A line passes through (9, –9) and (10, –5).Write an equation for the line in point-slope form. Rewrite the equation in standard form using integers.
I assume this is still part of your test? well
no i actually passed my test if your wondering
are you going to help me or not
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &(9\quad ,&9)\quad &(10\quad ,&-5) \end{array} \\\quad \\ slope = m= \cfrac{rise}{run} \implies \cfrac{y_2-y_1}{x_2-x_1} \\ \quad \\ y-y_1=m(x-x_1)\Leftarrow\textit{point-slope form, plug in your values there}\)
hmm -9... ok ... \(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &(9\quad ,&-9)\quad &(10\quad ,&-5) \end{array} \\\quad \\ slope = m= \cfrac{rise}{run} \implies \cfrac{y_2-y_1}{x_2-x_1} \\ \quad \\ y-y_1=m(x-x_1)\Leftarrow\textit{point-slope form, plug in your values there}\)
you've been at this junction already.... so you should know how to get the slope and use the point-slope form
ok im still confused how do i write in integers is what im asking
Ok, I couldn't read what jdoe said, so I'm going to start from scratch. I'm going to use linear form y=mx+b. To find m we use the slope equation
and it got m=4, so now we have y=4x+b, to solve for b we can just plug it one of our coordinate points from above. lets use 10, -5 -5=4(10)+b b=-45 now we have y=4x-45 to get into standard (Ax+By=c) we just rearrange it by moving x and y to the same side and have c to be positive. y=4x-45 y-4x=-45 divide by -1 4x-y=45 Tada! and integers just mean whole numbers like 1, 2, -3, 4 and no decimals or fractions like 4.5
A. y – 9 =4 (x + 9); –4x + y = 45 B.y + 9 =4 (x + 9); –4x + y = –45 C.y + 9 =4 (x – 9); –4x + y = –45 D. y – 9 =4 (x – 9); –4x + y = 45 im asking which one of these choices would fit that problem?
it would be C, 4x-y=45 can be rewritten as -4x+y=-45 and point slope is y-y1=m(x-x1) and x1=9 and y1=-9. When we plug that in, we get , y+y=4(x-9)
ok that makes perfect sense now :) lol thank you
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