Answer:
Let the required ratio be \[k:1\] and point on y-axis be (0, y) \[\therefore \,AP:PB=k:1\] Then, by section formula \[\frac{5-k}{k+1}=0\] \[5-k=0\] \[\Rightarrow k=5\] Hence, required ratio is \[5:1\] \[\therefore y=\frac{(-4)(5)+(-6)(1)}{5+1}\] \[\therefore y=\frac{-26}{6}=-\frac{13}{3}\] Hence, point on y-axis is \[\left( 0,\frac{-13}{3} \right)\]
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