Question

PLEASE HELP! The question asks to graph each line and to construct a perpendicular segment through the given point. It also asks to find the distance from the point to the line. The given equation and point is: 2x-3y=-9, (2,0). Please tell me all the steps to solve this and maybe the answer? I'm just a little confused on how to find the distance between two perpendicular lines.

Answer 1

what's the slope of 2x-3y=-9 ?

Answer 2

The slope would be 2/3x

Answer 3

well... if you solve for "y", you'd end up with \(\bf y = \cfrac{2}{3}x+3\)

so the slope will be just 2/3

so we want to find a perpendicular line that THAT ONE that passes through (2, 0) firstly

so, a perpendicular to that will have a NEGATIVE RECIPROCAL slope
what does that mean?

RECIPROCAL of that 3/2
NEGATIVIZE IT -3/2

so we know now that the line passes through ( 2, 0) and has a slope of -3/2

so, you can find the equation for that line by using the point-slope form
that is
\(\large{ slope=m=-\cfrac{3}{2}\qquad \qquad \begin{array}{lllll}
&x_1&y_1\\
&(2\quad ,&0)
\end{array}\\ \quad \\
y-y_1=m(x-x_1)}\)

Answer 4

Oh thank you! But from there on, how do you find the distance between those two perpendicular lines? I know how to find the distance if they were parallel cause they hve the same slope, but in this one they have opposite slopes... So how would I find it?

Answer 5

hmm what did you get for the equation of the line perpendicular to that?

Answer 6

I got y= -3/2x-3

Answer 7

once you get that 2nd equation
you'd end up with a SYSTEM OF EQUATIONS

the solution to a system of equations is where the graph of both lines in the equation meet

so you'd get a "x" and a "y" coordinate for that point, which will be the solution

so you'd need the distance between ( 2 , 0 ) and the SOLUTION to the system of equations

and for that you can use the distance formula, \(\bf \text{distance between 2 points}\\
d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)

Answer 8

ok... well... I got ....let's see \(\bf y-0=-\cfrac{3}{2}(x-2)\implies y = -\cfrac{3}{2}x+3\)

recall that minus times minus = plus

Answer 9

\(\bf y = \cfrac{2}{3}x+3\\ \quad \\
y = -\cfrac{3}{2}x+3\qquad \textit{notice the equivalence}\quad thus\\ \quad \\
\cfrac{2}{3}x+3=-\cfrac{3}{2}x+3\implies \cfrac{2}{3}x+\cfrac{3}{2}x=3-3\implies \cfrac{4x+9x}{6}=0\\ \quad \\
13x=0\implies \color{red}{x =0}\\ \quad \\ \quad \\

\textit{let's use that in either equation, say the 1st one}\\
y = \cfrac{2}{3}0+3\implies \color{red}{y = 3}\)

Answer 10

so, using the substitution method to solve the system of equations, that is, substituting "y" value of the 1st one, to the 2nd equation, we get x = 0

then using our x = 0 in the 1st one, we get y = 3

so the solution to the system of equations is ( 0, 3)

that's where the lines meet

Answer 11

Ok and then we use the distance formula. Thank you so much, you are literally a life saver!

Answer 12

so the graph looks about say

Answer 13

yw