Answer :

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(x_{1},y_{1}), B(x_{2},y_{2}) and C(x_{3},y_{3}).

Area of the triangle where the vertices are given is,

Area = 1/2 [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

And Area is given as 15 sq. units.

So,

1/2 [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = �15

⇒ [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = �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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