Question

Segment

Answer 1

Answer:

The possible first coordinates of point C are (-2.5,1.5)

The possible second coordinates of point C are (-9.5,1.5)

Step-by-step explanation:

we know that

Triangle ABC is a right isosceles triangle

so

Is a 45°-90°-45° triangle

AC=BC

we have

A(-6,-2), B(-6,5)

step 1

Find the length side of the hypotenuse AB

$AB=5-(-2)=7\ units$

step 2

Applying the Pythagoras Theorem

Find the length side of leg AC

$AB^{2}=AC^{2}+BC^{2}$

Remember that

AC=BC

substitute the given values

$7^{2}=AC^{2}+AC^{2}$

$49=2AC^{2}$

$AC^{2}=\frac{49}{2}$

$AC=\frac{7\sqrt{2}}{2}\ units$

step 3

Find the first possible coordinates of C

The point C is located at right of point A

Determine the x-coordinate of point C

The x-coordinate of point C must be equal to the x-coordinate of point A plus the horizontal distance between point A and point C

Let

ACx ------> the horizontal distance between point A and point C

The horizontal distance between point A and point C is equal to the distance AC multiplied by cos(45)

$ACx=(AC)cos(45\°)$

we have

$cos(45\°)=\frac{\sqrt{2}}{2}$

$AC=\frac{7\sqrt{2}}{2}\ units$

substitute

$ACx=(\frac{7\sqrt{2}}{2})\frac{\sqrt{2}}{2}=3.5\ units$

The x-coordinate of point C is

Cx=-6+3.5=-2.5

Determine the y-coordinate of point C

The y-coordinate of point C must be equal to the y-coordinate of point A plus the vertical distance between point A and point C

Let

ACy ------> the vertical distance between point A and point C

The vertical distance between point A and point C is equal to the distance AC multiplied by sin(45)

$ACy=(AC)sin(45\°)$

we have

$sin(45\°)=\frac{\sqrt{2}}{2}$

$AC=\frac{7\sqrt{2}}{2}\ units$

substitute

$ACy=(\frac{7\sqrt{2}}{2})\frac{\sqrt{2}}{2}=3.5\ units$

The y-coordinate of point C is

Cy=-2+3.5=1.5

therefore

The possible first coordinates of point C are (-2.5,1.5)

step 4

Find the second possible coordinate of C

The point C is located at left of point A

Determine the x-coordinate of point C

The x-coordinate of point C must be equal to the x-coordinate of point A minus the horizontal distance between point A and point C

Let

ACx ------> the horizontal distance between point A and point C

The horizontal distance between point A and point C is equal to the distance AC multiplied by cos(45)

$ACx=(AC)cos(45\°)$

we have

$cos(45\°)=\frac{\sqrt{2}}{2}$

$AC=\frac{7\sqrt{2}}{2}\ units$

substitute

$ACx=(\frac{7\sqrt{2}}{2})\frac{\sqrt{2}}{2}=3.5\ units$

The x-coordinate of point C is

Cx=-6-3.5=-9.5

Determine the y-coordinate of point C

The y-coordinate of point C must be equal to the y-coordinate of point A plus the vertical distance between point A and point C

Let

ACy ------> the vertical distance between point A and point C

The vertical distance between point A and point C is equal to the distance AC multiplied by sin(45)

$ACy=(AC)sin(45\°)$

we have

$sin(45\°)=\frac{\sqrt{2}}{2}$

$AC=\frac{7\sqrt{2}}{2}\ units$

substitute

$ACy=(\frac{7\sqrt{2}}{2})\frac{\sqrt{2}}{2}=3.5\ units$

The y-coordinate of point C is

Cy=-2+3.5=1.5

therefore

The possible second coordinates of point C are (-9.5,1.5)

see the attached figure to better understand the problem