Given points A(−3, 2), C(9, 10), and D(4, −2). Calculate the gradient of AC, find the equation of DE…

Question

Given points A(−3, 2), C(9, 10), and D(4, −2). Calculate the gradient of AC, find the equation of DE (parallel to CA), and determine the size of angle alpha (formed by AC and the y-axis).

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Here are the solutions to the questions:

Given points:
$A(-3, 2)$
$C(9, 10)$
$D(4, -2)$

2.1 Calculate the gradient of the line AC.

Use the gradient formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ with points $A(-3, 2)$ and $C(9, 10)$ .
$m_{AC} = \frac{10 - 2}{9 - (-3)}$ $m_{AC} = \frac{8}{9 + 3}$ $m_{AC} = \frac{8}{12}$ $m_{AC} = \frac{2}{3}$
The gradient of line AC is $\boxed{\frac{2}{3}}$ .

2.2 Determine the equation of line DE in the form $y = mx + c$ .
Since line DE is parallel to CA, their gradients are equal.
$m_{DE} = m_{CA} = \frac{2}{3}$
Use the point-slope form $y - y_1 = m(x - x_1)$ with point $D(4, -2)$ and $m_{DE} = \frac{2}{3}$ .
$y - (-2) = \frac{2}{3}(x - 4)$ $y + 2 = \frac{2}{3}x - \frac{8}{3}$
Solve for $y$ to get the equation in the form $y = mx + c$ .
$y = \frac{2}{3}x - \frac{8}{3} - 2$ $y = \frac{2}{3}x - \frac{8}{3} - \frac{6}{3}$ $y = \frac{2}{3}x - \frac{14}{3}$
The equation of line DE is $\boxed{y = \frac{2}{3}x - \frac{14}{3}}$ .

2.3 Calculate the size of $\alpha$ .
Identify $\alpha$ as $\angle CFO$ .
F is the y-intercept of line AC.
First, find the equation of line AC. We have $m_{AC} = \frac{2}{3}$ . Using point $A(-3, 2)$ :
$y - 2 = \frac{2}{3}(x - (-3))$ $y - 2 = \frac{2}{3}(x + 3)$ $y - 2 = \frac{2}{3}x + 2$ $y = \frac{2}{3}x + 4$
The y-intercept F is $(0, 4)$ .
Let $\theta$ be the angle of inclination of line AC with the positive x-axis.
$m_{AC} = \tan \theta$ $\frac{2}{3} = \tan \theta$ $\theta = \arctan\left(\frac{2}{3}\right)$ $\theta \approx 33.69^\circ$
Relate $\alpha$ to $\theta$ .
From the diagram, $\alpha$ is the angle between the line AC and the y-axis. In the right-angled triangle formed by the x-axis, y-axis, and line AC (let G be the x-intercept), $\alpha$ and $\theta$ are complementary angles.
$\alpha = 90^\circ - \theta$ $\alpha = 90^\circ - 33.69^\circ$ $\alpha = 56.31^\circ$
The size of $\alpha$ is $\boxed{56.3^\circ}$ .

2.4 B is a point in the first quadrant such that ABDE, in that order, forms a rectangle. Calculate, giving reasons, the:

2.4.1 Coordinates of M, the midpoint of BE.
In a rectangle, the diagonals bisect each other.
Therefore, the midpoint of diagonal BE is the same as the midpoint of diagonal AD.
Given points $A(-3, 2)$ and $D(4, -2)$ .
Use the midpoint formula $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ for AD.
$M = \left(\frac{-3 + 4}{2}, \frac{2 + (-2)}{2}\right)$ $M = \left(\frac{1}{2}, \frac{0}{2}\right)$ $M = \left(\frac{1}{2}, 0\right)$
The coordinates of M are $\boxed{\left(\frac{1}{2}, 0\right)}$ .

2.4.2 Length of diagonal BE.
In a rectangle, the diagonals are equal in length.
Therefore, the length of diagonal BE is equal to the length of diagonal AD.
Given points $A(-3, 2)$ and $D(4, -2)$ .
Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for AD.
$BE = AD = \sqrt{(4 - (-3))^2 + (-2 - 2)^2}$ $BE = \sqrt{(7)^2 + (-4)^2}$ $BE = \sqrt{49 + 16}$ $BE = \sqrt{65}$
The length of diagonal BE is $\boxed{\sqrt{65}}$ .

Given points A(-3, 2), C(9, 10), and D(4, -2). Calculate the gradient of AC, find the equation of DE (parallel to CA), and determine the size of angle alpha (formed by AC and the y-axis).

Step-by-step solution

Use the gradient formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ with points $A(-3, 2)$ and $C(9, 10)$ .

Since line DE is parallel to CA, their gradients are equal.

Use the point-slope form $y - y_1 = m(x - x_1)$ with point $D(4, -2)$ and $m_{DE} = \frac{2}{3}$ .

Solve for $y$ to get the equation in the form $y = mx + c$ .

Identify $\alpha$ as $\angle CFO$ .

Let $\theta$ be the angle of inclination of line AC with the positive x-axis.

Relate $\alpha$ to $\theta$ .

In a rectangle, the diagonals bisect each other.

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

In a rectangle, the diagonals are equal in length.

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

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More Mathematics Questions

Given points A(-3, 2), C(9, 10), and D(4, -2). Calculate the gradient of AC, find the equation of DE (parallel to CA), and determine the size of angle alpha (formed by AC and the y-axis).

Step-by-step solution

Use the gradient formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​ with points A(−3,2)A(-3, 2)A(−3,2) and C(9,10)C(9, 10)C(9,10).

Since line DE is parallel to CA, their gradients are equal.

Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​) with point D(4,−2)D(4, -2)D(4,−2) and mDE=23m_{DE} = \frac{2}{3}mDE​=32​.

Solve for yyy to get the equation in the form y=mx+cy = mx + cy=mx+c.

Identify α\alphaα as ∠CFO\angle CFO∠CFO.

Let θ\thetaθ be the angle of inclination of line AC with the positive x-axis.

Relate α\alphaα to θ\thetaθ.