Question

Find the co-ordinates of the points which divides the line formed by A (1,5) and B (7,2) into 3 similar parts .

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$\boxed{\begin{array}{c|c}\boxed {bf {command}} & \ boxed {\bf {output}} \\ \sf longmapsto &\longmapsto \\ \sf rightarrowtail &\rightarrowtail \\ \sf leadsto &\leadsto \\ \sf hookrightarrow &\hookrightarrow \\ \sf Rrightarrow &\Rrightarrow \end {array}}$

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use each command with "\"​

Answer 1

suppose those points are P(x,y) & Q(a,b)

now

Therefore point P divides AB internally in ratio 1 ∶ 2

$\longmapsto\: x = \frac{1(7) + 2(1)}{1 + 2} \\ \longmapsto \: x = \frac{9}{3} = 3 \\ \rightarrowtail \: y = \frac{1(2) + 2(5)}{2 + 1} \rightarrowtail \: y = 12 \div 2 = 6 \\ P(x,y) = P(3,6)$

Point Q divides AB internally in ratio 2 ∶ 1

$\Rrightarrow \: a = \frac{2(7) + 1(1)}{3} \\ \Rrightarrow \: a = 5 \\ \Rrightarrow \: b = \frac{2(2) + 1(5)}{3} \\ \Rrightarrow \: b = 3$

$Q(a,b) = Q(5,3)$

Answer 2

There exists two coordinates which divides the line formed by A (1,5) and B (7,2) into three similar parts.

Here,

$\boxed{\begin{array}{c|c} \bf x_{1} & \red1 \\ \bf y_{1} & \red5 \\ \bf x_{2} & \red7 \\ \bf y_{2} & \red1\end {array}}$

Let the first unknown coordinate be P(a,b)

$\rm P = \tt \frac{2(7) + 1(1)}{2 + 1}, \frac{2(2) + 1(5)}{2 + 1} \\ \\ \rm P = \tt\frac{14 + 1}{3}, \frac{4+ 5}{3} \\ \\ \rm P = \tt \bigg(\frac{15}{3} , \frac{9}{3}\bigg) \\ \\ \rm P = \tt(5,3)$

Let the second unknown coordinate be Q(c,d)

$\rm Q = \tt \frac{1(7) + 2(1)}{2 + 1}, \frac{1(2) + 2(5)}{1 + 2} \\ \\ \rm Q = \tt \frac{7 + 2}{3} , \frac{2 + 10}{3} \\ \\\rm Q = \tt \frac{9}{3} , \frac{12}{3} \\ \\ \rm Q = \tt(3 , 4)$

Thus,

P(5,3) and Q(3,4) divides the line formed by A (1,5) and B (7,2) into three similar parts.