Question

Find the equation of a straight line which has slope -5/4 and passing through the point (-1,2).​

Answer 1

Answer:

5x + 4y = 3

Step-by-step explanation:

use y - y1 = m(x - x1) ;

use values ( -1 , 2) as (x1 ,y1) and m as slope:

y - 2 = -5/4(x - ( -1)) ;

y - 2 = (-5x - 5)/4 ;

multiply 4 to other side.

4y - 8 = -5x - 5

5x + 4y = 3.

Answer 2

Given : Slope of line is -5/4 and it passes through the point (-1,2)

To find : Equation of the straight line

Solution :

We use the slope point form of straight line to find the equation of a straight line whose slope and one point is given.

Slope point form is given by,

=> ( y - y1 ) = m ( x - x1 )

Here

• x1 and y1 are the coordinates of point from where the given line passes.
• m is the slope or gradient of line.

By substituting x1 = -1, y1 = 2 and m = -5/4, we get :

=> ( y - 2 ) = -5/4 ( x - (-1 )

=> y - 2 = -5/4 ( x + 1 )

=> (y - 2)4 = -5(x + 1)

=> 4y - 8 = -5x - 5

=> 4y + 5x - 8 + 5 = 0

=> 4y + 5x - 3 = 0

This is the required equation of straight line.