Question

and passing through the point of intersection of the lines 2x -5y + 1 = 0 and x - 3y - 4 = 0.
[Ans : x + y + 32 = 0
Find the equation of the straight line making non-zero equal intercepts on the co-ordinate axes​

Answer 1

Given:

The point of intersection of the lines$2x -5y + 1 = 0$ and$x - 3y - 4 = 0$.

To find:

The equal intercepts on the co-ordinate axes.

Step-by-step explanation:

The line$2x-5y+1=0$

substituting the value $y=0$ in the equation,

$2x-5(0)+1=0$

$x=\frac{-1}{2}$

the co-ordinates of the point of intersection is $(\frac{-1}{2} ,0)$

$x=3y+4$

$\frac{-1}{2} =3y+4$

$3y=\frac{-1}{2} -4$

$3y=\frac{-9}{2}$

$y=\frac{-3}{2}$

The passing through the point $(\frac{-1}{2} ,\frac{-3}{2} )$

$y+\frac{3}{2}=(\frac{-3}{2} )(x+\frac{1}{2} )$

$x+y+32=0$.

Answer:

Therefore, equal intercepts on the co-ordinate axes $x+y+32=0$.

Answer 2

Given data:

• The required straight line makes non-zero equal intercepts on the coordinate axes.

• It also passes through the point of intersection of the lines $2x - 5y + 1 = 0$ and $x - 3y - 4 = 0$.

To find:

The equation of the required straight line

Step-by-step explanation:

Step 1. Assuming the required straight line

Let the equation of the required straight line be

$\quad\dfrac{x}{a}+\dfrac{y}{a}=1$

$\Rightarrow x + y = a$ ... ... (1)

Step 2. Finding the point of intersection of the given two lines

The given two lines are

$\quad\quad 2x - 5y + 1 = 0$ ... ... (2)

$\quad\quad x - 3y - 4 = 0$ ... ... (3)

From (3), we get

$\quad\quad x = 3y + 4$ ... ... (4)

Now we substitute $x = 3y + 4$ in (2). We get

$\quad 2 (3y + 4) - 5y + 1 = 0$

$\Rightarrow 6y + 8 - 5y + 1 = 0$

$\Rightarrow y + 9 = 0$

$\Rightarrow y = - 9$

Putting $y = - 9$ in (4), we get

$\quad x = 3 (- 9) + 4$

$\Rightarrow x = - 27 + 4$

$\Rightarrow x = - 23$

So, the point of intersection is $(- 23, - 9)$.

Step 3. Finding the value of a in (1)

The straight line (1) passes through the point $(- 23, - 9)$. Then

$\quad (- 23) + (- 9) = a$

$\Rightarrow a = - 32$

Step 4. Putting $a = - 32$ to get the required straight line

From (1), we finally have

$\quad x + y = - 32$

$\Rightarrow x + y + 32 = 0$

Final Answer:

The equation of the required straight line is $x + y + 32 = 0$.