Solution
(i)
$px + qy = p - q$ -----(i)
$qx - py = p + q$ -----(ii)
Multiplying equation (i) by $p$ and (ii) by $q$, we get
$p^2x + pqy = p^2 - pq$ -----(iii)
$q^2x - pqy = pq + q^2$ -----(iv)
Adding equations (iii) and (iv), we obtain
$p^2x + q^2y = p^2 + q^2$
$⇒ (p^2 + q^2)x = p^2 + q^2$
$⇒ x = 1$
Substituting $x = 1$ in equation (i), we get
$p + qy = p - q$
$qy = -q ⇒ y = -1$
Therefore $x = 1$ and $y = -1$
(ii)
$ax + by = c$ -----(i)
$bx + ay = 1 + c$ -----(ii)
Multiplying equation (i) by $a$ and equation (ii) by $b$, we obtain,
$a^2x + aby = ac$ -----(iii)
$b^2x + aby = b + bc$ -----(iv)
Subtracting equation (iv) from (iii), we get,
$a^2x - b^2x = ac - b - bc$
$(a^2 - b^2)x = ac - b - bc$
$⇒ x = \dfrac{ac - b - bc}{a^2 - b^2}$
Substituting value of $x$ in equation (i),
$ax + by = c$
$⇒ a × \dfrac{ac - b - bc}{a^2 - b^2} + by = c$
$⇒ \dfrac{a^2c - ab - abc}{a^2 - b^2} + by = c$
$⇒ by = c - \dfrac{a^2c - ab - abc}{a^2 - b^2}$
$⇒ by = \dfrac{a^2c - b^2c - a^2c + ab + abc}{a^2 - b^2}$
$⇒ by = \dfrac{- b^2c + ab + abc}{a^2 - b^2}$
$⇒ y = \dfrac{- bc + a + ac}{a^2 - b^2}$
Hence,
$x = \dfrac{ac - b - bc}{a^2 - b^2}$ $= \dfrac{c(a - b) - b}{a^2 - b^2}$
and
$y = \dfrac{- bc + a + ac}{a^2 - b^2}$ $= \dfrac{c(a - b) + a}{a^2 - b^2}$
(iii)
$\dfrac{x}{a} - \dfrac{y}{b} = 0$ -----(i)
$ax + by = a^2 + b^2$ -----(ii)
From equation (i), we get
$\dfrac{x}{a} - \dfrac{y}{b} = 0$
$bx - ay = 0$
$⇒ x = \dfrac{ay}{b}$ -----(iii)
Substituting value of $x$ in equation (ii), we obtain
$ax + by = a^2 + b^2$
$⇒a × \dfrac{ay}{b} + by = a^2 + b^2$
$⇒\dfrac{a^2y + b^2y}{b} = a^2 + b^2$
$⇒ y = b$
Substituting value of $y$ in equation (iii), we get
$x = \dfrac{ay}{b} = \dfrac{ab}{b}$
$⇒ x = a$
Hence, $x = a$ and $y = b$
(iv)
$(a - b)x + (a + b)y = a^2 - 2ab - b^2$ -----(i)
$(a + b)(x + y) = a^2 + b^2$ -----(ii)
Equation (ii) can be re-written as
$(a + b)(x + y) = a^2 + b^2$
$⇒ (a + b)x + (a + b)y = a^2 + b^2$ -----(iii)
Subtracting equation (iii) and (i), we get,
$(a - b)x + (a + b)y = a^2 - 2ab - b^2$
-
$(a + b)x + (a + b)y = a^2 + b^2$
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$(a - b)x - (a + b)x = -2ab - 2b^2$
$(ax - bx - ax - bx = -2ab - 2b^2$
$ - 2bx = -2ab - 2b^2$
$x = a + b$
Putting value of $x$ in equation (i), we get,
$(a - b)x + (a + b)y = a^2 - 2ab - b^2$
$⇒ (a - b) × (a + b) + (a + b)y = a^2 - 2ab - b^2$
$⇒ a^2 - b^2 + (a + b)y = a^2 - 2ab - b^2$
$⇒ (a + b)y = - 2ab$
$⇒ y = \dfrac{- 2ab}{a + b}$
Hence $x = a + b$ and $y = \dfrac{- 2ab}{a + b}$
(v)
$152x - 378y = -74$ -----(i)
$-378x + 152y = -604$ -----(ii)
Subtracting equation (ii) from (i), we obtain,
$152x - 378y = -74$
-
$-378x + 152y = -604$
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$530x - 530y = 530$
$⇒ x - y = 1$
$⇒ x = y + 1$ -----(iii)
Putting the value of $x$ in equation (i),
$152(y + 1) - 378y = -74$
$⇒ 152y + 152 - 378y = -74$
$⇒ -226y = -226$
$⇒ y = 1$
Putting value of y in equation (iii),
$x = y + 1 = 1 + 1 = 2$
Hence $x = 2$ and $y = 1$
8
ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.