Lecture 4 [24 Feb, 2020] MATH 127 (Prof Farid)

 MATH -127 (Coordinate Geometry)
Chapter-2 
(Pair of Straight Lines)

Question 3: Prove that  $6x^2 + 7 xy - 5y^2 - 22x 41 y - 8 = p$  represents a pair of straight lines. Find their point of intersection and angle between them.

Hint:   $$\alpha = \frac{fh - bg }{ab -h}= \frac{ bg - fh }{h^2 - ab}, ~~~~~\beta = \frac{gh - af}{ab - h^2 }= \frac{af - gh }{h^2 - ab}$$

Question 4: If the equation  $ax^2 + 3 xy - 2y^2 - 5x + 5y + c = 0$  represents a pair of straight lines perpendicular to eachother. Find the value of $a ~\textrm{and} ~c$

Hint:  $abc + 2fgh - af^2 - bg^2 - ch^2 = 0 ~~\textrm{and} ~~ a + b = 0,~ a = 2,~ c = -3$

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Condition for coincidence of two straight lines:

1.$f^2 - bc = 0$

2.$g^2 - ac = 0$

3.$h^2 - ab = 0$

4.$gh = af$

5.$hf = bg$

6.$fg = ch$

Bisectors of the angles between two straight lines:

  $ax^2 + 2hxy + by^2 + wgx + 2fy + c = 0~~~~~\bigg|~ (\alpha, \beta) ~\textrm{is the point of intersection}$  

Let $(\alpha$, $\beta)$ be the new origin,

$$\frac{(x - \alpha)^2 - (y - \beta)^2}{a - b} = \frac{(x - \alpha)(y - \beta)}{h}$$

If the equation is homogenous quadratic equation  $ax^2 + 2hxy + by^2 = 0$  The Bisectors are,

$$\frac{x^2 - y^2 }{a - b }= \frac{xy}{h}$$

Question 5: Find the equation of the bisectors of the angles between the lines represented by $6x^2 + 7xy - 5y^2 - 22x - 41 y - 8 = 0$

Solution: Let $(\alpha, \beta)$ be the point of intersection between the lines represented by the given equation (1).

Then, $$\alpha = \frac{bg - fh }{h^2 - ab}= 3 ~~~ \textrm{and}~~~ \beta = \frac{gh - af }{ab - h^2}= - 2$$

$\therefore$ The equation of the bisectors are,

$$\begin{align*} & \frac{(x - 3)^2 - (y + 2)^2}{6 + 5} = \frac{(x - 3) - (y + 3)}{\frac{7}{2}} \\ \implies & 7(x^2 - 6x + 9 - y^2 - 4y - 4)= 22 (xy + 2x - 3y - 6)\\ \implies & 7x^2 - 42 x + 63 - 7 y^2 - 28 y - 28 - 22xy - 44x + 132 = 0\\ \implies & 7 x^2 - 7 y^2 - 22x + 38y - 86x + 167 = 0 \end{align*}$$

Lines joining the origin to the points of intersection of a curve and a line.

$$\begin{align*} & (lx + my + n) = 0\\ \implies & \frac{l}{ - n }x + \frac{m}{- n}y = 1 \\ \implies & ax^2 + 2hxy + by^2 + 2(gx + fy) + c.1^2 = 0 \\ \implies & ax^2 + 2hxy + by + 2(gx + fy)\left((\frac{l}{- n}x + \frac{m}{- n}y)\right) + c \left(\frac{l}{- n }x + \frac{m}{- n}y \right) \end{align*}$$