Question

#3) What are the equation of the lines parallel to the line 3x+4y-12=0 and passing at a distance ±2 from the origin.

Answer 1

@icecreamman

Answer 2

@icecreamman

Answer 3

do you know how to write the question as a function of y?

Answer 4

like y =.....

Answer 5

yup

Answer 6

so in the equation... y=\[-\frac{ 3 }{ 4}x+3\]

Answer 7

the slope is - 3/4

Answer 8

so what should we do next?

Answer 9

im not really sure of what to do next but im gonna take a stab at this ok? um but i think parallel is slope the same and b as in y= mx+b is different right?

Answer 10

origin is (0,0)

Answer 11

sorry its been a while since ive done this kind of math... not really sure on what to do next

Answer 12

oh! D:

Answer 13

do you know someone who could help us here :D

Answer 14

sorry :(

Answer 15

are all your questions like these?

Answer 16

all in straight lines geometry

Answer 17

lemme see if i can help with those so just complie all those questions into one post and lemme see if i could help you

Answer 18

Find the equation of a line through (-8,6) with x- intercept twice the y- intercept.

Answer 19

Find the equation ,parallel,perpendicular of the lines 5x+12y+26=0 and passing at a distance numerically 2 units farther from the origin.

Answer 20

A circle of radius 4 touches the line 3x+4y=8. Find the locus of its center.

Answer 21

last is ,, The distance from a line to (6,2) is 3 units. The line is perpendicular to 4x +3y+8=0. Find its equation.

Answer 22

last question:

as the given line is perpendicular to the other line, the product of their slope will be equal to -1
slope of the given line is
4x-3y=8
-3y=8-4x
y=-8/3+4/3x
so slope equals 4/3

therefore the slope of the required line is -3/4
so using ponit slope formula the equation of the line is
y-y1= m(x-x1)+c (where x1,y1 is 6,2 & c is the distance that is 3)
y-2 =-3/4(x-6)+3
y-2 =(-3x+18+12)/4
4y-8 = -3x +30
4y = -3x+30+8
4y=-3x+38 this is the reuired line

Answer 23

5x + 12y + 26 = 0 ...re-writing the equation in slope-intercept form
12y = -5x - 26
y = (-5/12)x - (26/12)

1) If the distance is measured along the x axis then y = 0, (-5/12)x = 26/12 : x = 26/-5 = -5.2
Then we require the line to pass through (0,-7.2) and we can write -7.2 as 36/-5.
The equation thus becomes y = (-5/12)x - (36/12) or 5x + 12y + 36 = 0

2) If the distance is measured along the y axis then x = 0, y = -26/12 = -2⅙
Then we require the line to pass through (-4⅙,0) and we can write -4⅙ as -50/12.
The equation thus becomes, y = (-5/12)x - (50/12) or 5x + 12y + 50 = 0

Answer 24

wait im analyzing it