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In this video, we are obtaining the equation ...
of a line passing two points.
Previously, we dealt with straight lines ...
with positive and negative gradients.
They all had something in common:
they crossed the y-axis at a given point.
In this video, we are obtaining the equation ...
of a line passing two points. But what if we don't know the ...
y-intercept. Not a problem!
You can still work out the gradient,
can't you? It is the rise over the run as ...
we explained in detail in our previous video.
So the gradient of this line is 4/5.
So the equation of the line is y equals the y-intercept,
which we don't know, not yet…plus (4/5)x.
The y-intercept is clearly not zero,
otherwise the line would start at the point (0,0).
Our line starts at (2,3) so the equation is y=3+4/5(x-2) ...
because when x=2, y=3.
In this second line, y= the y-intercept + (2/5)x.
The y-intercept is clearly not zero.
Our line starts at (2,4) so the equation is y=4+2/5(x-2) ...
because when x=2, y=4.
Tis third line starts at (1,3) so the equation is y=3+2/5(x-1) ...
because when x=1, y=3.
This fourth line has a gradient of 8 over 16 or ½.
Since it starts at (4,8), the equation is y=8+(1/2)(x-4) ...
because when x=4, y=8.
Stop the video here and try this red line,
what is the equation?
Right! The gradient is -5/20 or -1/4.
We replace this into the equation.
Now, we know that the line passes ...
(8,8) so multiplying -1/4 by x and by -8 and collecting 8 + 2,
we obtain y=10-(1/4)x. Our line indeed crosses the ...
y-axis at 10, so we got that right!
In general, to write the equation of the ...
line passing two points, we calculate the gradient as ...
the rise over the run: y-b over x-a.
Rearranging (x-a) to the other side and adding b to both ...
sides, we can see that the equation ...
of the line is y=b+m(x-a) For more videos on linear algebra,
explore this playlist. Until next week,
happy learning!
