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Launch Area = (1, 2) Point A = (0, 3) Plug them into the slope formula
\(m = \dfrac{y_2-y_1}{x_2-x_1}\) \(m = \dfrac{2-3}{1-0}\) Simplify
Yep, divide
Yes, so -1 is the slope, now plug this in with any of the two points into point-slope form. \(y - y_1 = m(x - x_1)\) Where \(y_1\) is the y-value of the point, \(x_1\) is the x-value of the point, and \(m\) is the slope. So we have: \(y - 2 = -1(x - 1)\) Distribut -1 into the parenthesis: \(y - 2 = -x + 1\) Add 2 to both sides, what's 1 + 2?
Correct.
But it wants it in Standard Form, so we have to simplify it..can you add 1 + 2?
Yes, that is our equation in slope-intercept form, now add 'x' to both sides to convert to Standard form.
Yes..but it should be x + y = 3.
'Cause standard form is Ax + By = C
That's our equation in standard form..
Launching Point = (1, 2) Point B = (-3, 0) Plug it into the slope-intercept form.
point-slope form*
Oh, my bad..plug the two points into the slope formula first.
Nice work, so 1/2 is the slope.
Just plug any of the two points in with the slope. \(y - y_1 = m(x - x_1)\) Where \(y_1\) is the y-value of the point, \(x_1\) is the x-value of the point, and \(m\) is the slope.
Correct.
Now for C, find the slope between the two points.
I mean for #4
(1, 2) and (-1, -4)
Correct!
Now for #5: y-2=0.5(x-1) Distribute 0.5 into the parenthesis
Yes
Hold on.. 0.5 * -1 isn't 0.5
Yep, you got it..-0.5 So we have: y - 2 = 0.5x - 0.5
Add 2 to both sides.
There's only 1 minus sign..where did you get the 2nd one from?
It's fine..so finish simplifying. y - 2 = 0.5x - 0.5 Add 2 to both sides
Correct.
For #6, there aren't really any restrictions since this is space..and space is infinite.
Yep, that's all the questions.
No problem.
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