Given the points A (1,8),B(4,7) and O (0,0) find the coordinates of the point C at which these two lines intersect?

To find the coordinates of point C where the lines AB and OC intersect, we first need to find the equations of the lines AB and OC.

The equation of the line passing through points A(1,8) and B(4,7) can be found using the slope-intercept form:

Slope (m) = (y2 - y1) / (x2 - x1)
m = (7 - 8) / (4 - 1)
m = -1 / 3

Using point A(1,8) in the slope-intercept form equation:
y - y1 = m(x - x1)
y - 8 = (-1/3)(x - 1)
y = -1/3x + 8 + 1
y = -1/3x + 9

So, the equation of line AB is y = -1/3x + 9.

The equation of the line passing through points O(0,0) and C(x, y) can be found using the slope-intercept form:

Slope (m) = (y2 - y1) / (x2 - x1)
m = (y - 0) / (x - 0)
m = y / x

Using point O(0,0) in the slope-intercept form equation:
y = (y/x)x
y = x

So, the equation of line OC is y = x.

Now, we can find the coordinates of point C by solving the system of equations:
-1/3x + 9 = x

Multiplying by 3 to clear the fraction:
- x + 27 = 3x
27 = 4x
x = 27 / 4
x = 6.75

Substitute x back into the equation y = x:
y = 6.75

Therefore, the coordinates of point C are (6.75, 6.75).

The equation of line passing through the points (1, 8) and (4,7) is
$$y-8=\frac{7-8}{4-1}(x-1)$$

$$\Rightarrow y-8=-1(x-1)$$

$$\Rightarrow y-8=-x+1$$

$$\Rightarrow y=-x+9$$

The equation of line passing through the points (4,7) and (0,0) is

$$y-7=\frac{0-7}{0-4}(x-4)$$

$$\Rightarrow y-7=\frac{7}{4}(x-4)$$

$$\Rightarrow 4y-28=7x-28$$

$$\Rightarrow 4y=7x$$

$$\Rightarrow y=\frac{7}{4}x$$


Now solving the above two equations we obtain:

$$-x+9=\frac{7}{4}x$$

$$-x-\frac{7}{4}x=-9$$

$$\frac{-4x-7x}{4}=-9$$

$$\frac{-11x}{4}=-9$$

$$-11x=-36$$

$$x=\frac{-36}{-11}$$

$$x=\frac{36}{11}$$

Substituting the value of x in either equation 1 or 2 we get,

$$y=-(36/11)+9$$

$$y=-36/11+99/11$$

$$y=63/11$$

Hence,
$$C=(\frac{36}{11},\frac{63}{11})$$