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# GMAT: Math 16

87-91, pg. 163. Created by Sal Khan.

## Want to join the conversation?

• find the point on the axis that is equidistant from (-2,5) and (4,1) • i need to know more about the slope method used in problem 89 which math lesson should i watch exactly? • Well, there are lots of them, depending on what your confusion is.
One of the most important steps here was finding what the slope was according to the equation they gave us.
x = 3y -7
We use `y = mx + b` as the format that will immediately reveal the slope (m) of the line.
So, the fasted thing to do here, and the best, is to rearrange x = 3y -7 to solve for y.
3y - 7 = x
3y = x + 7
y = `1/3 `x + 7/3
Aha! slope is 1/3 (goes up 1 unit for every 3 units of x, or you can think of it going up 1/3 unit every time c increases by 1.)
OK, so let's look at the coordinates they gave us.(a, b) and (a + 3, b + k)
The definition of slope is `rise over run`, in other words, the change in y divided by the change in x.
The workhorse formula for finding slope = (y₂ - y₁)/(x₂ - x₁) That is what Sal used.

So if you have two points, call one Point 1 and the other Point 2. Then take the coordinates and label them for the two points:
Point 1 (x₁ , y₁) and Point 2 (x₂ , y₂)
x₁ = a
y₁ = b
x₂ = a + 3
y₂ = b + k
Plugging in, then (y₂ - y₁)/(x₂ - x₁)
= ((b + k) - b)/((a + 3) - a)
Clearing out the parentheses:
b + k - b is just k
a + 3 - a is just 3
so, the result for the two points is k/3
Wow, that makes it easy, because all we need is the number for k that makes
k/3 = 1/3
so k = 1
Actually, I saw that point a increased by 3 and point b increased by k, so I was able to skip most of the steps to the
k/3 = 1/3
so k = 1
And you will too after some practice: put slope-intercept into the search box or just slope and you will get lots of options for delving into this topic.
• Convert minutes to seconds • who can do that real quick i dont under stand it
(1 vote) • i always have a problem findind the point equidistance of (4,2) and (5,9)  