# If the vertices o

We have the following triangle,

Let the vertices of the triangle be A(1,-3), B(4,p) and C(-9,7) be denoted by A(x1,y1), B(x2,y2) and C(x3,y3).

Area of the triangle where the vertices are given is,

Area = 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

And Area is given as 15 sq. units.

So,

1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = �15

[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = �30

Substituting corresponding values in the equation, we get

[1(p – 7) + 4(7 – (-3)) + (-9)(-3 – p) = �30

[p – 7 + 40 + 27 + 9p = �30

60 + 10p = �30

10p = �30 – 60

p = ( �30 – 60)/10

So we have two solution, lets explore them.

I case: p = (30 – 60)/10

p = -30/10

p = -3

II case: p = (-30 – 60)/10

p = -90/10

p = -9

Hence, p has two values, they are -3 and -9.

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