MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Distance between Two Points on the Same Coordinate Axes

Prove that ...

Question

Prove that $x = \frac{a^{2} + a b + b^{2}}{a + b}, y = \frac{a b}{a + b}$ is solution of the equation $b x - a y = b^{2}$

Open in App

Solution

$\begin{matrix} x = \frac{a^{2} + a b + b^{2}}{(a + b)}, y = \frac{a b}{(a + b)} P u t t i n g V a l u e o f x a n d y i n b x - a y = b^{2} \Rightarrow b [\frac{a^{2} + a b + b^{2}}{a + b}] - \frac{a^{2} b}{(a + b)} = b^{2} \Rightarrow \frac{a^{2} b + a b^{2} + b^{3} - a^{2} b}{(a + b)} \Rightarrow \frac{a b^{2} + b^{3}}{(a + b)} \Rightarrow b^{2} \frac{(a + b)}{a + b} = b^{2} p r o v e d . H e n c e, V a l u e o f x a n d y s a t i s f y t h e e q u a t i o n b x + a y = b^{2} . \end{matrix}$

Suggest Corrections

thumbs-up

0

Similar questions

x = 1 + a + a^{2} + . . . . .

y = 1 + b + b^{2} . . . .

| a | < 1

| b | < 1

1 + a b + a^{2} b^{2} + . . . . = \frac{x y}{x + y - 1}

Q. Solve the following system of equations in

x

y

a x + b y = 1

b x + a y = \frac{(a + b)^{2}}{a^{2} + b^{2}} - 1

b x + a y = \frac{2 a b}{a^{2} + b^{2}}

Q. If x = cy + bz, y = az + cx, z = bx + ay (where x, y, z are not all zero) have a solution other than x = 0, y = 0, z = 0, then a, b and c are connected by the relation

x = \frac{a}{a + b}

y = \frac{b}{a - b}

then prove that

\frac{1}{x} + \frac{1}{y} = \frac{a^{2} + b^{2}}{a b}

a (x + y) + b (x - y) = a^{2} - a b + b^{2}

a (x + y) - b (x - y) = a^{2} + a b + b^{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Distance between Two Points on the Same Coordinate Axes

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon