Question

A $(6, - 3)$ , B $(0, 5)$ and C $(- 2, 1)$ are vertices of $△ A BC$ . Points P $(3, 1)$ and Q $(2, - 1)$ lie on sides AB and AC respectively. Check whether $\frac{A P}{PB} = \frac{A Q}{QC}$ .

Approach

Compute AP = $(3 - 6)^{2} + (1 + 3)^{2} = 5$ , PB = 5, AQ = $25$ , QC = $25$ . Thus $\frac{A P}{PB} = 1$ and $\frac{A Q}{QC} = 1$ . Hence the condition holds.

Step-by-step working

1
Compute AP = 5 using distance formula
0.5 marks
2
Compute PB = 5
0.5 marks
3
Compute AQ = $25$
0.5 marks
4
Compute QC = $25$
0.5 marks
5
Verify $\frac{A P}{PB} = \frac{A Q}{QC} = 1$
1 mark

Concepts used

section formularatiodistance

Approach

Step-by-step working

Concepts used

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