Geometry question! Please help ASAp

## find the distance from point p at (3,-4) to line one that crosses points (-5,2) and (-2, 6)

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# 4 Answers

Hi Kate,

Assuming, you want the shortest distance from the point (p) to the line that goes through the two points (-5,2) and (-2,6)...

first determine the equation of the line that goes through the 2 points.

y=(4/3)x + (26/3)... Using standard techniques for finding a line through 2 points

now determine the point on the line that is closed to the point (p). This would be the point on the line that is perpendicular to the point (p) which will have an equation with a slope of -1/(4/3) = (-3/4) and goes through the point (p). So this equation
is

y=-(3/4)x - (7/4)... again using the same technique from above given 2 points

Now the intersection of the two lines is at (-5,2), using a technique of solving two linear equations. Which happens to be one of the points of the first line that was given.

so the distance between point (p) and the closest point on the line (-5,2) is

d = (6

^{2}+8^{2})^{1/2}=10"Distance from a point to a line" is understood to refer to the distance along a segment perpendicular to the line, with one end in the line, and the other end at the given point.

So, first find the slope of the given line (rise/run), call it "m". Then the perpendicular to this must have slope -1/m.

Take that perpendicular slope, and generate the equation for the perpendicular line, using that slope and the given point.

Now solve for the point where the two lines cross, let's call it (x,y). In standard slope intercept form for the two lines, just set y(1) = y(2) = f1 = f2 where f1 and f2 are the mx + b expressions for each of the respective lines. Solve for x, then for
y.

Now you have 2 points: (3,-4) and (x,y). Use the Pythagorean Theorem to obtain the distance.

O.K.?

The distance between a line ax+by+c=0 and a point (x

_{0},y_{0}) not on the line is given byd=|ax

_{0}+by_{0}+c|/sqrt(a^{2}+b^{2})Find the equation of your line from its slope and y-intercept version and get 4x-3y+26=0, and the point is (3,-4), so that a=4, b=-3, c=26, x

_{0}=3, and y_{0}=-4. Therefore,d=|4*3+(-3)*(-4)+26|/sqrt(4

^{2}+3^{2}) = 50/5=10.First, let's consider the equation of that line.

slope = m = (2 - 6)/[(-5) - (-2)] = (-4)/(-3) = 4/3

Next, let's find the y intercept of the line:

y = mx + b

b = y - mx = y - (4x/3)

Since the point (-2,6) is on the line

b = 6 - (4/3)(-2)

b = (26/3)

so that

y = (4x/3) + (26/3)

When x = 3, y = 4 + (26/3) = (38/3)

If we measure the distance from P (3,-4) along the line x = 3, we get

(38/3) + 4 = (50/3)

Similarly, when y = -4

4x/3 = (-4) - (26/3)

4x = (-12) - (26) = -38

x = (-19/2)

so the distance from P to the line measured along the line y = -4 equals 12.5

The three points (3,-4), (-5, -4) and (-5, 2) define a right triangle the length of whose hypotenuse is the length for which we have been searching.

Length of side 1 = Δy = 2 - (-4) = 6

Length of side 2 = Δx = 3 - (-5) = 8

Finally, the length of the hypotenuse is given by the Pythagorean Theorem:

**L = [(6)**

^{2}+ (8)^{2}]^{½}= 10If we wish to find the distance along a line perpendicular to line 1 that passes through p

y = (-3x/4) + b

(-4) = (-9/4) + b

b = (-4) + (9/4) = -7/4

Line 2: y = [(-3x)/4] - (7/4)

The two lines intersect at (-5,2) (prove to yourself that this is so).