If the poins $ A(1,-2), B(2,3) C(a, 2) $ and $ D(-4,-3) $ form a parallelogram. find the value of $ a $ and height of the parallelogram taking AB as base.

AcademicMathematicsNCERTClass 10

Given:

The points $A (1, -2), B (2, 3), C (a, 2)$ and $D (-4, -3)$ form a parallelogram.

To do:

We have to find the value of $a$ and height of the parallelogram taking $AB$ as base.

Solution:

Draw a perpendicular from \( \mathrm{D} \) to \( \mathrm{AB} \) which meets \( \mathrm{AB} \) at \( \mathrm{P} \).

\( \mathrm{DP} \) is the height of the parallelogram. 

We know that, 

Diagonals of a parallelogram bisect each other.
This implies,

Mid-point of $AC =$ Mid-point of $BD$

The mid-point of a line segment joining points \( \left(x_{1}, y_{1}\right) \) and \( \left(x_{2}, y_{2}\right) \) is \( (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) \)

\( (\frac{1+a}{2}, \frac{-2+2}{2})=(\frac{2-4}{2}, \frac{3-3}{2}) \)

On comparing, we get,

\( \Rightarrow \frac{1+a}{2}=\frac{-2}{2}=-1 \)

\( \Rightarrow 1+a=-2 \)

\( \Rightarrow a=-3 \)

The required value of \( a \) is \( -3 \).

We know that,

A diagonal bisects a triangle into two triangles of equal area.

This implies,

Area of parallelogram $ABCD=$ Area of triangle $ABC+$ Area of triangle $ADC$. 

$=2\times$ Area of triangle $ABC$

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 \( ABC=\frac{1}{2}[1(3-2)+2(2+2)+(-3)(-2-3)] \)

\( =\frac{1}{2}[1(1)+(2)(4)+(-3)(-5)] \)

\( =\frac{1}{2}[1+8+15] \)

\( =\frac{1}{2} \times (24) \)

\( =12 \) sq. units.

Therefore,

The area of the parallelogram $ABCD=2\times 12=24$ sq. units.

Area of parallelogram $=$ Base $\times$ Height

Height $=$ Area $\div$ Base

$DP=\frac{24}{AB}$

By distance formula, we get,

$AB=\sqrt{(2-1)^2+(3+2)^2}$

$=\sqrt{1^2+5^2}$

$=\sqrt{1+25}$

$=\sqrt{26}$

Therefore,

$DP=\frac{24}{\sqrt{26}}$

$=\frac{24\times\sqrt{26}}{\sqrt{26}\times\sqrt{26}}$

$=\frac{24\sqrt{26}}{26}$

$=\frac{12\sqrt{26}}{13}$

The height of the parallelogram taking $AB$ as base is $\frac{12\sqrt{26}}{13}$.  

raja
Updated on 10-Oct-2022 13:28:51

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