In the diagram below, A(3;4), B(6;1) and C(−1;0) are vertices of a 25b3ABC with M(x;y) the midpoint of AC.

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Here's a step-by-step solution for Question 2:

Given points: $A(3;4)$ , $B(6;1)$ , $C(-1;0)$ .
$M(x;y)$ is the midpoint of $AC$ .

2.1 Determine the:

2.1.1 Length of AB

Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .
$AB = \sqrt{(6-3)^2 + (1-4)^2}$ $AB = \sqrt{(3)^2 + (-3)^2}$ $AB = \sqrt{9 + 9}$ $AB = \sqrt{18}$ $AB = \boxed{3\sqrt{2} \text{ units}}$
2.1.2 Length of AC
Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .
$AC = \sqrt{(-1-3)^2 + (0-4)^2}$ $AC = \sqrt{(-4)^2 + (-4)^2}$ $AC = \sqrt{16 + 16}$ $AC = \sqrt{32}$ $AC = \boxed{4\sqrt{2} \text{ units}}$
2.1.3 Gradients of AC and AB
Calculate the gradient of AC using the formula $m = \frac{y_2-y_1}{x_2-x_1}$ .
$m_{AC} = \frac{0-4}{-1-3} = \frac{-4}{-4} = 1$
Calculate the gradient of AB.
$m_{AB} = \frac{1-4}{6-3} = \frac{-3}{3} = -1$ $m_{AC} = \boxed{1}$ $m_{AB} = \boxed{-1}$
2.1.4 Coordinates of M
Use the midpoint formula $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ .
$M = \left(\frac{3+(-1)}{2}, \frac{4+0}{2}\right)$ $M = \left(\frac{2}{2}, \frac{4}{2}\right)$ $M = \boxed{(1; 2)}$
2.2 D is a point in the second quadrant such that ABCD is a parallelogram. Calculate the coordinates of D.
In a parallelogram, the diagonals bisect each other.
Therefore, the midpoint of $AC$ is the same as the midpoint of $BD$ . We know $M(1;2)$ is the midpoint of $AC$ . Let $D=(x_D; y_D)$ .
$M_x = \frac{x_B + x_D}{2} \implies 1 = \frac{6 + x_D}{2}$ $2 = 6 + x_D$ $x_D = -4$
Step 7
$M_y = \frac{y_B + y_D}{2} \implies 2 = \frac{1 + y_D}{2}$ $4 = 1 + y_D$ $y_D = 3$
The coordinates of $D$ are $\boxed{(-4; 3)}$ . This is in the second quadrant (negative x, positive y), as required.

2.3 Calculate the area of parallelogram ABCD.
From 2.1.3, we found $m_{AC} = 1$ and $m_{AB} = -1$ .
Since $m_{AC} \times m_{AB} = 1 \times (-1) = -1$ , the lines $AC$ and $AB$ are perpendicular.
This means $\triangle ABC$ is a right-angled triangle with the right angle at $A$ .
The area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times AB$ .
From 2.1.1, $AB = \sqrt{18}$ .
From 2.1.2, $AC = \sqrt{32}$ .
$\text{Area of } \triangle ABC = \frac{1}{2} \times \sqrt{32} \times \sqrt{18}$ $\text{Area of } \triangle ABC = \frac{1}{2} \times \sqrt{32 \times 18}$ $\text{Area of } \triangle ABC = \frac{1}{2} \times \sqrt{576}$ $\text{Area of } \triangle ABC = \frac{1}{2} \times 24$ $\text{Area of } \triangle ABC = 12 \text{ square units}$
The area of parallelogram ABCD is twice the area of $\triangle ABC$ .
$\text{Area of parallelogram ABCD} = 2 \times \text{Area of } \triangle ABC$ $\text{Area of parallelogram ABCD} = 2 \times 12$ $\text{Area of parallelogram ABCD} = \boxed{24 \text{ square units}}$

In the diagram below, A(3;4), B(6;1) and C(-1;0) are vertices of a 25b3ABC with M(x;y) the midpoint of AC.

Step-by-step solution

Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .

Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .

Calculate the gradient of AC using the formula $m = \frac{y_2-y_1}{x_2-x_1}$ .

Calculate the gradient of AB.

Use the midpoint formula $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ .

In a parallelogram, the diagonals bisect each other.

Step 7

From 2.1.3, we found $m_{AC} = 1$ and $m_{AB} = -1$ .

The area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times AB$ .

The area of parallelogram ABCD is twice the area of $\triangle ABC$ .

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More Mathematics Questions

In the diagram below, A(3;4), B(6;1) and C(-1;0) are vertices of a 25b3ABC with M(x;y) the midpoint of AC.

Step-by-step solution

Use the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​.

Use the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​.

Calculate the gradient of AC using the formula m=y2−y1x2−x1m = \frac{y_2-y_1}{x_2-x_1}m=x2​−x1​y2​−y1​​.

Calculate the gradient of AB.

Use the midpoint formula M=(x1+x22,y1+y22)M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)M=(2x1​+x2​​,2y1​+y2​​).

In a parallelogram, the diagonals bisect each other.

Step 7

From 2.1.3, we found mAC=1m_{AC} = 1mAC​=1 and mAB=−1m_{AB} = -1mAB​=−1.

The area of △ABC=12×base×height=12×AC×AB\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times AB△ABC=21​×base×height=21​×AC×AB.

The area of parallelogram ABCD is twice the area of △ABC\triangle ABC△ABC.

Still stuck on this one?

More Mathematics Questions

Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .

Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .

Calculate the gradient of AC using the formula $m = \frac{y_2-y_1}{x_2-x_1}$ .

Use the midpoint formula $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ .

From 2.1.3, we found $m_{AC} = 1$ and $m_{AB} = -1$ .

The area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times AB$ .

The area of parallelogram ABCD is twice the area of $\triangle ABC$ .