The points A(1,-2),B(2,3),C(k,-2),D(-4,-3)are the vertices of …

A(1, -2), B(2, 3), C(a, 2) and D(-4, -3)are the vertices of a parallelogram.
{tex}\Rightarrow{/tex}AB = CD and AD = BC
Consider AD = BC
{tex}\Rightarrow{/tex}AD² = BC²
{tex}\Rightarrow{/tex}(-4 -1)² + (-3 + 2)² = (a - 2)² + (2 - 3)²
{tex}\Rightarrow{/tex}(-5)² = (a - 2)²
{tex}\Rightarrow{/tex} a - 2 = -5
{tex}\Rightarrow{/tex} a = -3
Area of {tex}\Delta A B C = \frac { 1 } { 2 } | 1 ( 3 - 2 ) + 2 ( 2 + 2 ) - 3 ( - 2 - 3 ) |{/tex}
{tex}= \frac { 1 } { 2 } | 1 + 8 + 15 |{/tex}
{tex}= \frac { 1 } { 2 } \times 24{/tex}
= 12 sq. units.
Diagonal of a parallelogram divides it into two equal triangles.
Area of parallelogram ABCD = 2 {tex}\times{/tex}12=24 sq. units
{tex}\Rightarrow {/tex} AB {tex}\times{/tex}Height = 24
{tex}\Rightarrow {/tex} AB{tex}\times{/tex}Height = 24
{tex}\Rightarrow \left[ \sqrt { ( 2 - 1 ) ^ { 2 } + ( 3 + 2 ) ^ { 2 } } \right] \times \text { Height } = 24{/tex}
{tex}\Rightarrow [ \sqrt { 1 + 25 } ] \times \text { Height } = 24{/tex}
{tex}\Rightarrow \text { Height } = \frac { 24 } { \sqrt { 26 } } = \frac { 12 \times \sqrt { 2 } \times \sqrt { 2 } } { \sqrt { 13 } \times \sqrt { 2 } }{/tex}
{tex}= \frac { 12 \sqrt { 2 } } { \sqrt { 13 } }{/tex}