Question

Find the value of a straight line through the intersection of lines 7x+3y=10,5x-4y=1 and parallel to the line 13x+5y+12=0.

Answer 1

Given :

The equations of the two straight lines  through intersection of which the required line passes :

$L_1:7x + 3y =10$   -   (1)

$L_2 : 5x -4y = 1$     -  (2)

The  equation of the line to which the required line is parallel :

$L_3 : 13x + 5y + 12 = 0$    -(3)

To Find :

The equation of the line passing through the intersection of $L_1$ and $L_2$ and parallel to the $L_3$ = ?

Solution :

Let the required line be represented by $L$ , since it passes through the intersection of $L_1$ and $L_2$ , and its slope can be found from the $L_3$ i.e. slope of these two lines is same  .

So, here we  can use slope-point form .

Now for the point of intersection we have to solve the equations of $L_1$ and $L_2$ :

Multiplying equation (1) by 4 and equation (2) by 3 and adding we get :

 $28x + 12y$ + $15x - 12y =$ $40 + 3$

Or, 43x = 43

Or, x = 1

Now putting the value of x in equation (2) i.e. in the equation of 2nd line  :

5 -  4y = 1

Or, -4y = -4

Or, y =  1

So, the point of intersection is (1,1)

Now the slope from equation (3) will be :

$13x + 5y + 12y = 0\\y = \frac{-13}{5}x - \frac{-12}{5}$

So, slope is $\frac{-13}{5}$

Now the equation of L from slope-point form will be :

$(y -y_1)= m(x-x_1)$

Or, $(y -1) =\frac{-13}{5}(x-1)$

Or, 5y - 5 = -13x + 13

Or, 5y +13x = 18

So , the equation of required line is L : 5y +13x = 18