Question 34:

Show that ΔABC, where A(22, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 4) are similar triangles. [CBSE 2017]

Answer:

Disclaimer: Instead of 22 it should be $-$ 2

In ΔABC, the coordinates of the vertices are A(–2, 0), B(2, 0), C(0, 2).

$AB = \sqrt{{(2 + 2)}^{2} + {(0 - 0)}^{2}} = 4 CB = \sqrt{{(0 - 2)}^{2} + {(2 - 0)}^{2}} = \sqrt{8} = 2 \sqrt{2} AC = \sqrt{{(0 + 2)}^{2} + {(2 - 0)}^{2}} = \sqrt{8} = 2 \sqrt{2}$

In ΔPQR, the coordinates of the vertices are P(–4, 0), Q(4, 0), R(0, 4).

$PQ = \sqrt{{(4 + 4)}^{2} + {(0 - 0)}^{2}} = 8 QR = \sqrt{{(0 - 4)}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2} PR = \sqrt{{(0 + 4)}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2}$

Now, for ΔABC and ΔPQR to be similar, the corresponding sides should be proportional.

$So, \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{PR} \Rightarrow \frac{4}{8} = \frac{2 \sqrt{2}}{4 \sqrt{2}} = \frac{2 \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2}$
Thus, ΔABC is similar to ΔPQR.

Page No 320:

Question 1:

(i) Find the coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3.

(ii) Find the coordinates of the point which divides the join of A(−5, 11) and B(4, −7) in the ratio 7 : 2.

Answer:

(i) The end points of AB are A(−1, 7) and B(4, −3).
Therefore, (x₁ = −1, y₁ = 7) and (x₂ = 4, y₂ = −3)
Also, m = 2 and n = 3
Let the required point be P(x, y).
By section formula, we get:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{2 \times 4 + 3 \times (- 1)\}}{2 + 3}, y = \frac{\{2 \times (- 3) + 3 \times 7\}}{2 + 3} \Rightarrow x = \frac{8 - 3}{5}, y = \frac{- 6 + 21}{5} \Rightarrow x = \frac{5}{5}, y = \frac{15}{5} Therefore, x = 1 and y = 3$
Hence, the coordinates of the required point are (1, 3).

(ii) The end points of AB are A(−5, 11) and B(4, −7).
Therefore, (x₁ = −5, y₁ = 11) and (x₂ = 4, y₂ = −7).
Also, m = 7 and n = 2
Let the required point be P(x, y).
By section formula, we have:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{7 \times 4 + 2 \times (- 5)\}}{7 + 2}, y = \frac{\{7 \times (- 7) + 2 \times 11\}}{7 + 2} \Rightarrow x = \frac{28 - 10}{9}, y = \frac{- 49 + 22}{9} \Rightarrow x = \frac{18}{9}, y = - \frac{27}{9} Therefore, x = 2 and y = - 3$
Hence, the required point is P(2, −3).

Page No 320:

Question 2:

Find the coordinates of the points of trisection of the line segment joining the points A(7, –2) and B(1, –5). [CBSE 2017]

Answer:

Consider the figure.
Here points P and Q trisect AB.
Therefore, P divides AB into 1 : 2 and Q divides AB into 2 : 1.
Using section formula, coordinates of P are;
$P (x, y) = (\frac{1 \times 1 + 2 \times 7}{3}, \frac{1 \times - 5 + 2 \times - 2}{3}) P (x, y) = (\frac{15}{3}, \frac{- 9}{3}) = (5, - 3)$
Similarly, coordinates of Q are;
$Q (a, b) = (\frac{2 \times 1 + 1 \times 7}{3}, \frac{2 \times - 5 + 1 \times - 2}{3}) Q (a, b) = (\frac{9}{3}, \frac{- 12}{3}) = (3, - 4)$
Therefore, coordinates of points P and Q are (5, $-$ 3) and (3, $-$ 4) respectively.

The midpoint of AB is $(\frac{- 10 - 2}{2}, \frac{4 + 0}{2}) = P (- 6, 2)$ .
Let k be the ratio in which P divides CD. So
$(- 6, 2) = (\frac{k (- 4) - 9}{k + 1}, \frac{k (y) - 4}{k + 1}) \Rightarrow \frac{k (- 4) - 9}{k + 1} = - 6 and \frac{k (y) - 4}{k + 1} = 2 \Rightarrow k = \frac{3}{2}$
Now, substituting $k = \frac{3}{2}$ in $\frac{k (y) - 4}{k + 1} = 2$ , we get
$\frac{y \times \frac{3}{2} - 4}{\frac{3}{2} + 1} = 2 \Rightarrow \frac{3 y - 8}{5} = 2 \Rightarrow y = \frac{10 + 8}{3} = 6$
Hence, the required ratio is 3 : 2 and y = 6.

Page No 323:

Question 35:

A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is the midpoint of PQ then find the coordinates of P and Q. [CBSE 2017]

Answer:

Suppose the line intersects the y-axis at P(0, y) and the x-axis at Q(x, 0).

If the distance between the points A(4, p) and B(1, 0) is 5, then

(a) p = 4 only
(b) p = −4 only
(c) p = ±4 only
(d) p = 0

Answer:

(c) p = ±4 only
The given points are A(4, p) and B(1, 0) and AB = 5.
Then, $(x_{1} = 4, y_{1} = p) and (x_{2} = 1, y_{2} = 0)$
Therefore,
$A B = 5 \Rightarrow \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = 5 \Rightarrow \sqrt{{(1 - 4)}^{2} + {(0 - p)}^{2}} = 5 \Rightarrow {(- 3)}^{2} + {(- p)}^{2} = 25 \Rightarrow 9 + p^{2} = 25 \Rightarrow p^{2} = 16 \Rightarrow p = \pm \sqrt{16} \Rightarrow p = \pm 4$

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