- Demonstrate the ability to find the inclination of a line
- Demonstrate the ability to find the angle between two lines
- Demonstrate the ability to find the distance between a point and a line
#1:
Instructions: Find the inclination of the line θ.
$$a)\hspace{.1em}x + 5y=20$$
$$b)\hspace{.1em}11x - 5y=-35$$
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#2:
Instructions: Find the inclination of the line θ.
$$a)\hspace{.1em}x + y=3$$
$$b)\hspace{.1em}9x - y=-2$$
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#3:
Instructions: Find the acute angle θ between the two lines.
$$a)\hspace{.1em}x - 3y=6, 2x + 3y=-24$$
$$b)\hspace{.1em}5x - y=-1, x - 2y=16$$
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#4:
Instructions: Find the acute angle θ between the two lines.
$$a)\hspace{.1em}x + 3y=-24, 5x - 6y=6$$
$$b)\hspace{.1em}x - 4y=28, 15x + 4y=36$$
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#5:
Instructions: Find the distance between the point and the line.
$$a)\hspace{.1em}x - 4y=17, (-3, 4)$$
$$b)\hspace{.1em}x - 2y=-2, (-1, 3)$$
$$c)\hspace{.1em}ax + by + c=0, (x_1, y_1)$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.1em}θ ≈ 168.69°$$
$$b)\hspace{.1em}θ ≈ 65.56°$$
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#2:
Solutions:
$$a)\hspace{.1em}θ=135°$$
$$b)\hspace{.1em}θ ≈ 83.66°$$
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#3:
Solutions:
$$a)\hspace{.1em}θ ≈ 52.13°$$
$$b)\hspace{.1em}θ ≈ 52.13°$$
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#4:
Solutions:
$$a)\hspace{.1em}θ ≈ 58.24°$$
$$b)\hspace{.1em}θ ≈ 89.10°$$
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#5:
Solutions:
$$a)\hspace{.1em}d=\frac{36\sqrt{17}}{17}$$
$$b)\hspace{.1em}d=\sqrt{5}$$
$$c)\hspace{.1em}d=\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$ Our goal is to obtain the formula given above for the distance between the point (x1, y1) and the line ax + by + c = 0. Let's begin with a simple sketch. 
The shortest distance from the point (x1, y1) to the line ax + by + c = 0 is the perpendicular line segment joining the point and the line. 
To make things easier to understand, let's just convert the line segment into a line. This line passes through the point (x1, y1) and is perpendicular to ax + by + c = 0. The point of intersection is labeled (x2, y2).
At this point, let's stop and think about the steps.
- Write the equation ax + by + c = 0 in slope-intercept form
- Write the equation of the perpendicular line in slope-intercept form
- Solve the system of equations to find the point of intersection (x2, y2)
- Plug into the distance formula and simplify
- We want the distance between (x1, y1) and (x2, y2)
Step #1: Write the equation ax + by + c = 0 in slope-intercept form. $$ax + by + c=0$$ $$by=-ax - c$$ $$y=-\frac{a}{b}x - \frac{c}{b}$$ Step #2: Write the equation of the perpendicular line in slope-intercept form.
First off, we know that perpendicular lines have slopes whose product is -1. The slope from our line is given as -a/b. $$m \cdot -\frac{a}{b}=-1$$ $$m=\frac{b}{a}$$ Since we know a point (x1, y1) and the slope, we can use point-slope form. $$y - y_1=m(x - x_1)$$ $$y - y_1=\frac{b}{a}(x - x_1)$$ Solve for y to obtain slope-intercept form. $$y=\frac{b}{a}x - \frac{b}{a}x_1 + y_1$$ Step #3: Solve the system of equations to find the point of intersection (x2, y2).
Note: This is where things start to get tricky. If you are doing this on your own, double-check each step. It's very easy to make a simple mistake that carries through the whole process. $$1) y=-\frac{a}{b}x - \frac{c}{b}$$ $$2) y=\frac{b}{a}x - \frac{b}{a}x_1 + y_1$$ We will use the substitution method. Plug in for y in equation 1 (in other words, take the right side of equation 2, which y is equal to, and plug in for y in equation 1). $$\frac{b}{a}x - \frac{b}{a}x_1 + y_1=-\frac{a}{b}x - \frac{c}{b}$$ Here, you want to solve for x, this is eventually going to be labeled as x2 since it will be the x-value for the point of intersection. For now, let's just call it x. Note that you have an x1 here and you must be very careful about what x you are referring to at all times. $$\frac{b}{a}x + \frac{a}{b}x=- \frac{c}{b}+ \frac{b}{a}x_1 - y_1$$ $$\left(\frac{b}{a}+ \frac{a}{b}\right)x=- \frac{c}{b}+ \frac{b}{a}x_1 - y_1$$ $$x=\large{\frac{- \frac{c}{b}+ \frac{b}{a}x_1 - y_1}{\frac{b}{a}+ \frac{a}{b}}}$$ $$x=\frac{ab(- \frac{c}{b}+ \frac{b}{a}x_1 - y_1)}{ab(\frac{b}{a}+ \frac{a}{b})}$$ $$x=\frac{ab \cdot -\frac{c}{b}+ ab \cdot \frac{b}{a}x_1 - ab \cdot y_1}{ab \cdot \frac{b}{a}+ ab \cdot \frac{a}{b}}$$ $$x=\frac{-ac + b^2x_1 - aby_1}{a^2 + b^2}$$ We have found the x-value for the point of intersection. Again, we will label this as x2 shortly. For now, let's use this to plug in for x in either original equation. Since equation 1 is simpler, let's use that. $$1) y=-\frac{a}{b}x - \frac{c}{b}$$ $$y=-\frac{a}{b}\cdot \frac{-ac + b^2x_1 - aby_1}{a^2 + b^2}- \frac{c}{b}$$ $$y=\frac{-a(-ac + b^2x_1 - aby_1)}{b(a^2 + b^2)}- \frac{c}{b}$$ $$y=\frac{a^2c - ab^2x_1 + a^2by_1}{b(a^2 + b^2)}- \frac{c}{b}$$ $$y=\frac{a^2c - ab^2x_1 + a^2by_1}{b(a^2 + b^2)}- \frac{c(a^2 + b^2)}{b(a^2 + b^2)}$$ $$y=\frac{a^2c - ab^2x_1 + a^2by_1 - [c(a^2 + b^2)]}{b(a^2 + b^2)}$$ $$y=\frac{a^2c - ab^2x_1 + a^2by_1 - a^2c - b^2c}{b(a^2 + b^2)}$$ $$y=\frac{- ab^2x_1 + a^2by_1 - b^2c}{b(a^2 + b^2)}$$ $$y=\frac{b(- abx_1 + a^2y_1 - bc)}{b(a^2 + b^2)}$$ $$y=\frac{- abx_1 + a^2y_1 - bc}{a^2 + b^2}$$ At this point, we will rename x as x2 and y as y2. These are the coordinates from the point of intersection. In other words (x2, y2) is the solution for the system of equations. $$x_2=\frac{-ac + b^2x_1 - aby_1}{a^2 + b^2}$$ $$y_2=\frac{- abx_1 + a^2y_1 - bc}{a^2 + b^2}$$ Step #4: Plug into the distance formula and simplify. $$d=\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d=\sqrt{\left(\frac{-ac + b^2x_1 - aby_1}{a^2 + b^2}- x_1\right)^2 + \left(\frac{- abx_1 + a^2y_1 - bc}{a^2 + b^2}- y_1\right)^2}$$ Let's work on the first part separately. We will return to update the distance formula. $$\left(\frac{-ac + b^2x_1 - aby_1}{a^2 + b^2}- x_1\right)^2$$ Get a common denominator and simplify. $$=\left(\frac{-ac + b^2x_1 - aby_1}{a^2 + b^2}- \frac{x_1(a^2 + b^2)}{a^2 + b^2}\right)^2$$ $$=\left(\frac{-ac + b^2x_1 - aby_1 - [x_1(a^2 + b^2)]}{a^2 + b^2}\right)^2$$ $$=\left(\frac{-ac + b^2x_1 - aby_1 - a^2x_1 - b^2x_1}{a^2 + b^2}\right)^2$$ $$=\left(\frac{-ac - aby_1 - a^2x_1}{a^2 + b^2}\right)^2$$ Let's update our distance formula. $$d=\sqrt{\left(\frac{-ac - aby_1 - a^2x_1}{a^2 + b^2}\right)^2 + \left(\frac{- abx_1 + a^2y_1 - bc}{a^2 + b^2}- y_1\right)^2}$$ Let's work on the second part separately. We will return to update the distance formula. $$\left(\frac{- abx_1 + a^2y_1 - bc}{a^2 + b^2}- y_1\right)^2$$ Get a common denominator and simplify. $$=\left(\frac{- abx_1 + a^2y_1 - bc}{a^2 + b^2}- \frac{y_1(a^2 + b^2)}{a^2 + b^2}\right)^2$$ $$=\left(\frac{- abx_1 + a^2y_1 - bc - [y_1(a^2 + b^2)]}{a^2 + b^2}\right)^2$$ $$=\left(\frac{- abx_1 + a^2y_1 - bc - a^2y_1 - b^2y_1}{a^2 + b^2}\right)^2$$ $$=\left(\frac{- abx_1 - bc - b^2y_1}{a^2 + b^2}\right)^2$$ Let's update our distance formula. $$d=\sqrt{\left(\frac{-ac - aby_1 - a^2x_1}{a^2 + b^2}\right)^2 + \left(\frac{- abx_1 - bc - b^2y_1}{a^2 + b^2}\right)^2}$$ Factor out -a from the numerator of the first part and -b from the numerator of the second part. $$d=\sqrt{\left(\frac{-a(c + by_1 + ax_1)}{a^2 + b^2}\right)^2 + \left(\frac{-b (ax_1 + c + by_1)}{a^2 + b^2}\right)^2}$$ $$d=\sqrt{\left(\frac{-a(ax_1 + by_1 + c)}{a^2 + b^2}\right)^2 + \left(\frac{-b (ax_1 + by_1 + c)}{a^2 + b^2}\right)^2}$$ $$d=\sqrt{\frac{(-a)^2(ax_1 + by_1 + c)^2}{(a^2 + b^2)^2}+ \frac{(-b)^2(ax_1 + by_1 + c)^2}{(a^2 + b^2)^2}}$$ $$d=\sqrt{\frac{a^2(ax_1 + by_1 + c)^2 + b^2(ax_1 + by_1 + c)^2}{(a^2 + b^2)^2}}$$ Let's factor out (ax1 + by1 + c)2: $$d=\sqrt{\frac{(ax_1 + by_1 + c)^2 (a^2 + b^2)}{(a^2 + b^2)^2}}$$ Cancel a common factor of (a2 + b2) between the numerator and denominator. $$d=\sqrt{\frac{(ax_1 + by_1 + c)^2}{a^2 + b^2}}$$ $$d=\frac{\sqrt{(ax_1 + by_1 + c)^2}}{\sqrt{a^2 + b^2}}$$ For the numerator we can use the following rule: $$\sqrt{x^2}=|x|$$ If we apply this rule to the numerator: $$\sqrt{(ax_1 + by_1 + c)^2}=|ax_1 + by_1 + c|$$ This leads to the our formula: $$d=\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$
