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?
ok, here are hints.
1. Given a line l, get a point C(a,b)..
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.
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 = | / [sqrt(9/16 +1)]|
d = | / [sqrt(25/16)]|
d = 3 / (5/4)
d = 3 * (4/5)
d = 12/5 units. -------------same as above.
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:
I'll do this one for you:
Alternatively, you could have simply replaced in slope-intercept form:
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.