Answer :
To find the equation of the line [tex]\(AC\)[/tex] in the form [tex]\(p y + q x = r\)[/tex], we will follow these steps:
1. Determine the midpoint of diagonal [tex]\(BD\)[/tex].
2. Calculate the slope of diagonal [tex]\(BD\)[/tex].
3. Determine the slope of diagonal [tex]\(AC\)[/tex] using the property that [tex]\(AC\)[/tex] is perpendicular to [tex]\(BD\)[/tex].
4. Derive the equation of the line [tex]\(AC\)[/tex] in point-slope form.
5. Convert the equation to the desired form [tex]\(p y + q x = r\)[/tex].
### Step 1: Find the midpoint of [tex]\(BD\)[/tex]
The coordinates of points [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are [tex]\(B(10, 19)\)[/tex] and [tex]\(D(2, 7)\)[/tex], respectively.
The midpoint [tex]\(M\)[/tex] of diagonal [tex]\(BD\)[/tex] is calculated as:
[tex]\[ M_x = \frac{B_x + D_x}{2} = \frac{10 + 2}{2} = 6 \][/tex]
[tex]\[ M_y = \frac{B_y + D_y}{2} = \frac{19 + 7}{2} = 13 \][/tex]
So, the coordinates of midpoint [tex]\(M\)[/tex] are [tex]\((6, 13)\)[/tex].
### Step 2: Calculate the slope of [tex]\(BD\)[/tex]
The slope [tex]\(m_{BD}\)[/tex] of the line passing through points [tex]\(B(10, 19)\)[/tex] and [tex]\(D(2, 7)\)[/tex] is given by:
[tex]\[ m_{BD} = \frac{D_y - B_y}{D_x - B_x} = \frac{7 - 19}{2 - 10} = \frac{-12}{-8} = \frac{3}{2} \][/tex]
### Step 3: Determine the slope of [tex]\(AC\)[/tex]
Since the line [tex]\(AC\)[/tex] is perpendicular to [tex]\(BD\)[/tex], the slope [tex]\(m_{AC}\)[/tex] is the negative reciprocal of the slope of [tex]\(BD\)[/tex]:
[tex]\[ m_{AC} = -\frac{1}{m_{BD}} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \][/tex]
### Step 4: Derive the equation of [tex]\(AC\)[/tex] in point-slope form
We use the point-slope form of a linear equation:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is the point [tex]\(M(6, 13)\)[/tex] and [tex]\(m\)[/tex] is the slope [tex]\(-\frac{2}{3}\)[/tex].
Plug in the values:
[tex]\[ y - 13 = -\frac{2}{3}(x - 6) \][/tex]
### Step 5: Convert the equation to the form [tex]\(p y + q x = r\)[/tex]
First, simplify and convert to standard form:
[tex]\[ y - 13 = -\frac{2}{3}x + 4 \][/tex]
Multiply all terms by 3 to clear the fraction:
[tex]\[ 3(y - 13) = -2(x - 6) \][/tex]
[tex]\[ 3y - 39 = -2x + 12 \][/tex]
[tex]\[ 3y + 2x = 51 \][/tex]
We rearrange to match the desired form [tex]\(p y + q x = r\)[/tex]:
[tex]\[ 2x + 3y = 51 \][/tex]
Thus, the equation of the line [tex]\(AC\)[/tex] is:
[tex]\[ 6y + 10x = 170 \][/tex]
So, the values of [tex]\(p\)[/tex], [tex]\(q\)[/tex], and [tex]\(r\)[/tex] are:
[tex]\[ (p, q, r) = (6, 10, 170) \][/tex]
1. Determine the midpoint of diagonal [tex]\(BD\)[/tex].
2. Calculate the slope of diagonal [tex]\(BD\)[/tex].
3. Determine the slope of diagonal [tex]\(AC\)[/tex] using the property that [tex]\(AC\)[/tex] is perpendicular to [tex]\(BD\)[/tex].
4. Derive the equation of the line [tex]\(AC\)[/tex] in point-slope form.
5. Convert the equation to the desired form [tex]\(p y + q x = r\)[/tex].
### Step 1: Find the midpoint of [tex]\(BD\)[/tex]
The coordinates of points [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are [tex]\(B(10, 19)\)[/tex] and [tex]\(D(2, 7)\)[/tex], respectively.
The midpoint [tex]\(M\)[/tex] of diagonal [tex]\(BD\)[/tex] is calculated as:
[tex]\[ M_x = \frac{B_x + D_x}{2} = \frac{10 + 2}{2} = 6 \][/tex]
[tex]\[ M_y = \frac{B_y + D_y}{2} = \frac{19 + 7}{2} = 13 \][/tex]
So, the coordinates of midpoint [tex]\(M\)[/tex] are [tex]\((6, 13)\)[/tex].
### Step 2: Calculate the slope of [tex]\(BD\)[/tex]
The slope [tex]\(m_{BD}\)[/tex] of the line passing through points [tex]\(B(10, 19)\)[/tex] and [tex]\(D(2, 7)\)[/tex] is given by:
[tex]\[ m_{BD} = \frac{D_y - B_y}{D_x - B_x} = \frac{7 - 19}{2 - 10} = \frac{-12}{-8} = \frac{3}{2} \][/tex]
### Step 3: Determine the slope of [tex]\(AC\)[/tex]
Since the line [tex]\(AC\)[/tex] is perpendicular to [tex]\(BD\)[/tex], the slope [tex]\(m_{AC}\)[/tex] is the negative reciprocal of the slope of [tex]\(BD\)[/tex]:
[tex]\[ m_{AC} = -\frac{1}{m_{BD}} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \][/tex]
### Step 4: Derive the equation of [tex]\(AC\)[/tex] in point-slope form
We use the point-slope form of a linear equation:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is the point [tex]\(M(6, 13)\)[/tex] and [tex]\(m\)[/tex] is the slope [tex]\(-\frac{2}{3}\)[/tex].
Plug in the values:
[tex]\[ y - 13 = -\frac{2}{3}(x - 6) \][/tex]
### Step 5: Convert the equation to the form [tex]\(p y + q x = r\)[/tex]
First, simplify and convert to standard form:
[tex]\[ y - 13 = -\frac{2}{3}x + 4 \][/tex]
Multiply all terms by 3 to clear the fraction:
[tex]\[ 3(y - 13) = -2(x - 6) \][/tex]
[tex]\[ 3y - 39 = -2x + 12 \][/tex]
[tex]\[ 3y + 2x = 51 \][/tex]
We rearrange to match the desired form [tex]\(p y + q x = r\)[/tex]:
[tex]\[ 2x + 3y = 51 \][/tex]
Thus, the equation of the line [tex]\(AC\)[/tex] is:
[tex]\[ 6y + 10x = 170 \][/tex]
So, the values of [tex]\(p\)[/tex], [tex]\(q\)[/tex], and [tex]\(r\)[/tex] are:
[tex]\[ (p, q, r) = (6, 10, 170) \][/tex]
