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Write an equation of a line passing

through the point (6-3) and
perpendicular to Jk with ) (-2,7) and
K(6,5

User Aghidini
by
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1 Answer

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Final answer:

To find the equation of a line that is perpendicular to line JK and passes through the point (6,-3), we need to find the slope of line JK and use it to determine the slope of the perpendicular line. The slope of line JK is -1/4, and the slope of the perpendicular line is 4. Using the point-slope form of the equation, we can find the equation of the perpendicular line: y = 4x - 27.

Step-by-step explanation:

To find the equation of a line that is perpendicular to line JK and passes through the point (6,-3), we need to find the slope of line JK and use it to determine the slope of the perpendicular line.

First, let's find the slope of line JK using the formula: m = (y2 - y1) / (x2 - x1).

Substituting the coordinates of J(-2,7) and K(6,5), we get m = (5 - 7) / (6 - (-2)) = -2/8 = -1/4.

Since the slopes of perpendicular lines are negative reciprocals of each other, the slope of the perpendicular line is 4/1 = 4.

Now we have the slope and a point (6,-3), so we can use the point-slope form of the equation: y - y1 = m(x - x1).

Substituting the values, we get y - (-3) = 4(x - 6), which simplifies to y + 3 = 4x - 24.

Finally, rearranging the equation in slope-intercept form, we get y = 4x - 27.

User ZaneDarken
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