Question

(i) If the vertices of $△ABC$ be A(1, -3), B(4, p) and C(-9, 7) and its area is 15 square units, find the values of p.

(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is $(\frac{7}{2},y)$, find the value of y.

Answer 1

(i)
Let $A(x_1,y_1)=A(1,-3)$,
$B(x_2,y_2)=B(4,p)$ and
$C(x_3,y_3)=C(-9,7)$.

Now

$Area(\Delta ABC=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\Rightarrow 15=\frac{1}{2}[1(p–7)+4(7+3)–9(-3-p)]\Rightarrow 15=\frac{1}{2}[10p+60]\Rightarrow |10p+60|=30\Rightarrow 10p+60=-30or30\Rightarrow 10p=-90or-30\Rightarrow p=-9or-3$

Hence, p = -9 or p = -3

(ii)
Let $A(x_1,y_1)=A\left(\right(2,1\left)\right)$,
$B(x_2,y_2)=B(3,-2)$ and
$C(x_3,y_3)=C(\frac{7}{2},y)$.

Now

$Area(\Delta ABC=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\Rightarrow 5=\frac{1}{2}[2(-2-y)+3(y-1)+\frac{7}{2}(1+2)]\Rightarrow 5=\frac{1}{2}[-4-2y+3y-3+\frac{21}{2}]\Rightarrow 10=[y-7\frac{21}{2}]\Rightarrow 10=|\frac{2y-14+21}{2}|\Rightarrow 10=|\frac{2y+7}{2}|\Rightarrow 20=|2y+7|\Rightarrow 20=2y+7or-20=2y+7\Rightarrow 2y=20-7or2y=-20-7\Rightarrow 2y=13or2y=-27\Rightarrow y=\frac{13}{2}ory=\frac{-27}{2}$