Question

वह अनुपात ज्ञात कीजिए जिसमें रेखाखंड को मिलाने वाले बिंदु (-3,10)और (6,-8) को
(-1,6) के द्वारा विभाजित किया जाता है. ​

Answer 1

Answer:

2:7

Step-by-step explanation:

Coordinates of the ends of given segment AB are :

A(- 3, 10) and B(6, -8)

Write the components of the direct segment :

<($x_{2}$ - $x_{1}$), ($y_{2}$ - $y_{1}$)> = <(6 + 3), (- 8 - 10)> = <(9, - 18)>

Assume the ratio is a:b (but not the ratio of lengthc of AC and CB), then the components of segment AC, where C(-1,6)

<(a/b)×9, (a/b)×(-18)>

- 3 + 9(a/b) = - 1 ..... (1)

10 - 18(a/b) = 6 ..... (2)

Solve (1) and (2) for "a/b"

9(a/b) = 2 ⇒ a/b = 2/9

-18(a/b) = - 4 ⇒ a/b = 2/9

Clearly, that a = 2 and b = 9

Thus, the ratio is 2:9

Finally, the ratio of lengths of AC and CB is 2:7

(9-2 =7)

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It's much easier to do on the graph (see attachment)