Question

सरल रेखा 5x-3y+10=0 से अक्षो के बीच बने त्रिभुज का क्षेत्रफल होगा​

Answer 1

I think your question may be:

The area of ​​the triangle formed between the axes from the straight line 5x-3y + 10 = 0 will be

Given line is 5x-3y+10=0

This can be written as

5x-3y=-10

divide both sides by -10

$\frac{5x}{-10}+\frac{(-3y)}{-10}=1$

$\frac{x}{-2}+\frac{y}{10/3}=1$

comparing this with

$\boxed{\frac{x}{a}+\frac{y}{b}=1}$ we get

a=-2 and b=10/3

Area of the triangle formed between the axes from the straight line

$=\frac{1}{2}*base*height$

$=\frac{1}{2}*2*\frac{10}{3}$

$=\frac{10}{3}\,square\,units$