Question

Given the vertices of a triangle are A (-3,3), B (4,4), and C (x,y), line AC is perpendicular to line BC and angle BAC=45 degree, find the distance of AC

Answer 1

D. 5

Step-by-step explanation:

Given the vertices of the triangle: A (-3,3), B (4,4), and C (x,y).

The diagram illustrating this triangle is drawn below:

First, find the distance AB using the distance formula:

`\begin{gathered} Distance=√((x_2-x_1)^2+(y_2-y_1)^2) \\ AB=\sqrt[]{(4-(-3))^2+(4-3_{})^2} \\ =\sqrt[]{(4+3)^2+(1)^2} \\ =\sqrt[]{7^2+1^2} \\ =\sqrt[]{50} \\ AB=5\sqrt[]{2} \end{gathered}`

Next, we find the length of AC using trigonometric ratio:

`\begin{gathered} \cos \theta=\frac{\text{Adjacent}}{Hypotenuse} \\ \implies\cos A=(AC)/(AB) \\ \cos 45=\frac{AC}{5\sqrt[]{2}} \\ \frac{1}{\sqrt[]{2}}=\frac{AC}{5\sqrt[]{2}} \\ AC*\sqrt[]{2}=5\sqrt[]{2} \\ AC=5 \end{gathered}`

The length of line AC is 5 units.