Equation of the line passing through the point of intersection of the lines $2 x + 3 y - 5 = 0, 3 x + 2 y - 5 = 0$ and having slope $- 1$ is

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Solution

We have

Equation of lines

$2 x + 3 y - 5 = 0 . . . . . . (1)$

$3 x + 2 y - 5 = 0 . . . . . . (2)$

On solving equation $(1)$ and $(2)$ to,

$(x, y)$ = $(4, 1)$

Let the intersection point

$(x_{1}, y_{1})$ $=$ $(4, 1)$

Given that

The slope of line $(m) = - 1$

So,

Equation of line

$y - y_{1} = m (x - x_{1})$

$y - 1 = - 1 (x - 4)$

$y - 1 = - x + 4$

$x + y - 5 = 0$

Hence, this is the answer.

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Equation of the line passing through the point of intersection of the lines 2x+3y−5=0,3x+2y−5=0 and having slope −1 is

Equation of the line passing through the point of intersection of the lines $2 x + 3 y - 5 = 0, 3 x + 2 y - 5 = 0$ and having slope $- 1$ is