Math, asked by werwwer12345, 1 year ago

Write an equation of the line that is perpendicular to 3x + 9y = 7 and passes through the point (6, 4).

Answers

Answered by susheellattala

3x+9y=7 slope of the line be M1=-3/9=-1/3

Now let the other line perpendicular to it may have slope M2. Then

M1*M2=-1

Hence M2 is 3

Equation of the line havig slope 3 is

Y=3x+c

Now satisfy the point in the line to get the value of c

4=18+c

c=-14

So equation of line is y=3x-14

Mark my answer as BRAINLIEST

werwwer12345: that wasnt me there is another dude with no profile pic

susheellattala: Thanks a lot

werwwer12345: can you answer this to no one has answered it and its been 40 minutes Write an equation of the line that is parallel to -x + y = 5 and passes through the point (2, -5).
A) y = x - 7
B) y = x - 5
C) y = x - 3
D) y = -x - 3

susheellattala: See if line are parallel their slopes are same so slope will be 1

susheellattala: Slope=-coefficient of x/coefficient of y

susheellattala: -x+y=c now satisy the point in this 2,-5

susheellattala: We get -2-5=c. Therefore c=-7

susheellattala: So eq is -x+y=-7. Send x other side of equality y=x-7

susheellattala: Hope it helps

werwwer12345: ty

Answered by Ompanda

let x=6,y=4
3x+9y=7
3×6+9×4=7
18+36=7
54=7

Previous Question

Next Question