Question

The area of triangle formed by the line
$\frac{x}{a} + \frac{y}{b} = 1$
with the co-ordinate axis is

(a) ab
(b) 2ab
(c)
$\frac{1}{2} ab$
(d)
$\frac{1}{4} ab$
​

Answer 1

Here, Line ; x/a + y/b = 1

for forming triangle we have require three non - colinear vertices.

this line cut in X - axis ,

when, y = 0

x/a + 0/b = 1

x = a hence, at X-axis it cuts (a , 0)

similarly cuts in Y-axis

when, x = 0

0/a + y/b = 1

y = b hence, at (0, b) it cuts in Y-axis

now, we have three points (0, 0), (a , 0) and ( 0, b)

use co-ordinate Geometry formula for finding area of ∆ if points are given.

ar∆ = 1/2 [ 0 + a( b - 0) + 0 ]

= 1/2 ab

hence, area of ∆ = (ab)/2

Answer 2

Hi buddy!

Answer: (c) 1/2 ab.

Explanation:

Here, line = x/a = y/b = 1 => bx + ay = ab

Now, the X intercept will be:

bx + a·0 = ab

=> bx = ab

=> x = a

Or, X intercept = (a, 0).

Now, Y intercept:

b·0 + ay = ab

=> y = b

Or, Y intercept = (0, b).

Now, we know, area of ∆ = 1/2 bh

= 1/2 (ordinate)(abscissa)

= 1/2 ab