Question

if the vertices of a triangle are 1, 1, - 2, 7 and 3, - 3 then find its area​

Answer 1

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

$\frac{1}{2} * | [ x1(y2-y3) + x2(y3-y1) + x3(y1-y2) ] |$

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

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

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

$\frac{y2-y1}{x2-x1}$

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

=> $\frac{7-1}{-2-1} = \frac{-3-1}{3-1}$

=> $\frac{6}{-3} = \frac{-4}{2}$

=> $-2 = -2$

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

Please brainlist my answer, if helpful!

Answer 2

Answer:

0 sq. Unit.

Step-by-step explanation:

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