Question

Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Answer 1

Given lines

                  4x + 7y = 3       ...(1)
                  2x – 3y = – 1     ...(2)
(1) × 1 ⇒   4x + 7y = 3        ...(3)
(2) × 2 ⇒   4x – 6y =  – 2    ...(4)
              (–)   (+)     (+)        
(3) – (4) ⇒        13y   = 5

y = `5/13`

Substitute the value of y = `5/13` in (2)

`2x - 3 xx 5/13` = – 1

`2x - 15/13` = – 1

26x – 15 = – 13

26x = – 13 + 15

26x = 2

x = `2/26 = 1/13`

The point of intersection is `(1/13, 5/13)`

Let the x-intercept and y-intercept be “a”

Equation of a line is

`x/"a" + y/"b"` = 1

`x/"a" + y/"b"` = 1 ...(equal intercepts)

It passes through `(1/13, 5/13)`

`1/(13"a") + 5/(13"a")` = 1

`(1 + 5)/(13"a")` = 1

13a = 6

a = `6/13`

The equation of the line is

`x/(6/13) + y/(6/13)` = 1

`(13x)/6 + (13y)/6`  = 1

13x + 13y = 6

13x + 13y − 6 = 0