# using y - 3 = 3(x +1) what point does this line pass through, which is the basis of this equation? Show all steps. Thanks

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The problem provides the equation of a line. You need to remember that there exists several forms in which a equation of a line can be given. The provided form of equation follows the pattern of the point slope form. You need to compare the provided form of the equation to the point slope form, such that:

`y - y_0 = m(x - x_0)` (point slope form)

`y - 3 = 3(x + 1)` (provided form)

Comparing these forms yields that the given line passes through the point whose coordinates are `x_0 = -1` and `y_0 = 3` .

**Hence, observing the given equation to evaluate the point the provided line passes through yields that this point is `(-1,3)` .**

You should notice that if you draw a line, parallel to x axis, at `y = 3` , it intersects the red line at point, whose x coordinate is `x = -1` (drop a perpendicular line to x axis, from the intersection point between the line `y=3 ` and `y - 3 = 3(x + 1))` .

The equation of the line is y - 3 = 3(x +1).

If a line passes through the point (h, k) and its slope is m, the equation of the line is `(y - k)/(x - h) = m`

The given equation y - 3 = 3(x +1) can be rewritten as:

`(y - 3)/(x + 1) = 3`

From this it is clear that the slope of the line is 3 and it passes through the point (-1, 3)

The required point is (-1, 3)

since this is in y-y1=m(x-x1 form, we can say that the numbers in place of (x1 and y1, can give us a point the equation passes through

this in this case would be (-1,3).

y - 3 = 3(x +1)

This equation is in point-slope form:

y - y1 = m(x - x1) where m is the slope and (x1, y1) is a point. Since we need to find the point, just compare what you have to the generic form.

y1 = 3 and x1 = -1

So the point is (-1, 3).