]\[
y = -2(3.6) + 10 = 2.8
\][/tex]
The coordinates of point [tex]\( M \)[/tex] are [tex]\( (3.6, 2.8) \)[/tex].
### c. Find the x-intercept of line [tex]\( l_1 \)[/tex], point [tex]\( B \)[/tex], and calculate the area of triangle [tex]\( \triangle AMB \)[/tex].
1. Find point [tex]\( B \)[/tex]:
To find the [tex]\( x \)[/tex]-intercept of [tex]\( l_1 \)[/tex], set [tex]\( y = 0 \)[/tex]:
[tex]\[
2x - 10 = 0
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
2x = 10 \Rightarrow x = 5
\][/tex]
So, point [tex]\( B \)[/tex] is [tex]\( (5, 0) \)[/tex].
2. Calculate the area of triangle [tex]\( \triangle AMB \)[/tex]:
The vertices of triangle [tex]\( \triangle AMB \)[/tex] are [tex]\( A(-2, 0) \)[/tex], [tex]\( M(3.6, 2.8) \)[/tex], and [tex]\( B(5, 0) \)[/tex].
Use the formula for the area of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], [tex]\((x_3, y_3)\)[/tex]:
[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]
Substitute the values:
[tex]\[
\text{Area} = \frac{1}{2} \left| (-2)(2.8 - 0) + 3.6(0 - 0) + 5(0 - 2.8) \right|
\][/tex]
[tex]\[
\text{Area} = \frac{1}{2} \left| -5.6 + 0 - 14 \right|
\][/tex]
[tex]\[
\text{Area} = \frac{1}{2} \times 19.6 = 9.8
\][/tex]
The exact area of triangle [tex]\( \triangle AMB \)[/tex] is [tex]\( 9.8 \)[/tex] square units.
