1) Write the standard form of an equation of the line that passes through (3, 5) and has a slope of -3/4.Not Enough Information3/4x+y=29/43x+4y=29y = -3/4x +29/42)Write the point-slope form of an equation for a vertical line that passes through (-4, -7).0 = x + 70 = x - 40 = x + 40 = x - 73)Write the slope-intercept form of equation of the line that passes through (-2, 3) and has a slope of m = -3.3x+y = 93x+y = -9y = 3x+9y = -3x-34) Write the point-slope form of an equation for a horizontal line that passes through (-15, 11).y+11 = 0y-11 = 0y-15 = 0y+15 = 0

Question
Answer:
β—† Straight Lines β—†

[ 1 ]

Any equation with a slope m can be expressed in the form of :

y = mx + c

Here m = -3/4 ;
Line is passing through (3,5)

Substituting in std. equation ,

5 = -3/4(3) + c
---> 5 = -9/4 + c
---> c = 5 + 9/4 = 29/4

Hence ,

Equation of line is :-

y = -3x/4 + 29/4 Ans.

[ 2 ]

Line is passing through ( -4 , -7 )

Therefore x = -4 ; y = -7

+ It's a vertical line

Hence , equation of line -
x + 4 = 0 Ans.

[ 3 ]

Any equation with a slope m can be expressed in the form of :

y = mx + c

Here m = -3 ;
Line is passing through ( -2 , 3 )

Substituting in std. equation ,

3 = -3(-2) + c
---> 3 = 6 + c
---> c = -3

Hence ,

Equation of line is :-

y = -3x - 3

[ 4 ]

Line is passing through ( -15 , 11 )

Therefore x = -15 ; y = 11

+ It's a Horizontal line

Hence , equation of line -
y - 11 = 0 Ans.

Hope it helps you :)

β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†β—†
solved
general 11 months ago 5868