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Find the distance of the point (3,2) from the straight line whose slope is 5 and is passing through the point of intersection of lines x + 2y = 5 and x – 3y + 5 = 0___

User Rfusca
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Final answer:

The distance of the point (3,2) from the straight line with slope 5 that passes through the intersection of the lines x + 2y = 5 and x - 3y + 5 = 0 can be found by first determining the intersection point, creating the equation of the desired line, and then using the distance formula for a point to a line.

Step-by-step explanation:

The student is asking how to find the distance of a point from a straight line given the line's slope and a point it passes through. First, we need to identify the intersection point of the given lines, then write the equation of the line with a slope of 5 that passes through this intersection point.

Lastly, we'll find the perpendicular distance from the point (3,2) to this line using the distance formula for a point and a line.

To find the intersection point of the lines x + 2y = 5 and x – 3y + 5 = 0, we can solve these equations simultaneously:

  1. x + 2y = 5
  2. x - 3y + 5 = 0

From the second equation, x = 3y - 5.

Substituting this into the first equation gives us 3y - 5 + 2y = 5, leading to y = 2 and x = 1.

The intersection point is (1,2).

The equation of the line with a slope (m) of 5 passing through (1,2) can be written using the point-slope form:

y - 2 = 5(x - 1).

To find the perpendicular distance from the point (3,2) to the line, we'll use the formula
Distance = |Ax + By + C| / √(A² + B²),

where A, B, and C are coefficients from the straight line's general form Ax + By + C = 0.

Converting the line's equation to general form gives us -5x + y - 7 = 0. The distance is thus calculated as |(-5)(3) +

(2) - 7| / √((-5)² + 1²) = |(-15) + 2 - 7| / √(26) = | -20 | / √(26), which simplifies to 20 / √(26) or √26 after rationalizing the denominator.

User Ven Nilson
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