The points \(A(1,-2),\ B(2,3),\ C(a,2)\) and \(D(-4,-3)\) form a parallelogram. Find \(a\) and the height of the parallelogram taking \(AB\) as the base.
\(a=-3\) and the height (on base \(AB\)) is \(\dfrac{24}{\sqrt{26}}\).
Step 1: In a parallelogram, diagonals bisect each other. So, midpoint of \(AC\) = midpoint of \(BD\).
Step 2: Midpoint of \(AC\), where \(A(1,-2)\) and \(C(a,2)\):
\(M_{AC}=\left(\dfrac{1+a}{2},\dfrac{-2+2}{2}\right)=\left(\dfrac{1+a}{2},0\right)\).
Step 3: Midpoint of \(BD\), where \(B(2,3)\) and \(D(-4,-3)\):
\(M_{BD}=\left(\dfrac{2+(-4)}{2},\dfrac{3+(-3)}{2}\right)=(-1,0)\).
Step 4: Equate x-coordinates of the midpoints:
\(\dfrac{1+a}{2}=-1 \Rightarrow 1+a=-2 \Rightarrow a=-3\).
Step 5: To find the height on base \(AB\), first find the area using the area formula of a triangle (as shown in the method).
The parallelogram \(ABCD\) is made of two equal triangles, so:
\(\text{Area of parallelogram} = 2\times \text{Area of }\triangle ABD\).
Step 6: Find the area of \(\triangle ABD\) using coordinates:
\[\text{Area}(\triangle ABD)=\dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\]
Here \(A(1,-2),\ B(2,3),\ D(-4,-3)\):
\(\text{Area}(\triangle ABD)=\dfrac{1}{2}\left|1(3-(-3)) + 2((-3)-(-2)) + (-4)((-2)-3)\right|\)
\(=\dfrac{1}{2}\left|1\cdot 6 + 2\cdot(-1) + (-4)\cdot(-5)\right| =\dfrac{1}{2}\left|6-2+20\right| =\dfrac{1}{2}\cdot 24 =12\)
Step 7: Therefore,
\(\text{Area of parallelogram} = 2\times 12 = 24\) square units.
Step 8: Now find the base \(AB\):
\(AB=\sqrt{(2-1)^2+(3-(-2))^2}=\sqrt{1^2+5^2}=\sqrt{26}\).
Step 9: Use \(\text{Area} = \text{base} \times \text{height}\):
\(24 = \sqrt{26}\cdot h \Rightarrow h=\dfrac{24}{\sqrt{26}}\).
Final Answer: \(a=-3\) and height on base \(AB\) is \(\dfrac{24}{\sqrt{26}}\).