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A) A line passes through the point (2Ė5) and makes an angle of 135° with the positive

direction of the x-axis. Find its equation.
b) Find the equation of the perpendicular and its distance from point 4 (3Ė2) to the line
3x+4y=2=0.

1 Answer

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Final answer:

The equation of the line passing through the point (2, -5) and making an angle of 135° with the positive x-axis is y = x - 7. The equation of the perpendicular line to 3x + 4y = 2 passing through the point (3, -2) is y = 4/3x - 10/3.


Step-by-step explanation:

a) To find the equation of the line passing through the point (2, -5) and making an angle of 135° with the positive x-axis, we can use the slope-intercept form of a line, y = mx + b. The slope of the line can be found using the tangent of the angle, which is tan(135°) = -1. The negative reciprocal of the slope will be the slope of the perpendicular line, which is 1. Therefore, the equation of the line passing through (2, -5) with a slope of 1 is y + 5 = 1(x - 2), or y = x - 7.

b) To find the equation of the perpendicular line to 3x + 4y = 2 passing through the point (3, -2), we first need to find the slope of the given line. Rearranging the equation 3x + 4y = 2 in the slope-intercept form, we get y = -(3/4)x + 1/2. The negative reciprocal of the slope -3/4 will be the slope of the perpendicular line, which is 4/3. Using the point-slope form of a line, y - y₁ = m(x - x₁), we substitute the values (3, -2) and 4/3 for x₁, y₁, and m, respectively, to get the equation y + 2 = 4/3(x - 3), or y = 4/3x - 10/3.


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