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The four vertices of a quadrilaterals are $(1, 2), (-5, 6), (7, -4)$ and $(k, -2)$ taken in order. If the area of the quadrilateral is zero, find the value of $k$.
Given:
The four vertices of a quadrilaterals are $(1, 2), (-5, 6), (7, -4)$ and $(k, -2)$ taken in order.
The area of the quadrilateral is zero.
To do:
We have to find the value of $k$.
Solution:
Let $A (1, 2), B (-5, 6), C(7,-4)$ and $D (k, 2)$ be the vertices of a quadrilateral $ABCD$.
Join $A$ and $C$ to get two triangles $ABC$ and $ADC$.
This implies,
Area of quadrilateral $ABCD=$ Area of triangle $ABC+$ Area of triangle $ADC$.
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[1(6+4)+(-5)(-4-2)+7(2-6)] \)
\( =\frac{1}{2}[1(10)+(-5)(-6)+7(-4)] \)
\( =\frac{1}{2}[10+30-28] \)
\( =\frac{1}{2} \times 12 \)
\( =6 \) sq. units.
Area of triangle \( ADC=\frac{1}{2}[1(-4+2)+k(2+4)+7(-2-2)] \)
\( =\frac{1}{2}[1(-2)+k(6)+7(-4)] \)
\( =\frac{1}{2}[-2+6k-28] \)
\( =\frac{1}{2} \times (6k-30) \)
\( =3k-15 \) sq. units.
Therefore,
The area of the quadrilateral $ABCD=6+3k-15$ sq. units.
$0=3k-9$
$3k=9$
$k=\frac{9}{3}$
$k=3$
The value of $k$ is $3$.
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