R S Aggarwal and Veena Aggarwal Solutions for Exercise 3: Exercise 6C

If $A(–7, 5), B(–6, –7), C(–3, –8)$ and $D(2, 3)$ are the vertices of a quadrilateral $ABCD$ then find the area of the quadrilateral.

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

Show that the following points are collinear:
$A(-5, 1), B(5, 5)$ and $C(10, 7)$

For what value of $y$, are the points $P(1, 4), Q(3, y)$ and $R(-3, 16)$ are collinear ?

Find the value of $y$ for which the points $A(-3, 9), B(2, y)$ and $C(4, -5)$ are collinear.

Prove that the points $A(a, 0), B(0, b)$ and $C(1, 1)$ are collinear, if $\left(\frac{1}{a}+\frac{1}{b}\right) =1$.

Find the area of $\Delta ABC$ with vertices $A(0, -1), B(2, 1)$ and $C(0, 3)$. Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is $4:1$.

If $a\ne b\ne ca$, prove that $(a, a^2), (b, b^2), (0, 0)$ will not be collinear.