Given: [tex]$ABCD$[/tex] is a kite with [tex]$AB = AD$[/tex] and [tex]$CB = CD$[/tex]. [tex]$B$[/tex] is the point with coordinates [tex]$(10, 19)$[/tex] and [tex]$D$[/tex] is the point with coordinates [tex]$(2, 7)$[/tex].

Find an equation of the line [tex]$AC$[/tex]. Give your answer in the form [tex]$py + qx = r$[/tex] where [tex]$p$[/tex], [tex]$q$[/tex], and [tex]$r$[/tex] are integers.



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]