What is the equation of a line passing through (3, 4) and having a slope of -5? A) y = 5x + 19 B) y = -5x + 11 C) y = -5x + 19 D) y = -1/5x + 19 E) y = -1/5x + 11 @nikato @jdoe0001 @Mertsj
y-y1=m(x-x1)
Do you know the point slope form? If you do, plug in the given info in it and solve for y
No I don't, what's the form? Lol.
Oh it's that^. And m=slope (x1,y1) are your points
\(\bf \begin{array}{lllll} &x_1&y_1\\ &({\color{red}{ 3}}\quad ,&{\color{blue}{ 4}})\quad \end{array} \\\quad \\ slope = {\color{green}{ m}}= -5 \\ \quad \\ y-{\color{blue}{ y_1}}={\color{green} m}(x-{\color{red}{ x_1}})\qquad \textit{plug in your values and solve for "y"}\)
y-4=-5*x-3 ? Is that right?
\(\bf y-{\color{blue}{ y_1}}={\color{green} m}(x-{\color{red}{ x_1}}) \) <---- point-slope form
the form is y-y1=m(x-x1) m=slope=-5 y1=4 x1=3
y-4=-5* (x-3)
don't forget to distribute the -5
Ok
y=-5y + 19, right?
y=-5y + 19, right? \(\large \checkmark\)
Thanks :)
Join our real-time social learning platform and learn together with your friends!