Question

The four vertices of a quadrilateral are (1,2).(-5,6), (7,-4) and (k, - 2) taken in order. If the area
of the quadrilateral is 18 sq. units, then k is equal to
[ ]
b) 2
c) 3
d) 4​

Answer 1

$\sf{Answer}$

Given :-

• Four vertices of Quadrilateral is (1,2) , (-5,6) , (7,-4) and (k,-2)

• Area of quadrilateral is 18 sq units

To find :-

Value of k

Formula to know :-

Area of quadrilateral

$\tt\left{\dfrac{1}{2}\bigg|}\:\:\begin{array}{|cc|}\tt x_{1}&\tt y_{1}\\ \tt x_{2}&\tt y_{2}\end{array}\:+\:\begin{array}{|cc|}\tt x_{2}&\tt y_{2}\\ \tt x_{3}&\tt y_{3}\end{array}+\begin{array}{|cc|}\tt x_{3}&\tt y_{3}\\ \tt x_{4}&\tt y_{4}\end{array}\:+\:\begin{array}{|cc|}\tt x_{4}&\tt y_{4}\\ \tt x_{1}&\tt y_{1}\end{array}\ \ \right{\bigg|}$

Explanation :-

By using the above formula we can solve It is by using determinants concept

We have been used there formula ad -bc

$\sf{x_1 = 1}$ $\sf{y_1 = 2}$

$\sf{x_2= -5}$ $\sf{y_2 = 6}$

$\sf{ x_3 = 7}$ $\sf{y_3 = -4}$

$\sf{x_4 = k }$ $\sf{y_4 = -2}$

$|x| = x \:$

$\left | y \right | = \pm a$

Refer attachemnt for process

After simplification we get k = 9 , -3

Know more :-

Important terms :-

Centroid :- The point of intersection of three medians in a triangle is called Centroid

Incentre :- The point of conccurence of internal angular bisectors is called Incentre

Excentre :- The point of concurrence of 1 internal angular bisector 2 External angular bisectors is called excentre

Circumcentre :- The point of intersection of perpendicular bisectors of triangle is called circumcentre

Orthocentre :- The altitude of triangle are concurrent and their point of concurrence is called Orthocentre