Question

The straight line L1 passes through the points with coordinates (4, 6) and (12, 2) The straight line L2 passes through the origin and has gradient -3. The lines L1 and L2 intersect at point P. Find the coordinates of P.

Answer 1

Given:

L1 passes through (4, 6) and (12, 2)

L2 passes through origin (0,0) with gradient -3.

To find:

Coordinates Intersection point P of L1 and L2 = ?

Solution:

We need to first find the equations of L1 and L2 and then solve for x and y to find the coordinates of point P.

Let us use slope point form to find equation L1:

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

Slope of a line:  

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$x_2 = 12\\x_1 = 4\\y_2 = 2\\y_1 = 6$

$\therefore m = \dfrac{2-6}{12-4} =-\dfrac{1}{2}$

$(y-6)=-\dfrac{1}{2}(x-4)\\\Rightarrow 2y-12=-x+4\\\Rightarrow x+2y-16=0 ....... (L1)$

Using slope intercept form for L2:

$y=mx+c$

$y =-3x+c$

Putting (0, 0)

$\Rightarrow c =0$

So, $y=-3x ........ (L2)$

Putting y = -3x in L1:

$x+2(-3x)-16=0\\\Rightarrow -5x=16\\\Rightarrow x =-\dfrac{16}{5}$

$y = -3\times -\dfrac{16}{5}\\\Rightarrow y = \dfrac{48}{5}$

So, coordinates of point P is

$(-\dfrac{16}{5}, \dfrac{48}{5})$