Math / Geometry

QuestionFind the length of the altitude in $\triangle \mathrm{ABC}$ from B Given: $\mathrm{A}(0,2), \mathrm{B}(3,7), \mathrm{C}(5,3)$ Extra Credit: Find the parametric vector equation and standard

Studdy Solution

STEP 1

1. We are given the coordinates of points A, B, and C.
2. We need to find the length of the altitude from point B to line AC.
3. The altitude is the perpendicular distance from point B to line AC.
4. We will find the equation of line AC and use it to find the perpendicular distance from B.

STEP 2

1. Find the equation of line AC.
2. Calculate the perpendicular distance from point B to line AC.
3. Extra Credit: Find the parametric vector equation and standard form of line AC.

STEP 3

To find the equation of line AC, we need the slope of line AC. The slope $m$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Using points A(0, 2) and C(5, 3):
$m = \frac{3 - 2}{5 - 0} = \frac{1}{5}$

STEP 4

Using the point-slope form of a line $y - y_1 = m(x - x_1)$ , we can write the equation of line AC. Using point A(0, 2):
$y - 2 = \frac{1}{5}(x - 0)$ $y = \frac{1}{5}x + 2$

STEP 5

To find the perpendicular distance from point B(3, 7) to line AC, use the formula for the distance from a point to a line $Ax + By + C = 0$ :
$\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
Convert the line equation $y = \frac{1}{5}x + 2$ to standard form:
$y - \frac{1}{5}x - 2 = 0$ $-\frac{1}{5}x + y - 2 = 0$ $1x - 5y + 10 = 0$
Here, $A = 1$ , $B = -5$ , $C = 10$ .

STEP 6

Substitute point B(3, 7) into the distance formula:
$\text{Distance} = \frac{|1(3) - 5(7) + 10|}{\sqrt{1^2 + (-5)^2}}$ $= \frac{|3 - 35 + 10|}{\sqrt{1 + 25}}$ $= \frac{|-22|}{\sqrt{26}}$ $= \frac{22}{\sqrt{26}}$

STEP 7

Extra Credit: Find the parametric vector equation of line AC. The direction vector of AC is $\langle 5 - 0, 3 - 2 \rangle = \langle 5, 1 \rangle$ .
The parametric equations are:
$x = 0 + 5t$ $y = 2 + 1t$

STEP 8

Extra Credit: Convert the line equation $y = \frac{1}{5}x + 2$ to standard form:
$y - \frac{1}{5}x - 2 = 0$ $-\frac{1}{5}x + y - 2 = 0$ $1x - 5y + 10 = 0$
The length of the altitude from B to line AC is:
$\frac{22}{\sqrt{26}}$
The parametric vector equation of line AC is:
$x = 5t$ $y = 2 + t$
The standard form of line AC is:
$x - 5y + 10 = 0$

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QuestionFind the length of the altitude in $\triangle \mathrm{ABC}$ from B Given: $\mathrm{A}(0,2), \mathrm{B}(3,7), \mathrm{C}(5,3)$ Extra Credit: Find the parametric vector equation and standard

Studdy Solution

STEP 1

1. We are given the coordinates of points A, B, and C.
2. We need to find the length of the altitude from point B to line AC.
3. The altitude is the perpendicular distance from point B to line AC.
4. We will find the equation of line AC and use it to find the perpendicular distance from B.

STEP 2

1. Find the equation of line AC.
2. Calculate the perpendicular distance from point B to line AC.
3. Extra Credit: Find the parametric vector equation and standard form of line AC.

STEP 3

To find the equation of line AC, we need the slope of line AC. The slope $m$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Using points A(0, 2) and C(5, 3):
$m = \frac{3 - 2}{5 - 0} = \frac{1}{5}$

STEP 4

Using the point-slope form of a line $y - y_1 = m(x - x_1)$ , we can write the equation of line AC. Using point A(0, 2):
$y - 2 = \frac{1}{5}(x - 0)$ $y = \frac{1}{5}x + 2$

STEP 5

STEP 6

Substitute point B(3, 7) into the distance formula:
$\text{Distance} = \frac{|1(3) - 5(7) + 10|}{\sqrt{1^2 + (-5)^2}}$ $= \frac{|3 - 35 + 10|}{\sqrt{1 + 25}}$ $= \frac{|-22|}{\sqrt{26}}$ $= \frac{22}{\sqrt{26}}$

STEP 7

Extra Credit: Find the parametric vector equation of line AC. The direction vector of AC is $\langle 5 - 0, 3 - 2 \rangle = \langle 5, 1 \rangle$ .
The parametric equations are:
$x = 0 + 5t$ $y = 2 + 1t$

STEP 8

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QuestionFind the length of the altitude in △ABC\triangle \mathrm{ABC}△ABC from B Given: A(0,2),B(3,7),C(5,3)\mathrm{A}(0,2), \mathrm{B}(3,7), \mathrm{C}(5,3)A(0,2),B(3,7),C(5,3) Extra Credit: Find the parametric vector equation and standard

Studdy Solution

STEP 1

1. We are given the coordinates of points A, B, and C.2. We need to find the length of the altitude from point B to line AC.3. The altitude is the perpendicular distance from point B to line AC.4. We will find the equation of line AC and use it to find the perpendicular distance from B.

STEP 2

1. Find the equation of line AC.2. Calculate the perpendicular distance from point B to line AC.3. Extra Credit: Find the parametric vector equation and standard form of line AC.

STEP 3

To find the equation of line AC, we need the slope of line AC. The slope m m m is given by:m=y2−y1x2−x1 m = \frac{y_2 - y_1}{x_2 - x_1} m=x2​−x1​y2​−y1​​Using points A(0, 2) and C(5, 3):m=3−25−0=15 m = \frac{3 - 2}{5 - 0} = \frac{1}{5} m=5−03−2​=51​

STEP 4

Using the point-slope form of a line y−y1=m(x−x1) y - y_1 = m(x - x_1) y−y1​=m(x−x1​), we can write the equation of line AC. Using point A(0, 2):y−2=15(x−0) y - 2 = \frac{1}{5}(x - 0) y−2=51​(x−0) y=15x+2 y = \frac{1}{5}x + 2 y=51​x+2

STEP 5

STEP 6

STEP 7

Extra Credit: Find the parametric vector equation of line AC. The direction vector of AC is ⟨5−0,3−2⟩=⟨5,1⟩ \langle 5 - 0, 3 - 2 \rangle = \langle 5, 1 \rangle ⟨5−0,3−2⟩=⟨5,1⟩.The parametric equations are:x=0+5t x = 0 + 5t x=0+5t y=2+1t y = 2 + 1t y=2+1t

STEP 8

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Studdy solves anything!

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QuestionFind the length of the altitude in $\triangle \mathrm{ABC}$ from B Given: $\mathrm{A}(0,2), \mathrm{B}(3,7), \mathrm{C}(5,3)$ Extra Credit: Find the parametric vector equation and standard

1. We are given the coordinates of points A, B, and C.
2. We need to find the length of the altitude from point B to line AC.
3. The altitude is the perpendicular distance from point B to line AC.
4. We will find the equation of line AC and use it to find the perpendicular distance from B.

1. Find the equation of line AC.
2. Calculate the perpendicular distance from point B to line AC.
3. Extra Credit: Find the parametric vector equation and standard form of line AC.

To find the equation of line AC, we need the slope of line AC. The slope $m$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Using points A(0, 2) and C(5, 3):
$m = \frac{3 - 2}{5 - 0} = \frac{1}{5}$

Using the point-slope form of a line $y - y_1 = m(x - x_1)$ , we can write the equation of line AC. Using point A(0, 2):
$y - 2 = \frac{1}{5}(x - 0)$ $y = \frac{1}{5}x + 2$

Extra Credit: Find the parametric vector equation of line AC. The direction vector of AC is $\langle 5 - 0, 3 - 2 \rangle = \langle 5, 1 \rangle$ .
The parametric equations are:
$x = 0 + 5t$ $y = 2 + 1t$