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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.

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Respuesta :

Answer:

a) X would be marked 2 cm away from point A towards the point B

b) X would be marked 1.5 cm away from point P towards the point Q

Step-by-step explanation:

Data provided in the question:

Length of AB = 6 cm

Length of PQ = 6 cm

a) AX : AB 1 : 3

This 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]

or

AX = [tex]\frac{6}{3}\ cm[/tex]

or

AX = 2 cm

Hence,

X would be marked 2 cm away from point A towards the point B

b) PX: XQ 1 :3

This means

[tex]\frac{PX}{XQ}=\frac{1}{3}[/tex]

or

⇒ 3PX = XQ   .............(1)

also,

PX + XQ = PQ

or

PX + XQ = 6 cm

substituting the value of XQ from (1)

PX + 3PX = 6 cm

or

4PX = 6 cm

or

PX = [tex]\frac{6}{4}\ cm[/tex]

or

PX = 1.5 cm

Hence,

X would be marked 1.5 cm away from point P towards the point Q

Answer:

Step-by-step explanation:

Answer:

a) X would be marked 2 cm away from point A towards the point B

b) X would be marked 1.5 cm away from point P towards the point Q

Step-by-step explanation:

Data provided in the question:

Length of AB = 6 cm

Length of PQ = 6 cm

a) AX : AB 1 : 3

This means

on substituting the value of AB, we get

or

AX =

or

AX = 2 cm

Hence,

X would be marked 2 cm away from point A towards the point B

b) PX: XQ 1 :3

This means

or

⇒ 3PX = XQ   .............(1)

also,

PX + XQ = PQ

or

PX + XQ = 6 cm

substituting the value of XQ from (1)

PX + 3PX = 6 cm

or

4PX = 6 cm

or

PX =

or

PX = 1.5 cm

Hence,

X would be marked 1.5 cm away from point P towards the point Q

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