Step 1:

\

The points are $\"\"$ and $\"\"$ .

\

(i)

\

The slope of two pints is $\"\"$ .

\

The points are $\"\"$ and $\"\"$ .

\

The slope of $\"\"$ is $\"\"$

\

$\"\"$

\

The slope perpendicular to $\"\"$ is $\"\"$ .

\

$\"\"$

\

Substitute $\"\"$ in $\"\"$ .

\

$\"\"$

\

The slope perpendicular to $\"\"$ is $\"\"$ .

\

Step 2: \ \

\

The mid point of $\"\"$ is

\

$\"\"$

\

The line equation passes through the point $\"\"$ and slope is $\"\"$ .

\

The slope intercept form $\"\"$ . \ \

\

Substitute $\"\"$ and slope is $\"\"$ .

\

$\"\"$

\

The line equation is $\"\"$ . \ \

\

(ii) \ \

\

The point is $\"\"$ . \ \

\

The line equation is $\"\"$ .

\

Substitute $\"\"$ in $\"\"$ .

\

$\"\"$

\

The $\"\"$ satisfies the equation. \ \

\

so, $\"\"$ lies on the $\"\"$ .

\

The points of triangle are $\"\"$ , $\"\"$ and $\"\"$ .

\

The base of triangle is $\"\"$ .

\

$\"\"$ \ \

\

The height of the triangle is the distance between mid point of $\"\"$ and $\"\"$ .

\

The mid point of $\"\"$ is

\

$\"\"$

\

The height of the triangle is \ \

\

$\"\"$ \ \

\

The area of triangle is $\"\"$ . \ \

\

Substitute $\"\"$ and $\"\"$ . \ \

\

$\"\"$

\

The area of triangle is $\"\"$ square units. \ \

\

Solution: \ \

\

The area of triangle is $\"\"$ square units.

\

\ \