A line cuts the x-axis at A(4,0) and the y-axis at B(0,3). Find the perdpendicular distance from O to L where O is the orgin. I took by the pythagors theorem 4² = 2.5² + ans²

2.5 is half of the lenght of the line. Am i correct to do it this way?

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- Aug 20th 2008, 01:41 AMhelloyingpythagorean
A line cuts the x-axis at A(4,0) and the y-axis at B(0,3). Find the perdpendicular distance from O to L where O is the orgin. I took by the pythagors theorem 4² = 2.5² + ans²

2.5 is half of the lenght of the line. Am i correct to do it this way? - Aug 20th 2008, 02:46 AMkalagota
- Aug 20th 2008, 02:54 AMChop Suey
Yes, indeed. Kalagota implied that the point you found the distance from to the origin may not be on the line that contains point (0, 0) and perpendicular to the line in question. Do you get it now?

- Aug 20th 2008, 03:08 AMkalagota
ok, here are hints.

1. Given a line, get a point C(a,b)..**l**

using the point-slope form, you will be able to solve that point C, (i mean the coordinates of C) by solving simultaneously the two equations derived using the points C and the slope of the line AB.. (note that line AB contains C)

2. After solving for the coordinates of C, then you can now solve the distance from O to C. - Aug 20th 2008, 04:03 AMticbol
One way of doing this is by geometry.

Let P be the point on line AB where the line from O intersects AB.

Let u = line segment OP.

Right triangle BOA is a 3-4-5 triangle, or hypotanuse AB = 5 units long.

((That can be shown by the Pythagorean theorem.))

OP is perpendicular to AB

OB is perpendicular to OA

Therefore angle BOP is congruient to angle BAO.

((That is, angles whose sides are perpendicular each to each are congruent.))

In right triangle BPO,

cos (angle BOP) = u/3

In right triangle BOA,

cos(angle BAO) = 4/5

Since angle BOP = angle BAO, then,

cos(angle BOP) = cos(angle BAO)

u/3 = 4/5

u = 3*4/5 = 12/5 units long. --------answer.

----------------------------------------

Another way is by "distance of a point to a line"

The "line" is AB.

Its equation, using the slope-intercept form, or y = mx +b, is

y = (-3/4)x +3

since its slope is -3/4 and its y-intercept is 3.

In Ax +By +C = 0 form, that is

(-3/4)x -y +3 = 0

The "point" is the origin (0,0).

The formula for the distance of a point to a line is

d = |[A(x1) +B(y1) +C] / [sqrt(A^2 +B^2)]|

So, since (x1,y1) = (0,0) here,

d = |[(-3/4)*0 +(1)*0 +3] / [sqrt((-3/4)^2 +1^2)]|

d = |[3] / [sqrt(9/16 +1)]|

d = |[3] / [sqrt(25/16)]|

d = 3 / (5/4)

d = 3 * (4/5)

d = 12/5 units. -------------same as above. - Aug 22nd 2008, 03:44 AMhelloying
- Aug 22nd 2008, 04:49 AMChop Suey
You have two points, (4,0) and (0,3), that are on the line in question. Now, to find the equation of the line, we must first find the slope (denoted by m):

$\displaystyle m = \frac{3-0}{0-4} = -\frac{3}{4}$

Now, to find the equation of the line (let's name it x), plug in the slope and any of the given points in this equation:

$\displaystyle y - y_1 = m(x-x_1)$

I'll do this one for you:

$\displaystyle y - 0 = -\frac{3}{4}(x-4)$

$\displaystyle y = -\frac{3}{4}x + 3$

Alternatively, you could have simply replaced in slope-intercept form:

$\displaystyle y = mx + b$

where b is the y-intercept. But I needed to show you the steps. You have the equation of the line x now. What's next? Find the equation of the line that is perpendicular to line x. We'll name it y.

You have to know that, to get any line y that is perpendicular to line x, line y should have a slope such that if I multiply it with the slope of line x, I get -1.

Now you know how to find the slope of the perpendicular line, and you know one point on that line (don't forget it passes through the origin). Find the equation like I did before.

Once you have the two equations, solve them simultaneously. Find the intersection point of the two lines. Then you know what to do next. ;) - Aug 22nd 2008, 06:16 AMkalagota
chops has given you all the necessary steps..

just feel free to ask what you dont understand.. :)