Find the lengths of a triangle whose two sides lie on the coordinate axes and the other side passes through the point (1,1)
To find the lengths of a triangle whose two sides lie on the coordinate axes and the other side passes through the point (1,1), we need to determine the coordinates of the two points where the triangle intersects the axes.
Let's consider the side of the triangle that passes through the point (1,1). We can find the equation of this line using the point-slope form:
(y - y₁) = m(x - x₁),
where (x₁, y₁) is the given point on the line and m is the slope of the line.
Since the line passes through (1,1), we have:
(y - 1) = m(x - 1).
To find the slope, we need another point on the line. One of the other endpoints of the triangle lies on the x-axis, so its y-coordinate would be 0. Let's call this point (x₂, 0).
Substituting (x₂, 0) into the equation, we get:
(0 - 1) = m(x₂ - 1),
-1 = m(x₂ - 1).
Simplifying, we have:
m = -1/(x₂ - 1).
So, the equation of the line passing through (1,1) and (x₂,0) is:
(y - 1) = -1/(x₂ - 1)(x - 1).
To find the x-coordinate of the point of intersection on the x-axis, we substitute y = 0:
0 - 1 = -1/(x₂ - 1)(x - 1),
-1 = -1/(x₂ - 1)(x - 1).
Cross multiplying, we get:
1 = (x₂ - 1)(x - 1).
Expanding the equation, we have:
1 = x₂x - x - x₂ + 1.
Simplifying further:
x₂x - x - x₂ = 0.
To find the x-coordinate of the point of intersection on the x-axis, we need to solve this quadratic equation.