# Determine the equation of the line that passes through the points (10, 0) and (-1, 10)

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### 3 Answers

We have the points:

(10, 0) and (-1, 10)

The equation for the line is:

y-y1 = m (x-x1) where m is the slope:

But we know that:

m = (y2-y1)/(x2-x1) = 10/-11= -10/11

==> y- 0 = (-10/11) (x- 10)

==> y= -(10/11)x + 100/11

Multiply by 11:

==> 11y= -10x + 100

==> **10x + 11y -100 = 0**

We'll find the equation of the line that passes through the given points, using the formula:

(x2-x1)/(x-x1) = (y2-y1)/(y-y1)

Now, we'll substitute the coordinates of the given points, into the formula above:

(-1-10)/(x-10) = (10-0)/(y-0)

-11/(x-10) = 10/(y)

We'll cross multiply:

-11y = 10(x-10)

-11y = 10x - 100

We'll divide by -11:

y = -10x/11 + 100/11

**So, the equation of the line is: y = -10x/11 + 100/11**

To find the equation of the line passing through (10,0) and (-1 ,10).

Solution:

Since the line is passing through (10,0), the line makes an intercept of 10 on y axis. So the equation is of the form :

y = mx+c where c is the intercept = 10. So y = mx+10 . We determine m from the fact that (-1, 10) is a point on y = mx+10 and hence the coordinates (-1,10) should satisfy y = mx+10:

-1 = m*10 +10.

-1 = 10m +10. Or 10m = -1-10 = -11.

Therefore m = -11/10.

So the equation y = mx+10 becomes y = -(11/10)x+10. Or Multiplying by 10 we get:

10y =-11x+100. Or

11x+10y -100 = 0