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line l contains points (-2,4) and (1,-9). Point P has coordinates (14,-6) answer

Sagot :

We are given a line and a point, I assume that we want to see if the point is on the line or not.

We will see that the point does not belong to the line.

First, let's find the equation of the line.

A general line can be written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

If the line passes through points (x₁, y₁) and (x₂, y₂), then the slope can be computed as:

[tex]a = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Here we know that our line passes through the points (-2, 4) and (1, -9). Then the slope is:

[tex]a = \frac{4 - (-9)}{-2 - 1} = \frac{13}{-3} = -\frac{13}{3}[/tex]

Then our line is something like:

[tex]y = -\frac{13}{3} *x + b[/tex]

To find the value of b, we can use one of the two points that we know that are on the line.

For example, the point (-2, 4) means that when x = -2, we must have y = 4.

Replacing that in the equation we get:

[tex]4 = -\frac{13}{3} *(-2) + b\\\\4 = \frac{26}{3} + b\\\\4 - \frac{26}{3} = b\\\\\frac{-14}{3} = b[/tex]

Now we can write the line equation as:

[tex]y = -\frac{13}{3} *x - \frac{14}{3}[/tex]

Now we need to see if the point (14, -6) belongs to the line.

To see this, we evaluate the line in x = 14.

[tex]y = -\frac{13}{3} *14 - \frac{14}{3} = -\frac{196}{3}[/tex]

This is different than -6, so the point does not belong to the line.

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