Respuesta :
The equation in slope-intercept form for the line that passes through the point ( -1 , -2 ) and is perpendicular to the line − 4 x − 3 y = − 5 is [tex]y = \frac{3}{4}x - \frac{5}{4}[/tex]
Solution:
The slope intercept form is given as:
y = mx + c ----- eqn 1
Where "m" is the slope of line and "c" is the y - intercept
Given that the line that passes through the point ( -1 , -2 ) and is perpendicular to the line − 4 x − 3 y = − 5
Given line is perpendicular to − 4 x − 3 y = − 5
− 4 x − 3 y = − 5
-3y = 4x - 5
3y = -4x + 5
[tex]y = \frac{-4x}{3} + \frac{5}{3}[/tex]
On comparing the above equation with eqn 1, we get,
[tex]m = \frac{-4}{3}[/tex]
We know that product of slope of a line and slope of line perpendicular to it is -1
[tex]\frac{-4}{3} \times \text{ slope of line perpendicular to it}= -1\\\\\text{ slope of line perpendicular to it} = \frac{3}{4}[/tex]
Given point is (-1, -2)
Now we have to find the equation of line passing through (-1, -2) with slope [tex]m = \frac{3}{4}[/tex]
Substitute (x, y) = (-1, -2) and m = 3/4 in eqn 1
[tex]-2 = \frac{3}{4}(-1) + c\\\\-2 = \frac{-3}{4} + c\\\\c = - 2 + \frac{3}{4}\\\\c = \frac{-5}{4}[/tex]
[tex]\text{ substitute } c = \frac{-5}{4} \text{ and } m = \frac{3}{4} \text{ in eqn 1}[/tex]
[tex]y = \frac{3}{4} \times x + \frac{-5}{4}\\\\y = \frac{3}{4}x - \frac{5}{4}[/tex]
Thus the required equation of line is found
