Answer :
To solve this question, we will need to use concepts from coordinate geometry.
Given that ABCD is a rhombus and we have the coordinates of A which is point (5,11), we also know the equation of diagonal DB is y = x + 5.
In rhombus ABCD, the diagonals cross at a 90-degree angle and bisect each other. This means if we know the slope of one diagonal, the slope of the other will be the negative reciprocal of it. Since the equation of diagonal DB is y = x + 5, we can see that its slope, [tex]\( m_{DB} \)[/tex], is 1 (because the coefficient of x is 1).
Therefore, the slope of diagonal AC, [tex]\( m_{AC} \)[/tex], will be the negative reciprocal of the slope of diagonal DB, which is -1.
Now we need to find the midpoint of diagonal DB which will also be the midpoint of diagonal AC. Since AB is a rhombus, both diagonals bisect each other. We will find the midpoint by setting the x and y coordinates equal to each other since we know diagonal DB lies on the line y = x + 5.
Let’s call [tex]\( M(x_m, y_m) \)[/tex] the midpoint. We know:
[tex]\[ y_m = x_m + 5 \][/tex]
Since [tex]\( A(5, 11) \)[/tex] is one endpoint of diagonal AC, and the midpoint of AC will be the same as the midpoint of DB, we can find the x-coordinate of the midpoint, [tex]\( x_m \)[/tex], by solving:
[tex]\[ 2x_m + 5 = 11 \][/tex]
Subtract 5 from both sides of the equation:
[tex]\[ 2x_m = 6 \][/tex]
Divide by 2:
[tex]\[ x_m = 3 \][/tex]
Knowing [tex]\( x_m \)[/tex], we can now substitute back to find [tex]\( y_m \)[/tex]:
[tex]\[ y_m = x_m + 5 = 3 + 5 = 8 \][/tex]
So the midpoint [tex]\( M \)[/tex] is at coordinates (3, 8).
Now, we have the slope of diagonal AC, [tex]\( m_{AC} = -1 \)[/tex], and a point it passes through, which is the midpoint [tex]\( M(3, 8) \)[/tex]. We can use this to find the equation of diagonal AC in the form [tex]\( y = mx + c \)[/tex].
Using the point-slope form of a line equation:
[tex]\[ y - y_m = m_{AC}(x - x_m) \][/tex]
Substitute the known values for [tex]\( m_{AC} \)[/tex], [tex]\( x_m \)[/tex], and [tex]\( y_m \)[/tex]:
[tex]\[ y - 8 = -1(x - 3) \][/tex]
Expand and simplify:
[tex]\[ y = -x + 3 + 8 \][/tex]
[tex]\[ y = -x + 11 \][/tex]
So the equation of diagonal AC is [tex]\( y = -x + 11 \)[/tex], which is in the desired form of [tex]\( y = mx + c \)[/tex].
Here we find that [tex]\( m = -1 \)[/tex] and [tex]\( c = 11 \)[/tex], where both [tex]\( m \)[/tex] and [tex]\( c \)[/tex] are integers.
Given that ABCD is a rhombus and we have the coordinates of A which is point (5,11), we also know the equation of diagonal DB is y = x + 5.
In rhombus ABCD, the diagonals cross at a 90-degree angle and bisect each other. This means if we know the slope of one diagonal, the slope of the other will be the negative reciprocal of it. Since the equation of diagonal DB is y = x + 5, we can see that its slope, [tex]\( m_{DB} \)[/tex], is 1 (because the coefficient of x is 1).
Therefore, the slope of diagonal AC, [tex]\( m_{AC} \)[/tex], will be the negative reciprocal of the slope of diagonal DB, which is -1.
Now we need to find the midpoint of diagonal DB which will also be the midpoint of diagonal AC. Since AB is a rhombus, both diagonals bisect each other. We will find the midpoint by setting the x and y coordinates equal to each other since we know diagonal DB lies on the line y = x + 5.
Let’s call [tex]\( M(x_m, y_m) \)[/tex] the midpoint. We know:
[tex]\[ y_m = x_m + 5 \][/tex]
Since [tex]\( A(5, 11) \)[/tex] is one endpoint of diagonal AC, and the midpoint of AC will be the same as the midpoint of DB, we can find the x-coordinate of the midpoint, [tex]\( x_m \)[/tex], by solving:
[tex]\[ 2x_m + 5 = 11 \][/tex]
Subtract 5 from both sides of the equation:
[tex]\[ 2x_m = 6 \][/tex]
Divide by 2:
[tex]\[ x_m = 3 \][/tex]
Knowing [tex]\( x_m \)[/tex], we can now substitute back to find [tex]\( y_m \)[/tex]:
[tex]\[ y_m = x_m + 5 = 3 + 5 = 8 \][/tex]
So the midpoint [tex]\( M \)[/tex] is at coordinates (3, 8).
Now, we have the slope of diagonal AC, [tex]\( m_{AC} = -1 \)[/tex], and a point it passes through, which is the midpoint [tex]\( M(3, 8) \)[/tex]. We can use this to find the equation of diagonal AC in the form [tex]\( y = mx + c \)[/tex].
Using the point-slope form of a line equation:
[tex]\[ y - y_m = m_{AC}(x - x_m) \][/tex]
Substitute the known values for [tex]\( m_{AC} \)[/tex], [tex]\( x_m \)[/tex], and [tex]\( y_m \)[/tex]:
[tex]\[ y - 8 = -1(x - 3) \][/tex]
Expand and simplify:
[tex]\[ y = -x + 3 + 8 \][/tex]
[tex]\[ y = -x + 11 \][/tex]
So the equation of diagonal AC is [tex]\( y = -x + 11 \)[/tex], which is in the desired form of [tex]\( y = mx + c \)[/tex].
Here we find that [tex]\( m = -1 \)[/tex] and [tex]\( c = 11 \)[/tex], where both [tex]\( m \)[/tex] and [tex]\( c \)[/tex] are integers.
