The points L, M and N are such that LMN is a straight line - Edexcel - GCSE Maths - Question 6 - 2022 - Paper 2

Let the length of segment LM be represented as 2x and MN as 3x, where x is a common multiplier.

We can use the coordinates of points L and M to find the distance LM:

$d = ext{LM} = ext{sqrt}( (x_2 - x_1)^2 + (y_2 - y_1)^2 )$

Substituting the coordinates of L and M: $ext{LM} = ext{sqrt}( (4 - (-3))^2 + (9 - 1)^2 ) = ext{sqrt}( (4 + 3)^2 + (9 - 1)^2 )$ $= ext{sqrt}( 7^2 + 8^2 ) = ext{sqrt}( 49 + 64 ) = ext{sqrt}(113)$

Since LM is 2x, we have: $2x = ext{sqrt}(113)$ Thus, x = rac{ ext{sqrt}(113)}{2}

Now, we need to find the coordinates of point N, which divides the segment LM in the ratio 2 : 3. We can use the section formula:

The coordinates of N can be expressed as: $(x_N, y_N) = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right)$ Where m and n are the ratios which are 2 and 3 respectively. Substituting the known values: $(x_N, y_N) = \left(\frac{2 \cdot 4 + 3 \cdot (-3)}{2 + 3}, \frac{2 \cdot 9 + 3 \cdot 1}{2 + 3}\right)$ Simplifying: $(x_N, y_N) = \left(\frac{8 - 9}{5}, \frac{18 + 3}{5}\right) = \left(\frac{-1}{5}, \frac{21}{5}\right)$

Thus, the coordinates of N are: N = \left(\frac{-1}{5}, \frac{21}{5}\right).