Question

Segment

AB
is the hypotenuse of the right isosceles ΔABC with A(–6, –2) and B(–6, 5). Find all possible coordinates of C.

Answer 1

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

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

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)=(49)/(2)`

`AC=(7√(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\°)=(√(2))/(2)`

`AC=(7√(2))/(2)\ units`

substitute

`ACx=((7√(2))/(2))(√(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\°)=(√(2))/(2)`

`AC=(7√(2))/(2)\ units`

substitute

`ACy=((7√(2))/(2))(√(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\°)=(√(2))/(2)`

`AC=(7√(2))/(2)\ units`

substitute

`ACx=((7√(2))/(2))(√(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\°)=(√(2))/(2)`

`AC=(7√(2))/(2)\ units`

substitute

`ACy=((7√(2))/(2))(√(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