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What is the x-intercept of a line which passes through point (2,1) and is perpendicular to a line 3x + 4y + 8 =0?

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Final answer:

To determine the x-intercept of a line perpendicular to 3x + 4y + 8 = 0, which passes through (2,1), we find the negative reciprocal of the original line's slope, use the point-slope form to get the equation of the new line, and then set y to zero to solve for the x-intercept, which is (2.75, 0).

Step-by-step explanation:

To find the x-intercept of a line perpendicular to the given line 3x + 4y + 8 = 0, we first need to determine the slope of the given line. Rearrange the equation into slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept. Dividing by 4, the equation of the given line becomes y = -​​(3/4)x - 2. Therefore, the slope of the given line is -3/4.

Since the line we seek is perpendicular to this one, it will have a slope that is the negative reciprocal of -3/4, which is 4/3. Next, we use the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is a point on the line (in this case (2,1)) and m is the slope (4/3). Substituting these values into the formula gives us y - 1 = (4/3)(x - 2).

To find the x-intercept, we set y to zero and solve for x: 0 - 1 = (4/3)(x - 2); simplifying, x = 2 + (3/4); x = 2.75. Thus, the x-intercept of the line is (2.75, 0).

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