AB and PQ are lines of length 6 cm a) AX: AB 1:3 Mark where point X would be on the line, with a cross. A B b) PX: XQ 1 :3 Mark where point X would be on the line, with a cross.
Accepted Solution
A:
Answer:a) X would be marked 2 cm away from point A towards the point Bb) X would be marked 1.5 cm away from point P towards the point QStep-by-step explanation:Data provided in the question:Length of AB = 6 cmLength of PQ = 6 cma) AX : AB 1 : 3This means[tex]\frac{AX}{AB}=\frac{1}{3}[/tex]on substituting the value of AB, we get[tex]\frac{AX}{6\ cm}=\frac{1}{3}[/tex]orAX = [tex]\frac{6}{3}\ cm[/tex]orAX = 2 cmHence,X would be marked 2 cm away from point A towards the point Bb) PX: XQ 1 :3This means[tex]\frac{PX}{XQ}=\frac{1}{3}[/tex]or⇒ 3PX = XQ .............(1)also,PX + XQ = PQorPX + XQ = 6 cmsubstituting the value of XQ from (1)PX + 3PX = 6 cmor4PX = 6 cmorPX = [tex]\frac{6}{4}\ cm[/tex]orPX = 1.5 cmHence,X would be marked 1.5 cm away from point P towards the point Q