The vertices of a ?ABC are A(5, 5), B(1, 5), and C(9, 1). A line is drawn to intersect AB and AC at P and Q respectively, such that

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Solution

Q.
The vertices of a ΔABC are A(5, 5), B(1, 5), and C(9, 1). A line is drawn to intersect AB and AC at P and Q respectively, such that $\frac{A P}{A B} = \frac{A Q}{A C} = \frac{3}{4}$

Find the length of line segment PQ and coordinates of point P.

(i) Coordinates of P are (2, 5).

$(i i) L e n g t h o f P Q i s 3 \sqrt{5} u n i t s .$

$(i i i) L e n g t h o f P Q i s 6 \sqrt{5} u n i t s .$

(iv) Coordinates of P are (5, 2).

Choose the correct option from the following:

1 (i) and (ii) are correct.

2 (ii) and (iv) are correct

3 (i) and (iii) are correct

4 (iii) and (iv) are correct

Ans.
(1)

We have,

$\frac{A P}{A B} = \frac{A Q}{A C} = \frac{3}{4}$

$\Rightarrow \frac{A P}{A P + P B} = \frac{A Q}{A Q + Q C} = \frac{3}{4}$

$\Rightarrow \frac{A P + P B}{A P} = \frac{4}{3} and \frac{A Q + Q C}{A Q} = \frac{4}{3} (Take reciprocal)$

$\Rightarrow 1 + \frac{P B}{A P} = \frac{4}{3} and 1 + \frac{Q C}{A Q} = \frac{4}{3}$

$\Rightarrow \frac{P B}{A P} = \frac{1}{3} and \frac{Q C}{A Q} = \frac{1}{3}$

Again taking reciprocal, we get:

$\frac{A P}{P B} = \frac{3}{1} and \frac{A Q}{Q C} = \frac{3}{1}$

So, P and Q divide AB and AC, respectively, in the same ratio 3 : 1. Thus, the coordinates of P and Q are

$P (\frac{3 \times 1 + 1 \times 5}{3 + 1}, p \frac{3 \times 5 + 1 \times 5}{3 + 1}) = (\frac{3 + 5}{4}, \frac{15 + 5}{4}) = (2, 5) a n d$

$Q (\frac{3 \times 9 + 1 \times 5}{3 + 1}, \frac{3 \times 1 + 1 \times 5}{3 + 1}) = (\frac{27 + 5}{4}, \frac{3 + 5}{4}) = (8, 2)$

Now, $P Q = \sqrt{(2 - 8)^{2} + (5 - 2)^{2}} = \sqrt{36 + 9} = \sqrt{45} = 3 \sqrt{5} units .$

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