Name: Sum and Product of the Slopes of Pair of Straight Lines
Uploaded: 2019-07-19
Duration: 5 min 24 s
Description: A combined equation of a pair of straight lines is obtained by multiplying the equations of two straight lines. Let's learn about their sum and product of sl...

Question

(i) If the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , represents two straight lines, prove that the square of the distance of their point of intersection from the origin is $\frac{c (a + b) - f^{2} - g^{2}}{a b - h^{2}}$ .

(ii) Further, if the lines are perpendicular, prove that this distance $= \frac{(f^{2} + g^{2})}{(h^{2} + b^{2})}$ .

Accepted Answer

(i) We are given the equation as

$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$

As we know the general equation of a straight line is $a x + b y + c = 0$ .Let the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ represent the line $a x + b y + c = 0$ , where take $a = l, b = m, c = n$ .

$l_{1} x + m_{1} y + n_{1} = 0 . . . (1)$

$l_{2} x + m_{2} y + n_{2} = 0 . . . (2)$

Now, as the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ represent the lines, we can write

$(l_{1} x + m_{1} y + n_{1}) (l_{2} x + m_{2} y + n_{2}) = a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c$

Then solving L.H.S, we get

$l_{1} l_{2} x^{2} + l_{1} m_{2} x y + l_{1} n_{2} x + m_{1} l_{2} x y + m_{1} m_{2} y^{2} + m_{1} m_{2} y + n_{1} l_{2} x + n_{1} m_{2} y + n_{1} n_{2}$

$\Rightarrow (l_{1} l_{2}) x^{2} + (l_{1} m_{2} + m_{1} l_{2}) x y + (m_{1} m_{2}) y^{2} + x (l_{1} n_{2} + n_{1} l_{2}) + y (m_{1} n_{2} + n_{1} m_{2}) + n_{1} n_{2}$

Now comparing above equation with equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , we get the values

$a = l_{1} l_{2}, 2 h = l_{1} m_{2} + m_{1} l_{2}, b = m_{1} m_{2}, 2 g = l_{1} n_{2} + n_{1} l_{2}, 2 f = m_{1} n_{2} + n_{1} m_{2}, c = n_{1} n_{2}$

By cross multiplication solving eq (1) and eq(2), we get

$\frac{x}{m_{1} n_{2} - m_{2} n_{1}} = \frac{y}{l_{2} n_{1} - l_{1} n_{2}} = \frac{1}{l_{1} m_{2} - l_{2} m_{1}}$

Then, for $x$

$\Rightarrow \frac{x}{m_{1} n_{2} - m_{2} n_{1}} = \frac{1}{l_{1} m_{2} - l_{2} m_{1}}$

$\Rightarrow x = \frac{m_{1} n_{2} - m_{2} n_{1}}{l_{1} m_{2} - l_{2} m_{1}}$

Then, for $y$

$\Rightarrow \frac{y}{l_{2} n_{1} - l_{1} n_{2}} = \frac{1}{l_{1} m_{2} - l_{2} m_{1}}$

$\Rightarrow y = \frac{l_{2} n_{1} - l_{1} n_{2}}{l_{1} m_{2} - l_{2} m_{1}}$

Now, the point of intersection we get is $P [\frac{m_{1} n_{2} - m_{2} n_{1}}{l_{1} m_{2} - l_{2} m_{1}}, \frac{l_{2} n_{1} - l_{1} n_{2}}{l_{1} m_{2} - l_{2} m_{1}}]$ .

Then the square of the distance of their point of intersection $P$ from the origin $O$ is

$O P = \sqrt{{(x - x_{1})}^{2} + {(y - y_{1})}^{2}}$

$\Rightarrow O P = \sqrt{{(\frac{m_{1} n_{2} - m_{2} n_{1}}{l_{1} m_{2} - l_{2} m_{1}} - 0)}^{2} + {(\frac{l_{2} n_{1} - l_{1} n_{2}}{l_{1} m_{2} - l_{2} m_{1}} - 0)}^{2}}$

$\Rightarrow {(O P)}^{2} = \frac{{(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(l_{2} n_{1} - l_{1} n_{2})}^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} . . . (3)$

We know that formula ${(a - b)}^{2} = a^{2} + b^{2} - 2 a b$ is arrived from

${(a - b)}^{2} = {(a + b)}^{2} - 4 a b$

$\Rightarrow {(a - b)}^{2} = a^{2} + b^{2} + 2 a b - 4 a b$

$\Rightarrow {(a - b)}^{2} = a^{2} + b^{2} - 2 a b$

Now applying the formula ${(a - b)}^{2} = {(a + b)}^{2} - 4 a b$ in eq(3), we get

$\Rightarrow {(O P)}^{2} = \frac{{(m_{1} n_{2} + m_{2} n_{1})}^{2} - 4 m_{1} m_{2} n_{1} n_{2} + {(l_{2} n_{1} + l_{1} n_{2})}^{2} - 4 l_{2} n_{1} l_{1} n_{2}}{{(l_{1} m_{2} + l_{2} m_{1})}^{2} - 4 l_{1} m_{2} l_{2} m_{1}}$

Then Putting the values $a = l_{1} l_{2}, 2 h = l_{1} m_{2} + m_{1} l_{2}, b = m_{1} m_{2}, 2 g = l_{1} n_{2} + n_{1} l_{2}, 2 f = m_{1} n_{2} + n_{1} m_{2}, c = n_{1} n_{2}$ in above equation, we get

${(O P)}^{2} = \frac{{(2 f)}^{2} - 4 b c + 4 g^{2} - 4 a c}{{(2 h)}^{2} - 4 a b}$

$\Rightarrow {(O P)}^{2} = \frac{4 f^{2} - 4 b c + 4 g^{2} - 4 a c}{4 h^{2} - 4 a b}$

$\Rightarrow {(O P)}^{2} = \frac{- 4 (c (b + a) - f^{2} - g^{2})}{- 4 (a b - h^{2})}$

$\Rightarrow {(O P)}^{2} = \frac{c (a + b) - f^{2} - g^{2}}{a b - h^{2}}$

Therefore, the square of the distance of their point of intersection from the origin is $\frac{c (a + b) - f^{2} - g^{2}}{a b - h^{2}}$ .

(ii) Now, if the given pair of lines are perpendicular then

$a + b = 0$

$\Rightarrow a = - b . . . (4)$

Then, putting eq(4) in equation ${(O P)}^{2} = \frac{c (a + b) - f^{2} - g^{2}}{a b - h^{2}}$ , we get

${(O P)}^{2} = \frac{c (- b + b) - f^{2} - g^{2}}{(- b) b - h^{2}}$

$\Rightarrow {(O P)}^{2} = \frac{0 - f^{2} - g^{2}}{- b^{2} - h^{2}}$

$\Rightarrow {(O P)}^{2} = \frac{- (f^{2} + g^{2})}{- (h^{2} + b^{2})}$

$\Rightarrow {(O P)}^{2} = \frac{f^{2} + g^{2}}{h^{2} + b^{2}}$

Therefore, if the lines are perpendicular the distance is $\frac{f^{2} + g^{2}}{h^{2} + b^{2}}$ .

Important Questions on Pair of Straight Lines

Learn and Prepare for any exam you want !

(i) If the equationax2+2hxy+by2+2gx+2fy+c=0, represents two straight lines, prove that the square of the distance of their point of intersection from the origin is ca+b-f2-g2ab-h2. (ii) Further, if the lines are perpendicular, prove that this distance =f2+g2h2+b2.

Important Questions on Pair of Straight Lines

Learn and Prepare for any exam you want !