Question

write an equation in slope intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation.

(0,0) y=-3x+2

(5,0) y+1=2(x-3)

Answer 1

@Data_LG2

Answer 2

so now, they are looking for the "perpendicular line"
remember what i said before about their slope?
the slope of a perpendicular line is the \(\sf negative\ reciprocal\) of the given line.

so the first one, what is the slope?

Answer 3

-3x

Answer 4

right, just take the coefficient. the slope is -3

now what is the negative reciprocal of -3?

Answer 5

3

Answer 6

close ! how about it's reciprocal?

we have 3 and underneath it, we have the imaginary 1
so to get the reciprocal, we have to flip them
look here: (give me a sec my drawing tool is not working)

Answer 7

1/3

Answer 8

yes right, (lol i don't need the drawing tool then xD )

the slope of the required line is 1/3
so now we have \(\sf y=\frac{1}{3}x+b\)
like before, we have to solve for \(\sf b\)

Answer 9

b=0

Answer 10

@Catlover5925

Answer 11

can you help me with the second one plz :3

Answer 12

i can try, what is the question??

Answer 13

^^same question at the top just the second equation now

Answer 14

ummm i am not sure @Data_LG2 said he would be right back...sorry

Answer 15

its ok.

Answer 16

@mathmath333

Answer 17

(0,0) y=-3x+2
\(\large\tt \begin{align} \color{black}{ y=-3x+2\\~\\
}\end{align}\)
the line passing through (0,0) and perpendicular with y=-3x+2 have its slope
as negative inverse of the line y=-3x+2

so
\(m=\dfrac{1}{3}\)
\(\large\tt \begin{align} \color{black}{( y-y_1)=m(x-x_1)\\~\\
( y-0)=\dfrac{1}{3}(x-0)\\~\\
y=\dfrac{1}{3}x\\~\\
}\end{align}\)
https://www.desmos.com/calculator/kt7mmqvgif