(k+1)x+(k+2)y+5=0 Find The Point Of Concurrency

Jasonbradley

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Jun 16, 2024

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Finding the Point of Concurrency for a Family of Lines

The equation (k+1)x + (k+2)y + 5 = 0 represents a family of lines where k is a parameter. To find the point of concurrency for this family, we need to determine a single point that lies on all lines represented by this equation, regardless of the value of k.

Here's how we can approach this problem:

1. Rearrange the equation:

   Let's rewrite the equation in slope-intercept form (y = mx + c):

   (k+2)y = -(k+1)x - 5 y = (-k-1)/(k+2) x - 5/(k+2)

   This form clearly shows the slope (m) and y-intercept (c) of the line in terms of the parameter k.

2. Consider the slopes:

   Notice that the slope of the line depends on the value of k. As k varies, the slopes of the lines will also change. This implies that the lines in this family are not parallel.

3. Find a common point:

   To find a point that lies on all lines, we need to find a point where the slope is undefined (or infinite). This occurs when the denominator of the slope term is zero:

   k + 2 = 0 k = -2

   Substitute k = -2 back into the original equation:

   (-2 + 1)x + (-2 + 2)y + 5 = 0 -x + 5 = 0 x = 5

4. Determine the y-coordinate:

   Now that we have x = 5, we can substitute it back into the original equation (with any value of k) to find the corresponding y value. Let's use k = 0 for simplicity:

   (0 + 1) * 5 + (0 + 2)y + 5 = 0 5 + 2y + 5 = 0 2y = -10 y = -5

5. The point of concurrency:

   Therefore, the point of concurrency for the family of lines represented by (k+1)x + (k+2)y + 5 = 0 is (5, -5). This point lies on all lines in the family, regardless of the value of k.

In summary, by rearranging the equation and analyzing the slopes, we found that the lines represented by this equation have a point of concurrency at (5, -5).