Question

if vertices of triangle are (1,1),(-2,7),(3,-3) then it's area is​

Answer 1

$\bold{AnsweR:}$

Area of the triangle formed is zero since the lines are concurrent

Step-by-step explanation:

The formula for the area of a triangle is

$\bold{\frac{1}{2}*|[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]| }$

Where (x1,y1); (x2,y2) and (x3,y3) are the vertices of the triangle respectively

Substituting the given values of the vertices in the formula and solving, we get:

$\bold{\frac{1}{2}*|[1(7+3)+(-2)(-3-1)+3(1-7)]| }$

When we solve this, we get the area as zero

This impies that the points lie on the same line and are concurrent

Therefore the points do not form a triangle.

Another way to confirm this is by finding the slopes of the lines joining the points

Formula for finding the slope of a line is:

$\bold{\frac{y_2-y_1}{x_2-x_1} }$

Where (x1,y1) and (x2,y2) are the points given

If the given 3 points are concurrent, then the slopes would be equal to each other.

This implies:

$\bold{\frac{7-1}{-2-1}=\frac{-3-1}{3-1} }\\\\\bold{\frac{6}{-3}=\frac{-4}{2} }\\\\\bold{-2=-2}$

Since the slopes are equal, the given points are concurrent and the area of the triangle formed is zero .

Answer 2

Answer:

Area of the triangle formed is zero since the lines are concurrent

Step-by-step explanation:

The formula for the area of a triangle is

\bold{\frac{1}{2}*|[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]| }

2

1

∗∣[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]∣

Where (x1,y1); (x2,y2) and (x3,y3) are the vertices of the triangle respectively

Substituting the given values of the vertices in the formula and solving, we get:

\bold{\frac{1}{2}*|[1(7+3)+(-2)(-3-1)+3(1-7)]| }

2

1

∗∣[1(7+3)+(−2)(−3−1)+3(1−7)]∣

When we solve this, we get the area as zero

This impies that the points lie on the same line and are concurrent

Therefore the points do not form a triangle.

Another way to confirm this is by finding the slopes of the lines joining the points

Formula for finding the slope of a line is:

\bold{\frac{y_2-y_1}{x_2-x_1} }

x

2

−x

1

y

2

−y

1

Where (x1,y1) and (x2,y2) are the points given

If the given 3 points are concurrent, then the slopes would be equal to each other.

This implies:

\begin{gathered}\bold{\frac{7-1}{-2-1}=\frac{-3-1}{3-1} }\\\\\bold{\frac{6}{-3}=\frac{-4}{2} }\\\\\bold{-2=-2}\end{gathered}

−2−1

7−1

=

3−1

−3−1

−3

6

=

2

−4

−2=−2

Since the slopes are equal, the given points are concurrent and the area of the triangle formed is zero .