NCERT Solutions
Class 11 Maths
Straight Lines

Ex.10.1 Q.1
Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (-4, -2).
Also, find its area.
Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5) and D (-4, -2).
Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as
To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (∆ABC) + area (∆ACD)
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is ( )|x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) |
Therefore, area of ∆ABC = ( )|-4(7 + 5) + 0(-5 - 5) + 5(5 - 7) |
= ( )|-4 × 12 – 5 × 2|
= ( )|-48 – 10|
= ( )|-58|
=
= 29 unit2
Area of ∆ACD = ( )|-4(-5 + 2) + 5(-2 - 5) + (-4) (5 + 5) |
= ( )|4 × 3 – 5 × 7 – 4 × 10|
= ( )|12 -35 – 40|
= ( )|-63|
= unit2
Thus, area (ABCD) = 29 + =
unit2