Name: Family of Straight Lines: Example
Uploaded: 2019-07-19
Duration: 3 min 23 s
Description: Do you know how to find the equation of a straight line passing through the point of intersection of a family of two lines and also through a point? Let's find out.

Question

Prove that the line

x (a + 2 b) + y (a - 3 b) = a - b

passes through a fixed point for different values of

a

and

b

. Also find the fixed point.

Accepted Answer

The given line is $x (a + 2 b) + y (a - 3 b) = a - b .$

We can also write the given line as $a x + 2 b x + a y - 3 b y - a + b = 0$

$\Rightarrow a (x + y - 1) + b (2 x - 3 y + 1) = 0$

$\Rightarrow (x + y - 1) + \frac{b}{a} (2 x - 3 y + 1) = 0 .$

The above equation is in the form $L_{1} + λ L_{2} = 0,$ where

$L_{1} \equiv x + y - 1 = 0, L_{2} \equiv 2 x - 3 y + 1 = 0$ and $λ = \frac{b}{a} .$

We know that, the equation of the family of lines

$L_{1} + λ L_{2} = 0, λ \in R$ passes through the point of intersection of the lines $L_{1} = 0$ and $L_{2} = 0$ for different values of $λ \in R .$

Now to find the point of intersection of the lines $x + y - 1 = 0$ and $2 x - 3 y + 1 = 0,$

put $x = 1 - y$ from the first line to $2 x - 3 y + 1 = 0$ to get

$2 (1 - y) - 3 y + 1 = 0$

$\Rightarrow 2 - 2 y - 3 y + 1 = 0$

$\Rightarrow 3 - 5 y = 0$

$\Rightarrow y = \frac{3}{5}$ and $x = 1 - y = 1 - \frac{3}{5} = \frac{2}{5} .$

Thus, the point through which the two lines pass is $(\frac{2}{5}, \frac{3}{5}) .$

Prove that the line $x (a + 2 b) + y (a - 3 b) = a - b$ passes through a fixed point for different values of $a$ and $b$ . Also find the fixed point.

Important Questions on Straight Lines

Learn and Prepare for any exam you want !

Prove that the line xa+2b+ya-3b=a-b passes through a fixed point for different values of a and b. Also find the fixed point.

Important Questions on Straight Lines

Learn and Prepare for any exam you want !

Prove that the line $x (a + 2 b) + y (a - 3 b) = a - b$ passes through a fixed point for different values of $a$ and $b$ . Also find the fixed point.