Question

The straight line 4x+5y-20=0 cuts the x-axis at A and y-axis at B. Find x intercept and y intercept. Hence find the area of triangla AOB​

Answer 1

Correct Question -

The straight line 4x+5y-20=0 cuts the x-axis at A and y-axis at B.

Find the x intercept and the y intercept.

Hence find the area of triangle ∆ AOB .

Solution -

In the above Question, the following information is given ...

The straight line 4x+5y-20=0 cuts the x-axis at A and y-axis at B.

We have to find -

• the x intercept and the y intercept

• the area of triangle ∆ AOB

We know that -

Equation of a Straight line in Intercept Form -

$\dfrac{x}{A} + \dfrac{y}{B} = 1 \\ \\ \sf{ Where \::- } \\ \\ \sf{ A \: is \: the \: x \: intercept .} \\ \\ \sf{ B \: is \: the \: y \: intercept. }$

The given Equation of the straight line is :

$4x + 5y - 20 = 0 \\ \\ = > 4x + 5y = 20 \\ \\ = > \dfrac{4x + 5y}{20} = 1 \\ \\ = > \dfrac{4x}{20} + \dfrac{5y}{20} = 1$ $= > \dfrac{x}{5} + \dfrac{y}{4} = 1$ $\sf{ Comparing \: With \: The \: General \: Equation \: Of \: A \: Line \: in \: Intercept \: Form \::- } \\ \\ => a = 5 \\ \\ => b = 4 \\ \\ \sf{ Hence \: - } \\ \\ \sf{ X \: Intercept \: = 5 } \\ \\ \sf{ Y \: Intercept \: = \: 4 } .......... [ A _{1} ] \\ \\ \sf{ Now \: In \: Triangle \: AOB \: :- } \\ \\ = > \: OA = 5 \\ \\ => \: OB = 4 \\ \\ => Area = \dfrac{1}{2} \times [ Base ] \times [ Height ] \\ \\ => \dfrac{1}{2} \times [ OA ] \times [ OB ] \\ \\ => \dfrac{1}{2} \times 5 \times 4 \\ \\ => 10 \: { unit } ^ 2 \: ........... [ A _{2} ]$