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Find K so that the line through(1,-2) and (k,1) is parallel to 5x+6y=12

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To find the value of k so that the line passing through the points (1, -2) and (k, 1) is parallel to the line 5x + 6y = 12, we need to determine the slope of the given line and then use this slope to find the value of k.

The equation of a line can be written in slope-intercept form as y = mx + b, where m is the slope of the line and b is the y-intercept.

To determine the slope of the given line 5x + 6y = 12, we need to rewrite the equation in slope-intercept form. Subtracting 5x from both sides gives 6y = -5x + 12, and dividing both sides by 6 gives y = (-5/6)x + 2.

Comparing this equation to y = mx + b, we can see that the slope of the given line is -5/6.

Since we want the line passing through (1, -2) and (k, 1) to be parallel to the given line, it should have the same slope of -5/6.

The slope formula is given by (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are the coordinates of two points on the line.

Using the points (1, -2) and (k, 1), we can write the slope formula as (-2 - 1) / (1 - k) = -5/6.

Simplifying this equation, we get -3 / (1 - k) = -5/6.

To solve for k, we can cross-multiply to get -3 * 6 = -5 * (1 - k).

This simplifies to -18 = -5 + 5k.

Adding 5 to both sides, we have -18 + 5 = 5k.

Simplifying further, -13 = 5k.

Finally, dividing both sides by 5, we find that k = -13/5.

Therefore, the value of k that makes the line passing through (1, -2) and (k, 1) parallel to the line 5x + 6y = 12 is k = -13/5.

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User Joebeeson
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