Question

The line segment joining the points A (3, -4) and B (1, 2) is trisected at the points P(p, -2) and Q $(\frac{5}{3},q)$. Find the values of p and q.

Answer 1

We know that a ratio m:n divides with coordinates $\therefore P(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Here trisection points are P(p, -2) and Q ($\frac{5}{3}$,q) and points are A (3, -4) and B (1, 2). Trisection can occur in either 1:2 or 2:1 ratio. But we don't know which point (P or Q) trisects in which ratio.

lets think P divides in m:n ratio and so

$y=\frac{m{y}_2+n{y}_1}{m+n}-2=\frac{2m+(-4)n}{m+n})-2(m+n)=2m-4n-2m-2n=2m-4n-4m=-2n\frac{m}{n}=\frac{2}{4}=\frac{1}{2}$

So P divides in the ratio 1:2

now $p=\frac{m{x}_2+n{x}_1}{m+n}=\frac{1\times 1+2\times 3}{3}=\frac{1+6}{3}=\frac{7}{3}$

now for Q,

$\frac{5}{3}=\frac{m{x}_2+n{x}_1}{m+n}\frac{5}{3}=\frac{m+3n}{m+n}5m+5n=3m+9n2m=4nfracmn=\frac{4}{2}=\frac{2}{1}$

So Q divides in the ratio 2:1

So,

$q=\frac{m{y}_2+n{y}_1}{m+n}=\frac{2\times 2+1\times -4}{3}=\frac{4-4}{3}=0$

value of $p=\frac{7}{3}$ and q = 0