Verified answer

Appropriate Question:

The quadratic equation whose roots are the x and y intercepts of the line passing through (1, 1) and making a triangle of area A with the axes, may be

(a) x² + Ax + 2A=0

(b) x²- 2x + 2A = 0

(c) x²- 2Ax + 2A = 0

(d) x²- 2x - 2A = 0

Answer:

[tex]\boxed{\bf\: (c) \: \: {x}^{2} - 2Ax + 2 A= 0 \: } \\ [/tex]

Step-by-step explanation:

Let assume that intercept made by line on x axis and y axis be a and b respectively.

So, Equation of line which makes an intercept of a and b units on respective axis is

[tex] \sf \:\dfrac{x}{a} + \dfrac{y}{b} = 1 - - - (1)\\ [/tex]

Now, It is given that, area of triangle formed by this line with coordinate axis is A.

So, we get

[tex] \sf \: A = \dfrac{1}{2} \times a \times b \\ [/tex]

[tex]\implies\sf\:ab = 2A - - - (2) \\ [/tex]

As it is given that, this line passes through (1, 1)

So, we get

[tex] \sf \:\dfrac{1}{a} + \dfrac{1}{b} = 1 \\ [/tex]

[tex] \sf \:\dfrac{b + a}{ab} = 1 \\ [/tex]

[tex] \sf \:\dfrac{b + a}{2A} = 1 \\ [/tex]

[tex]\implies\sf\:a + b = 2A - - - (3) \\ [/tex]

Now, The required quadratic equation having roots as a and b is

[tex] \sf \: {x}^{2} - (a + b)x + ab = 0 \\ [/tex]

On substituting the values from equation (2) and (3), we get

[tex] \sf \: {x}^{2} - (2A)x + 2 A= 0 \\ [/tex]

Hence,

[tex]\implies\sf\:\boxed{\bf\: {x}^{2} - 2Ax + 2 A= 0 \: } \\ [/tex]