Answer:
M = (17 /2 , 13 / 6)
Explanation:
Let us draw a sketch to understand the situation better.
If the diagonal length segments are in a ratio of 5: 1, then so are corresponding vertical and horizontal segments.
This tells us that the vertical segments L2 and L1 must also be in ratio 5: 1; therefore, we have
[tex]\frac{L_1}{L_2}=\frac{3-My}{M_y-(-2)_{}}=\frac{1}{5}[/tex][tex]\frac{3-My}{M_y+2_{}}=\frac{1}{5}[/tex]Cross multiplication gives
[tex](3-M_y)(5)=(M_y+2_{})(1)[/tex][tex]15-5M_y=M_y+2[/tex]Subtracting 15 from both sides gives
[tex]\begin{gathered} -5M_y=M_y+2-15 \\ -5M_y=M_y-13 \end{gathered}[/tex]subtracting My from both sides gives
[tex]-6M_y=-13[/tex]Finally, dividing both sides by -6 gives
[tex]\boxed{M_y=\frac{13}{6}\text{.}}[/tex]Now for the horizontal x-axis, the lengths k2 and k1 are also in a ratio of 5: 1; therefore,
[tex]\frac{k_1}{k_2}=\frac{10-M_x}{M_x-1}=\frac{1}{5}_{}[/tex]cross multiplication gives
[tex](10-M_x)(5)=(M_x-1)(1)[/tex][tex]50-5M_x=M_x-1[/tex]Adding 1 to both sides gives
[tex]51-5M_x=M_x[/tex]adding 5Mx to both sides gives
[tex]51=6M_x[/tex]dividing both sides by 6 gives
[tex]M_x=\frac{51\div3}{6\div3}=\frac{17}{2}[/tex]Hence, the coordinates of the point are
[tex]\boxed{M=(\frac{17}{2},\frac{13}{6})}[/tex]
