The equations of the sides $AB$ , $BC$ and $CA$ of the triangle $ABC$ are $3 = Ax + 13$ , $y = - 1$ and $y = 1 - 2x$ respectively. $6)$ Find the coordinates of $A$ , $B$ and $BC$ $0$ . $BC$ $1$ IF $BC$ $2$ is a parallelogram, find the coordinates of $BC$ $3$ .

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Q. The equations of the sides

AB

BC

and

CA

of the triangle

ABC

are

3 = Ax + 13

y = - 1

and

y = 1 - 2x

respectively.

6)

Find the coordinates of

A

B

and

BC

0

BC

1

BC

2

is a parallelogram, find the coordinates of

BC

3

Find Point A Intersection: To find the coordinates of point A, we need to find the intersection of lines AB and CA. The equation of AB is $3 = A x + 13$ , and the equation of CA is $y = 1 - 2x$ .
Solve for x in AB: First, we solve for x in the equation of AB: $3 = Ax + 13$ . Since $A$ is not given, we assume it to be $1$ (as it's the coefficient of $x$ ). So, $3 = x + 13$ , which gives us $x = -10$ .
Substitute $x$ into CA: Now we substitute $x = -10$ into the equation of CA: $y = 1 - 2(-10)$ , which gives us $y = 1 + 20$ , so $y = 21$ .
Coordinates of Point A: The coordinates of point A are $(-10, 21)$ .
Find Point B Intersection: To find the coordinates of point B, we need to find the intersection of lines AB and BC. The equation of BC is $y = -1$ .
Substitute $y$ into $AB$ : Since the equation of $BC$ is $y = -1$ , and it intersects $AB$ , we substitute $y = -1$ into the equation of $AB$ : $3 = x + 13$ , which gives us $x = -10$ .
Coordinates of Point B: The coordinates of point B are $(-10, -1)$ .
Find Point C Intersection: To find the coordinates of point C, we need to find the intersection of lines BC and CA. We already have the equation of BC as $y = -1$ , and the equation of CA is $y = 1 - 2x$ .
Set y-values equal: We set the y-values equal to each other to find the x-coordinate of C: $-1 = 1 - 2x$ , which gives us $2x = 2$ , so $x = 1$ .
Substitute $x$ into BC: Now we substitute $x = 1$ into the equation of BC: $y = -1$ , which confirms the $y$ -coordinate of $C$ .
Coordinates of Point C: The coordinates of point C are $(1, -1)$ .
Find Vector $AB$ : If $ABCD$ is a parallelogram, then $AD$ and $BC$ are parallel and equal in length. Since we have the coordinates of $A$ and $B$ , we can find the vector $AB$ .
Calculate Vector AB: The vector AB is given by the coordinates of B minus the coordinates of A: $\mathbf{AB} = (-10, -1) - (-10, 21)$ , which simplifies to $\mathbf{AB} = (0, -22)$ .
Find Coordinates of D: To find the coordinates of D, we add the vector $AB$ to the coordinates of $C$ : $D = C + AB$ , which gives us $D = (1, -1) + (0, -22)$ .
Find Coordinates of D: To find the coordinates of D, we add the vector $AB$ to the coordinates of $C$ : $D = C + AB$ , which gives us $D = (1, -1) + (0, -22)$ .The coordinates of $D$ are $(1, -1) + (0, -22) = (1, -23)$ .