Question

find the area of triangle whose vertices are P(3/2,1),Q(4,2),R(4,-1/2)
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Answer 1

Given:

P(3/2,1),Q(4,2),R(4,-1/2)

To find:

The area of the triangle of the given vertices

Solution:

We have given the vertices of the triangle as:

P(3/2,1),Q(4,2),R(4,-1/2)

and we have the direct formula to get the area of the triangle formed the coordinates

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

here we have given the coordinates as:

P(3/2,1),Q(4,2),R(4,-1/2)

now, substitute the value of the coordinates as per the sequence of the points.

we get the area as:

$\frac{1}{2}[\frac{3}{2}(2+\frac{1}{2})-4(-\frac{1}{2}-1)-4(1-2)]$

$\frac{1}{2}[\frac{3}{2}(\frac{5}{2})-4(-\frac{3}{2})-4(-1)]$

$\frac{1}{2}[\frac{15}{4}+6+4]$

$\frac{1}{2}[\frac{15}{4}+10]$

$\frac{15}{8}+5$

1.875 + 5

6.875

Hence The area of the triangle of the given vertices is 6.875 sq unit