Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 22 - Brief Review Of Cartesian System Of Rectangular Co Ordinates

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Page No 22.12:

Question 1:

If the line segment joining the points P (x₁, y₁) and Q (x₂, y₂) subtends an angle α at the origin O, prove that
OP · OQ cos α = x₁ x₂ + y₁, y₂

Answer:

From the figure,
$O P^{2} = {x_{1}}^{2} + {y_{1}}^{2}$
$O Q^{2} = {x_{2}}^{2} + {y_{2}}^{2}$
$P Q^{2} = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
Using cosine formula in $∆ O P Q$ , we get:
$P Q^{2} = O P^{2} + O Q^{2} - 2 O P \cdot O Q \cos α$
$\Rightarrow {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} = {x_{1}}^{2} + {y_{1}}^{2} + {x_{2}}^{2} + {y_{2}}^{2} - 2 O P \cdot O Q \cos α$
$\Rightarrow {x_{2}}^{2} + {x_{1}}^{2} - 2 x_{1} x_{2} + {y_{2}}^{2} + {y_{1}}^{2} - 2 y_{1} y_{2} = {x_{1}}^{2} + {y_{1}}^{2} + {x_{2}}^{2} + {y_{2}}^{2} - 2 O P \cdot O Q \cos α$
$\Rightarrow - 2 x_{1} x_{2} - 2 y_{1} y_{2} = - 2 O P \cdot O Q \cos α$
$\Rightarrow O P \cdot O Q \cos α = x_{1} x_{2} + y_{1} y_{2}$

Page No 22.13:

Question 2:

The vertices of a triangle ABC are A (0, 0), B (2, −1) and C (9, 2). Find cos B.

Answer:

We know that $\cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$ , where a = BC, b = CA and c = AB are the lengths of the sides of $∆$ ABC.
Thus,
$a = B C = \sqrt{{(2 - 9)}^{2} + {(- 1 - 2)}^{2}} = \sqrt{49 + 9} = \sqrt{58}$
$b = A C = \sqrt{{(0 - 9)}^{2} + {(0 - 2)}^{2}} = \sqrt{81 + 4} = \sqrt{85}$
$c = A B = \sqrt{{(2 - 0)}^{2} + {(- 1 - 0)}^{2}} = \sqrt{4 + 1} = \sqrt{5}$

Using cosine formula in $∆ A B C$ , we get:
$\cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$

$\Rightarrow \cos B = \frac{58 + 5 - 85}{2 \sqrt{58} \times \sqrt{5}} = - \frac{11}{\sqrt{290}}$

Page No 22.13:

Question 3:

Four points A (6, 3), B (−3, 5), C (4, −2) and D (x, 3x) are given in such a way that $\frac{Δ D B C}{Δ A B C} = \frac{1}{2}$ . Find x.

Answer:

We know that the area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})$ is given by:

$Area = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\}$
$∴ Area of ∆ D B C = \frac{1}{2} |- 3 (- 2 - 3 x) + 4 (3 x - 5) + x (5 + 2)|$
$\Rightarrow Area of ∆ D B C = 7 (2 x - 1)$

$∴ Area of ∆ A B C = \frac{1}{2} |6 (5 + 2) - 3 (- 2 - 3) + 4 (3 - 5)|$
$= \frac{49}{2}$

It is given that $\frac{Δ D B C}{Δ A B C} = \frac{1}{2}$ .
$∴ \frac{7 (2 x - 1) \times 2}{49} = \frac{1}{2}$

$∴ x = \frac{11}{8}$

Page No 22.13:

Question 4:

The points A (2, 0), B (9, 1), C (11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

Answer:

The given points are A (2, 0), B (9, 1), C (11, 6) and D (4, 4).
Let us find the length of all the sides of the quadrilateral ABCD.

$A B = \sqrt{{(2 - 9)}^{2} + {(0 - 1)}^{2}} = \sqrt{50} = 5 \sqrt{2}$

$B C = \sqrt{{(11 - 9)}^{2} + {(6 - 1)}^{2}} = \sqrt{29}$

$C D = \sqrt{{(4 - 11)}^{2} + {(4 - 6)}^{2}} = \sqrt{49 + 4} = \sqrt{53}$

$A D = \sqrt{{(4 - 2)}^{2} + {(4 - 0)}^{2}} = \sqrt{4 + 16} = 2 \sqrt{5}$

$∵ A B \neq B C \neq C D \neq A D$ , quadrilateral ABCD is not a rhombus.

Page No 22.13:

Question 5:

Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8).

Answer:

The coordinates of the in-centre of a triangle whose vertices are $A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3})$ are $(\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c})$ , where a = BC, b = AC and c = AB.
Let A(−36, 7), B(20, 7) and C(0, −8) be the coordinates of the vertices of the given triangle.
Now,
$a = B C = \sqrt{{(20 - 0)}^{2} + {(7 + 8)}^{2}} = 25$
$b = A C = \sqrt{{(0 + 36)}^{2} + {(- 8 - 7)}^{2}} = 39$
$c = A B = \sqrt{{(20 + 36)}^{2} + {(7 - 7)}^{2}} = 56$

Thus, the coordinates of the in-centre of the given triangle are:

$(\frac{25 \times (- 36) + 39 \times 20 + 0}{25 + 39 + 56}, \frac{25 \times 7 + 39 \times 7 + 56 (- 8)}{25 + 39 + 56})$

= $(\frac{- 120}{120}, 0)$

= $(- 1, 0)$

Hence, the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8) is $(- 1, 0)$ .

Page No 22.13:

Question 6:

The base of an equilateral triangle with side 2a lies along the y-axis, such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Answer:

Let ABC be an equilateral triangle, where BC = 2a. Let A(x, 0) be the third vertex of $∆$ ABC.

In equilateral triangle ABC,
AB = BC = AC
$\Rightarrow$ ${AB}^{2} {= BC}^{2} {= AC}^{2}$

$\Rightarrow a^{2} + x^{2} = {(2 a)}^{2} (∵ B C = 2 a) \Rightarrow x^{2} = 3 a^{2} \Rightarrow x = \pm \sqrt{3} a$

So, the vertices of the triangle are $(0, - a), (0, a) and (\sqrt{3} a, 0) or (0, - a), (0, a) and (- \sqrt{3} a, 0)$ .

Page No 22.13:

Question 7:

Find the distance between P (x₁, y₁) and Q (x₂, y₂) when (i) PQ is parallel to the y-axis (ii) PQ is parallel to the x-axis.

Answer:

The given points are $P (x_{1}, y_{1}) and Q (x_{2}, y_{2})$ .

Distance between P and Q is:

$P Q = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

(i) When PQ is parallel to the y-axis:

In this case, $x_{1} = x_{2}$ .

$∴ P Q = \sqrt{{(x_{1} - x_{1})}^{2} + {(y_{1} - y_{2})}^{2}} = |y_{1} - y_{2}|$

(ii) When PQ is parallel to the x-axis:

In this case, $y_{1} = y_{2}$ .

$∴ P Q = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{1})}^{2}} = |x_{1} - x_{2}|$

Page No 22.13:

Question 8:

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer:

Let C(x, 0) be a point on the x-axis, which is equidistant from the points A(7, 6) and B(3, 4).

$∴$ AC = BC

$\Rightarrow A C^{2} = B C^{2}$

$\Rightarrow {(7 - x)}^{2} + {(6 - 0)}^{2} = {(3 - x)}^{2} + {(4 - 0)}^{2} \Rightarrow 49 + x^{2} - 14 x + 36 = 9 + x^{2} - 6 x + 16 \Rightarrow 85 - 14 x = 25 - 6 x \Rightarrow 60 = 8 x \Rightarrow \frac{15}{2} = x$

Thus, the point on the x-axis, which is equidistant from the points (7, 6) and (3, 4) is $(\frac{15}{2}, 0)$ .

Page No 22.18:

Question 1:

Find the locus of a point equidistant from the point (2, 4) and the y-axis.

Answer:

Let P(h, k) be the point which is equidistant from the point (2, 4) and the y-axis.
The distance of point P(h, k) from the y-axis is h.

$∴ h = \sqrt{{(h - 2)}^{2} + {(k - 4)}^{2}} \Rightarrow h^{2} - 4 h + 4 + k^{2} - 8 k + 16 = h^{2} \Rightarrow k^{2} - 4 h - 8 k + 20 = 0$

Hence, the locus of (h, k) is $y^{2} - 4 x - 8 y + 20 = 0$ .

Page No 22.18:

Question 2:

Find the equation of the locus of a point which moves such that the ratio of its distances from (2, 0) and (1, 3) is 5 : 4.

Answer:

Let A(2, 0) and B(1, 3) be the given points. Let P (h, k) be a point such that PA:PB = 5:4

$∴ \frac{P A}{P B} = \frac{5}{4} \Rightarrow \frac{\sqrt{{(h - 2)}^{2} + {(k - 0)}^{2}}}{\sqrt{{(h - 1)}^{2} + {(k - 3)}^{2}}} = \frac{5}{4}$

Squaring both sides, we get:

$16 (h^{2} - 4 h + 4 + k^{2}) = 25 (h^{2} - 2 h + 1 + k^{2} - 6 k + 9) \Rightarrow 9 h^{2} + 9 k^{2} + 64 h - 50 h - 150 k - 64 + 250 = 0 \Rightarrow 9 h^{2} + 9 k^{2} + 14 h - 150 k + 186 = 0$

Hence, the locus of (h, k) is $9 x^{2} + 9 y^{2} + 14 x - 150 y + 186 = 0$ .

Page No 22.18:

Question 3:

A point moves so that the difference of its distances from (ae, 0) and (−ae, 0) is 2a. Prove that the equation to its locus is
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , where b² = a² (e² − 1).

Answer:

Let $A (- a e, 0) and B (a e, 0)$ be the given points. Let P(h, k) be a point such that $P A - P B = 2 a$ .

$∴ \sqrt{{(h + a e)}^{2} + {(k - 0)}^{2}} - \sqrt{{(h - a e)}^{2} + {(k - 0)}^{2}} = 2 a \Rightarrow \sqrt{{(h + a e)}^{2} + k^{2}} = 2 a + \sqrt{{(h - a e)}^{2} + k^{2}}$

Squaring both sides, we get:

$h^{2} + a^{2} e^{2} + 2 a e h + k^{2} = 4 a^{2} + h^{2} + a^{2} e^{2} - 2 a e h + k^{2} + 4 a \sqrt{{(h - a e)}^{2} + k^{2}} \Rightarrow a e h = 2 a^{2} - a e h + 2 a \sqrt{{(h - a e)}^{2} + k^{2}} \Rightarrow e h - a = \sqrt{{(h - a e)}^{2} + k^{2}} (∵ a \neq 0)$

Squaring both sides again, we get:

$e^{2} h^{2} + a^{2} - 2 a e h = h^{2} + a^{2} e^{2} - 2 a e h + k^{2} \Rightarrow e^{2} h^{2} + a^{2} = h^{2} + a^{2} e^{2} + k^{2} \Rightarrow a^{2} (e^{2} - 1) = h^{2} (e^{2} - 1) - k^{2} \Rightarrow \frac{h^{2}}{a^{2}} - \frac{k^{2}}{a^{2} (e^{2} - 1)} = 1$

Hence, the locus of (h, k) is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, where b^{2} = a^{2} (e^{2} - 1)$ .

Page No 22.18:

Question 4:

Find the locus of a point such that the sum of its distances from (0, 2) and (0, −2) is 6.

Answer:

Let P(h, k) be a point. Let the given points be $A (0, 2) and B (0, - 2)$ .
According to the given condition,

AP + BP = 6

$\Rightarrow$ $\sqrt{{(h - 0)}^{2} + {(k - 2)}^{2}} + \sqrt{{(h - 0)}^{2} + {(k + 2)}^{2}} = 6$

$\Rightarrow$ $\sqrt{h^{2} + {(k - 2)}^{2}} = 6 - \sqrt{h^{2} + {(k + 2)}^{2}}$

Squaring both sides, we get:

$\Rightarrow$ $h^{2} + {(k - 2)}^{2} = 36 + h^{2} + {(k + 2)}^{2} - 12 \sqrt{h^{2} + {(k + 2)}^{2}}$

$\Rightarrow$ $h^{2} + k^{2} + 4 - 4 k = 36 + h^{2} + k^{2} + 4 + 4 k - 12 \sqrt{h^{2} + {(k + 2)}^{2}}$

$\Rightarrow$ $3 \sqrt{h^{2} + {(k + 2)}^{2}} = 9 + 2 k$

$\Rightarrow$ $9 (h^{2} + k^{2} + 4 + 4 k) = 81 + 4 k^{2} + 36 k$ (Squaring both sides)

$\Rightarrow 9 h^{2} + 9 k^{2} + 36 + 36 k = 81 + 4 k^{2} + 36 k$

$\Rightarrow$ $9 h^{2} + 5 k^{2} - 45 = 0$

Hence, the locus of (h, k) is $9 x^{2} + 5 y^{2} - 45 = 0$ .

Page No 22.18:

Question 5:

Find the locus of a point which is equidistant from (1, 3) and the x-axis.

Answer:

Let P(h, k) be a point that is equidistant from A(1, 3) and the x-axis.

Now, the distance of the point P(h, k) from the x-axis is k.
$∴$ AP = k

$\Rightarrow A P^{2} = k^{2}$

$\Rightarrow {(h - 1)}^{2} + {(k - 3)}^{2} = k^{2} \Rightarrow h^{2} - 2 h + 1 + k^{2} - 6 k + 9 = k^{2} \Rightarrow h^{2} - 2 h - 6 k + 10 = 0$

Hence, the locus of (h, k) is $x^{2} - 2 x - 6 y + 10 = 0$ .

Page No 22.18:

Question 6:

Find the locus of a point which moves such that its distance from the origin is three times its distance from the x-axis.

Answer:

Let P(h, k) be a point. Let O(0, 0) be the origin.
So, the distance of point P(h, k) from the x-axis is k.

$∴ O P = 3 k$

$\Rightarrow O P^{2} = {(3 k)}^{2}$

$\Rightarrow {(h - 0)}^{2} + {(k - 0)}^{2} = {(3 k)}^{2} \Rightarrow h^{2} + k^{2} = 9 k^{2} \Rightarrow h^{2} = 8 k^{2}$

Hence, the locus of (h, k) is $x^{2} = 8 y^{2}$ .

Page No 22.18:

Question 7:

A (5, 3), B (3, −2) are two fixed points; find the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 units.

Answer:

Let P(h, k) be a point. Let the given points be A(5, 3) and B(3, $-$ 2).

$∴ Area of ∆ A B P = \frac{1}{2} |\{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\}| \Rightarrow 9 = \frac{1}{2} |\{5 (- 2 - k) + 3 (k - 3) + h (3 + 2)\}| \Rightarrow |5 h - 2 k - 19| = 18 \Rightarrow 5 h - 2 k - 19 = 18 or 5 h - 2 k - 19 = - 18 \Rightarrow 5 h - 2 k - 37 = 0 or 5 h - 2 k - 1 = 0$

Hence, the locus of (h, k) is $5 x - 2 y - 37 = 0 or 5 x - 2 y - 1 = 0$ .

Page No 22.18:

Question 8:

Find the locus of a point such that the line segments with end points (2, 0) and (−2, 0) subtend a right angle at that point.

Answer:

Let the given points be $A (2, 0) and B (- 2, 0)$ . Let P(h, k) be a point such that $∠ A P B = 90^{\circ}$ .

Thus, $∆$ APB is a right angled triangle.

$∴ A B^{2} = A P^{2} + B P^{2}$

$∴ {(2 + 2)}^{2} + 0 = {(h - 2)}^{2} + k^{2} + {(h + 2)}^{2} + k^{2} \Rightarrow 16 = h^{2} + 4 - 4 h + k^{2} + h^{2} + 4 + 4 h + k^{2} \Rightarrow h^{2} + k^{2} = 4$

Hence, the locus of (h, k) is x²+ y² = 4.

Page No 22.18:

Question 9:

If A (−1, 1) and B (2, 3) are two fixed points, find the locus of a point P, so that the area of ∆PAB = 8 sq. units.

Answer:

Let the coordinates of P be (h, k).
Let the given points be $A (- 1, 1) and B (2, 3)$ .

$∴ Area of ∆ P A B = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| \Rightarrow 8 \times 2 = |- 1 (3 - k) + 2 (k - 1) + h (1 - 3)| \Rightarrow 16 = |- 3 + k + 2 k - 2 - 2 h| \Rightarrow 16 = |2 h - 3 k + 5| \Rightarrow 2 h - 3 k + 5 = 16 or 2 h - 3 k + 5 = - 16 \Rightarrow 2 h - 3 k - 11 = 0 or 2 h - 3 k + 21 = 0$

Hence, the locus of (h, k) is $2 x - 3 y - 11 = 0 or 2 x - 3 y + 21 = 0$ .

Page No 22.18:

Question 10:

A rod of length l slides between two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.

Answer:

Let the two perpendicular lines be the coordinate axes. Let AB be a rod of length l and the coordinates of A and B be (a, 0) and (0, b) respectively.
As the rod AB slides, the values of a and b change. Let P(h, k) be a point on AB.

Here, BP:AP = 1:2 .

$∴ h = \frac{a + 0}{3}, k = \frac{0 + 2 b}{3} \Rightarrow a = 3 h, b = \frac{3 k}{2}$ ... (1)

The length of the given rod is l.

$∴ A B = l \Rightarrow \sqrt{a^{2} + b^{2}} = l \Rightarrow a^{2} + b^{2} = l^{2}$

Using equation (1), we get:

$\Rightarrow 9 h^{2} + {(\frac{3 k}{2})}^{2} = l^{2} \Rightarrow h^{2} + \frac{k^{2}}{4} = \frac{l^{2}}{9}$

Hence, the locus of (h, k) is $x^{2} + \frac{y^{2}}{4} = \frac{l^{2}}{9}$ .

Page No 22.18:

Question 11:

Find the locus of the mid-point of the portion of the line x cos α + y sin α = p which is intercepted between the axes.

Answer:

The given line is $x \cos α + y \sin α = p$ .
We need to find the intersection of the above line with the coordinate axes.
Let us put x = 0, and y = 0, respectively.
Thus,
at x = 0, $0 + y \sin α = p \Rightarrow y = p cosec α$

at y = 0, $x \cos α + 0 = p \Rightarrow x = p s ec α$

So, the points on the axes are $A (p \sec α, 0) and B (0, p cosec α)$ .

Let P(h, k) be the mid-point of the line AB.

$∴ h = \frac{p \sec α + 0}{2} and k = \frac{0 + p cosec α}{2} \Rightarrow \cos α = \frac{p}{2 h} and \sin α = \frac{p}{2 k}$

We know that $\sin^{2} α + \cos^{2} α = 1$ .

$∴ {(\frac{p}{2 h})}^{2} + {(\frac{p}{2 k})}^{2} = 1 \Rightarrow \frac{1}{h^{2}} + \frac{1}{k^{2}} = \frac{4}{p^{2}}$

Hence, the locus of (h, k) is $\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{4}{p^{2}}$ .

Page No 22.18:

Question 12:

If O is the origin and Q is a variable point on y² = x, find the locus of the mid-point of OQ.

Answer:

Let the coordinates of Q be (a, b), which lies on the parabola $y^{2} = x$ .

$\Rightarrow b^{2} = a$ ... (1)

Let P(h, k) be the mid-point of OQ.

Now,

$h = \frac{0 + a}{2} and k = \frac{0 + b}{2} \Rightarrow a = 2 h and b = 2 k$

Putting a = 2h and b = 2k in equation (1), we get:

${(2 k)}^{2} = 2 h \Rightarrow 2 k^{2} = h$

Hence, the locus of the mid-point of OQ is $2 y^{2} = x$ .

Page No 22.21:

Question 1:

What does the equation (x − a)² + (y − b)² = r² become when the axes are transferred to parallel axes through the point (a − c, b)?

Answer:

Substituting $x = X + a - c, y = Y + b$ in the given equation, we get:
${(X + a - c - a)}^{2} + {(Y + b - b)}^{2} = r^{2} \Rightarrow {(X - c)}^{2} + Y^{2} = r^{2} \Rightarrow X^{2} + Y^{2} - 2 c X = r^{2} - c^{2}$

Hence, the transformed equation is $X^{2} + Y^{2} - 2 c X = r^{2} - c^{2}$ .

Page No 22.21:

Question 2:

What does the equation (a − b) (x² + y²) −2abx = 0 become if the origin is shifted to the point $(\frac{a b}{a - b}, 0)$ without rotation?

Answer:

Substituting $x = X + \frac{a b}{a - b}, y = Y + 0$ in the given equation, we get:

$(a - b) [{(X + \frac{a b}{a - b})}^{2} + Y^{2}] - 2 a b \times (X + \frac{a b}{a - b}) = 0 \Rightarrow (a - b) (X^{2} + \frac{a^{2} b^{2}}{{(a - b)}^{2}} + \frac{2 a b X}{a - b} + Y^{2}) - 2 a b X - \frac{2 a^{2} b^{2}}{a - b} = 0 \Rightarrow (a - b) (X^{2} + Y^{2}) + \frac{a^{2} b^{2}}{a - b} + 2 a b X - 2 a b X - \frac{2 a^{2} b^{2}}{a - b} = 0 \Rightarrow (a - b) (X^{2} + Y^{2}) - \frac{a^{2} b^{2}}{a - b} = 0 \Rightarrow {(a - b)}^{2} (X^{2} + Y^{2}) = a^{2} b^{2}$

Hence, the transformed equation is ${(a - b)}^{2} (X^{2} + Y^{2}) = a^{2} b^{2}$ .

Page No 22.21:

Question 3:

Find what the following equations become when the origin is shifted to the point (1, 1).
(i) x² + xy − 3x − y + 2 = 0
(ii) x² − y² − 2x +2y = 0
(iii) xy − x − y + 1 = 0
(iv) xy − y² − x + y = 0

Answer:

(i) Substituting $x = X + 1, y = Y + 1$ in the given equation, we get:

${(X + 1)}^{2} + (X + 1) (Y + 1) - 3 (X + 1) - (Y + 1) + 2 = 0 \Rightarrow X^{2} + 2 X + 1 + X Y + X + Y + 1 - 3 X - 3 - Y - 1 + 2 = 0 \Rightarrow X^{2} + X Y = 0$

Hence, the transformed equation is $x^{2} + x y = 0$ .

(ii) Substituting $x = X + 1, y = Y + 1$ in the given equation, we get:

${(X + 1)}^{2} - {(Y + 1)}^{2} - 2 (X + 1) + 2 (Y + 1) = 0 \Rightarrow X^{2} + 2 X + 1 - Y^{2} - 2 Y - 1 - 2 X - 2 + 2 Y + 2 = 0 \Rightarrow X^{2} - Y^{2} = 0$

Hence, the transformed equation is $x^{2} - y^{2} = 0$ .

(iii) Substituting $x = X + 1, y = Y + 1$ in the given equation, we get:

$(X + 1) (Y + 1) - (X + 1) - (Y + 1) + 1 = 0 \Rightarrow X Y + X + Y + 1 - X - 1 - Y - 1 + 1 \Rightarrow X Y = 0$

Hence, the transformed equation is xy = 0.

(iv) Substituting $x = X + 1, y = Y + 1$ in the given equation, we get:

$(X + 1) (Y + 1) - {(Y + 1)}^{2} - (X + 1) + (Y + 1) = 0 \Rightarrow X Y + X + Y + 1 - Y^{2} - 1 - 2 Y - X - 1 + Y + 1 = 0 \Rightarrow X Y - Y^{2} = 0$

Hence, the transformed equation is $x y - y^{2} = 0$ .

Page No 22.21:

Question 4:

To what point should the origin be shifted so that the equation x² + xy − 3x − y + 2 = 0 does not contain any first degree term and constant term?

Answer:

Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.

Substituting x = X + h and y = Y + k in the equation x² + xy − 3x − y + 2 = 0, we get:

${(X + h)}^{2} + (X + h) (Y + k) - 3 (X + h) - (Y + k) + 2 = 0 \Rightarrow X^{2} + 2 h X + h^{2} + X Y + k X + h Y + h k - 3 X - 3 h - Y - k + 2 = 0 \Rightarrow X^{2} + X Y + X (2 h + k - 3) + Y (h - 1) + h^{2} + h k - 3 h - k + 2 = 0$

For this equation to be free from the first-degree terms and constant term, we must have

$2 h + k - 3 = 0, h - 1 = 0, h^{2} + h k - 3 h - k + 2 = 0 \Rightarrow h = 1, k = 1, h^{2} + h k - 3 k - h + 2 = 0$

Also, h =1 and k = 1 satisfy the equation $h^{2} + h k - 3 k - h + 2 = 0$ .

Hence, the origin should be shifted to the point (1, 1).

Page No 22.21:

Question 5:

Verify that the area of the triangle with vertices (2, 3), (5, 7) and (− 3 − 1) remains invariant under the translation of axes when the origin is shifted to the point (−1, 3).

Answer:

Let A(2, 3), B(5, 7) and C(− 3 − 1) represent the vertices of the triangle.

$∴ Area of ∆ A B C = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| = \frac{1}{2} |2 (7 + 1) + 5 (- 1 - 3) - 3 (3 - 7)| = \frac{1}{2} |16 - 20 + 12| = 4$

Since the origin is shifted to the point (−1, 3), the vertices of the $∆$ ABC will be

$A^{'} (2 + 1, 3 - 3), B^{'} (5 + 1, 7 - 3), and C^{'} (- 3 + 1, - 1 - 3) or A^{'} (3, 0), B^{'} (6, 4), and C^{'} (- 2, - 4)$

Now, area of $∆$ A'B'C' :

$\frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| = \frac{1}{2} |3 (4 + 4) + 6 (- 4 - 0) - 2 (0 - 4)| = 4$

Hence, area of the triangle remains invariant.

Page No 22.21:

Question 6:

Find what the following equations become when the origin is shifted to the point (1, 1).
(i) x² + xy − 3y² − y + 2 = 0
(ii) xy − y² − x + y = 0
(iii) xy − x − y + 1 = 0
(iv) x² − y² − 2x + 2y = 0

Answer:

(i) The given equation is x² + xy − 3y² − y + 2 = 0.

Substituting $x = X + 1, y = Y + 1$ in the given equation, we get:

${(X + 1)}^{2} + (X + 1) (Y + 1) - 3 {(Y + 1)}^{2} - (Y + 1) + 2 = 0 \Rightarrow X^{2} + 1 + 2 X + X Y + X + Y + 1 - 3 Y^{2} - 3 - 6 Y - Y - 1 + 2 = 0 \Rightarrow X^{2} + X Y - 3 Y^{2} + 3 X - 6 Y = 0$

Hence, the transformed equation is $x^{2} + x y - 3 y^{2} + 3 x - 6 y = 0$ .

(ii) The given equation is xy − y² − x + y = 0.

Substituting $x = X + 1, y = Y + 1$ in the given equation, we get:

$(X + 1) (Y + 1) - {(Y + 1)}^{2} - (X + 1) + (Y + 1) = 0 \Rightarrow X Y + X + Y + 1 - Y^{2} - 2 Y - 1 - X - 1 + Y + 1 = 0 \Rightarrow X Y - Y^{2} = 0$

Hence, the transformed equation is $x y - y^{2} = 0$ .

(iii) The given equation is xy − x − y + 1 = 0.

Substituting $x = X + 1, y = Y + 1$ in the given equation, we get:

$(X + 1) (Y + 1) - (X + 1) - (Y + 1) + 1 = 0 \Rightarrow X Y + X + Y + 1 - X - 1 - Y - 1 + 1 = 0 \Rightarrow X Y = 0$

Hence, the transformed equation is $x y = 0$ .

(iv) The given equation is x² − y² − 2x + 2y = 0.

Substituting $x = X + 1, y = Y + 1$ in the given equation, we get:

${(X + 1)}^{2} - {(Y + 1)}^{2} - 2 (X + 1) + 2 (Y + 1) = 0 \Rightarrow X^{2} + 2 X + 1 - Y^{2} - 2 Y - 1 - 2 X - 2 + 2 Y + 2 = 0 \Rightarrow X^{2} - Y^{2} = 0$

Hence, the transformed equation is $x^{2} - y^{2} = 0$ .

Page No 22.21:

Question 7:

Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first degree terms:
(i) y² + x² − 4x − 8y + 3 = 0
(ii) x² + y² − 5x + 2y − 5 = 0
(iii) x² − 12x + 4 = 0

Answer:

Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.

(i) Substituting x = X + h and y = Y + k in the equation y² + x² − 4x − 8y + 3 = 0, we get:

${(Y + k)}^{2} + {(X + h)}^{2} - 4 (X + h) - 8 (Y + k) + 3 = 0 \Rightarrow Y^{2} + 2 k Y + k^{2} + X^{2} + 2 h X + h^{2} - 4 X - 4 h - 8 Y - 8 k + 3 = 0 \Rightarrow X^{2} + Y^{2} + X (2 h - 4) + Y (2 k - 8) + k^{2} + h^{2} - 4 h - 8 k + 3 = 0$

For this equation to be free from the terms containing X and Y, we must have

$2 h - 4 = 0, 2 k - 8 = 0 \Rightarrow h = 2, k = 4$

Hence, the origin should be shifted to the point (2, 4).

(ii) Substituting x = X + h and y = Y + k in the equation x² + y² − 5x + 2y − 5 = 0, we get:

${(X + h)}^{2} + {(Y + k)}^{2} - 5 (X + h) + 2 (Y + k) - 5 = 0 \Rightarrow X^{2} + 2 h X + h^{2} + Y^{2} + 2 k Y + k^{2} - 5 X - 5 h + 2 Y + 2 k - 5 = 0 \Rightarrow X^{2} + Y^{2} + X (2 h - 5) + Y (2 k + 2) + k^{2} + h^{2} - 5 h + 2 k - 5 = 0$

For this equation to be free from the terms containing X and Y, we must have

$2 h - 5 = 0, 2 k + 2 = 0 \Rightarrow h = \frac{5}{2}, k = - 1$

Hence, the origin should be shifted to the point $(\frac{5}{2}, - 1)$ .

(iii) Substituting x = X + h and y = Y + k in the equation x² − 12x + 4 = 0, we get:

${(X + h)}^{2} - 12 (X + h) + 4 = 0 \Rightarrow X^{2} + 2 h X + h^{2} - 12 X - 12 h + 4 = 0 \Rightarrow X^{2} + X (2 h - 12) + h^{2} - 12 h + 4 = 0$

For this equation to be free from the terms containing X and Y, we must have

$2 h - 12 \Rightarrow h = 6$

Hence, the origin should be shifted to the point $(6, k), k \in R$ .

Page No 22.21:

Question 8:

Verify that the area of the triangle with vertices (4, 6), (7, 10) and (1, −2) remains invariant under the translation of axes when the origin is shifted to the point (−2, 1).

Answer:

Let the vertices of the given triangle be A(4, 6), B(7, 10) and C(1,− 2).

$∴ Area of triangle A B C = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| \Rightarrow Area of triangle A B C = \frac{1}{2} |4 (10 + 2) + 7 (- 2 - 6) + 1 (6 - 10)| \Rightarrow Area of triangle A B C = \frac{1}{2} |48 - 56 - 4| = 6$

As the origin is shifted to the point (−2, 1), the vertices of the triangle ABC will be

$A^{'} (4 + 2, 6 - 1), B^{'} (7 + 2, 10 - 1) and C^{'} (1 + 2, - 2 - 1) or A^{'} (6, 5), B^{'} (9, 9) and C^{'} (3, - 3)$

Now, area of triangle A'B'C':

$\frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| = \frac{1}{2} |6 (9 + 3) + 9 (- 3 - 5) + 3 (5 - 9)| = 6$

So, in both the cases, the area of the triangle is 6 sq. units.

Hence, area of the triangle remains invariant.

Page No 22.21:

Question 1:

If the points (a cosα, a sinα) and (a cosβ, a sinβ) are at a distance k $\sin (\frac{α - β}{2})$ apart, then k = __________.

Answer:

Given (a cos α, a sin α) and (a cos β, a sin β) are at a distance k $\sin (\frac{α - β}{2})$ a part
Using distance formula,

$\sqrt{{(a \cos β - a \cos α)}^{2} + {(a \sin β - a \sin α)}^{2}} = \sqrt{a^{2} {(\cos β - \cos α)}^{2} + a^{2} {(\sin β - \sin α)}^{2}} = a \sqrt{\cos^{2} β + \cos^{2} α - 2 \cos α \cos β + \sin^{2} α + \sin^{2} β - 2 \sin α \sin β} = a \sqrt{1 + 1 - 2 \cos α \cos β - 2 \sin α \sin β} = a \sqrt{2 (1 - (\cos α \cos β - \sin α \sin β))} = \sqrt{2} a \sqrt{1 - \cos (α - β)} = \sqrt{2} a \sqrt{2 \sin^{2} (\frac{α - β}{2})} = 2 a \sin (\frac{α - β}{2}) i . e K = 2 a$

Page No 22.21:

Question 2:

If two vertices of a triangle are (6, 4), (2, 6) and its centroid is (4, 6), then the coordinates of its third vertex are _________.

Answer:

Let us suppose the co-ordinates of third vertex is given by (x, y)
then centroid = $\frac{sum of respective co - ordinates}{3}$
i.e for (6, 4), (2, 6) and (x, y) centroid is (4, 6)

$i . e 4 = \frac{6 + 2 + x}{3}$
i.e 12 = 8 + x
i.e x = 4 and 6 = $\frac{4 + 6 + y}{3}$
i.e 18 = 10 + y
i.e y = 8
∴ Co-ordinate of third vertex is (4, 8)

Page No 22.21:

Question 3:

The coordinates of the orthocentre of the triangle with vertices $(2, \frac{\sqrt{3} - 1}{2}), (\frac{1}{2}, - \frac{1}{2})$ and $(2, - \frac{1}{2})$ are ________.

Answer:

Given vertices are
$(2, \frac{\sqrt{3} - 1}{2}), (\frac{1}{2}, - \frac{1}{2})$ and $(2, - \frac{1}{2})$

Here slope of BC = $\frac{- \frac{1}{2} + \frac{1}{2}}{2 - \frac{1}{2}} = 0$
Slope of AC = $\frac{\frac{\sqrt{3} - 1}{2} + \frac{1}{2}}{2 - 2} = \infty$
i.e Slope of BC is O and slope of AC is ∞
⇒ BC and AC are perpendicular to each other at angle C
i.e ∠C = 90^o
Since orthocentre of right angled triangle is at C point
∴ Co-ordinate of orthocentre at $(2, - \frac{1}{2})$

Page No 22.22:

Question 4:

The coordinates of the in centre of the triangle whose vertices are (0, 0), (4, 0) and (0, 3) are __________.

Answer:

In a triangle with vertices (0, 0), (4, 0) and (0, 3)

Since co-ordinate of the in-centre of the triangle ABC are
$(\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}) where a = B C b = A C c = A B$
$Here a = B C = \sqrt{{(4 - 0)}^{2} + {(0 - 0)}^{2}} = 4 b = A C = \sqrt{4^{2} + 9} = \sqrt{25} = 5 c = A B = \sqrt{0 + 9} = 3$
∴ Co-ordinates of in-centre of the triangle ABC
$= \{(\frac{4 \times 0 + 5 \times 0 + 3 \times 4}{12}), (\frac{4 \times 3 + 5 \times 0 + 3 \times 0}{12})\} = \{(\frac{12}{12}, \frac{12}{12})\}$
i.e co-ordinates of in-centre = (1, 1)

Page No 22.22:

Question 5:

A(–3, 0), B(4,–1) and C(5, 2) are vertices of ΔABC. The length of altitude through A is __________.

Answer:

In ΔABC, Let D be such that AD defines altitude of ΔABC.
A = (–3, 0), B = (4, –1) and C = (5, 2)

$Area of ∆ A B C = \frac{1}{2} |\begin{matrix} - 3 & 0 & 1 \\ 4 & - 1 & 1 \\ 5 & 2 & 1 \end{matrix}|$ applying R₂ → R₂ – R₁ and R₃ → R₃ – R_{1

$i . e \frac{1}{2} |\begin{matrix} - 3 & 0 & 1 \\ 7 & - 1 & 0 \\ 8 & 2 & 0 \end{matrix}| i . e \frac{1}{2} (14 + 8) = \frac{22}{2} = 11$}
Since area of ΔABC = $\frac{1}{2}$ × Base × Altitude

$Here Base = B C = \sqrt{{(5 - 4)}^{2} + {(2 + 1)}^{2}} = \sqrt{1 + 9} = \sqrt{10} \Rightarrow 11 = \frac{1}{2} \times \sqrt{10} \times A D i . e \frac{22}{\sqrt{10}} is the length of altitude A D .$

Page No 22.22:

Question 6:

A(a, 1), B(b, 3) and C(4, c) are the vertices of ΔABC. If its centroid lies on x–axis, then __________.

Answer:

In ΔABC,
A = (a, 1)
B = (b, 3)
C = (4, c)
Since centroid lies an x-axis
⇒ y-cordinate of centroid is 0

i.e 0 = $\frac{1 + 3 + c}{3}$    (since for A(x₁, y₁)  B(x₂, y₂)  C(x₃, y₃)
i.e c = –4            (centroid is given by $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

Page No 22.22:

Question 7:

The distance between the circumcentre and centroid of a triangle whose vertices are (6, 0), (0, 6) and (6, 6), is _______.

Answer:

In ∆ABC
A = (6, 0)
B = (0, 6)
C = (6, 6)

Clearly ∆ABC is a right-angled triangle at C and in right-angled triangle,
Circum-centre is mid point of hypotenuse
Since AB is hypotenuse and mid point of AB is (3, 3) $[(\frac{6 + 0}{2}, \frac{0 + 6}{2})]$
Hence circumcentre is (3, 3) and centroid is $(\frac{(6 + 0 + 6)}{3}, \frac{(0 + 6 + 6)}{3})$
i.e (4, 4)
∴ Distance between circumcentre (3, 3) and centroid (4, 4) is
$\begin{array}{rcl} \sqrt{{(4 - 3)}^{2} + {(4 - 3)}^{2}} & = & \sqrt{1 + 1} \\ = & \sqrt{2} \end{array}$

Page No 22.22:

Question 8:

The distance between the orthocentre and circumcentre of the triangle whose vertices are at $(- \frac{1}{2}, \frac{\sqrt{3}}{2}), (- \frac{1}{2}, - \frac{\sqrt{3}}{2})$ and (1, 0),
is __________.

Answer:

In ∆ABC
$A = (\frac{- 1}{2}, \frac{\sqrt{3}}{2}) B = (\frac{- 1}{2}, \frac{- \sqrt{3}}{2}) C = (1, 0) A B = \sqrt{{(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2})}^{2} + {(\frac{- 1}{2} + \frac{1}{2})}^{2}} i . e A B = \sqrt{{(\sqrt{3})}^{2}} = \sqrt{3} B C = \sqrt{{(\frac{\sqrt{3}}{2})}^{2} + {(\frac{3}{2})}^{2}} i . e B C = \sqrt{\frac{3}{4} + \frac{9}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3} and A C = \sqrt{{(\frac{\sqrt{3}}{2})}^{2} + {(1 + \frac{1}{2})}^{2}} = \sqrt{3}$

Since AB = BC = AC
i.e ∆ABC is an equilateral triangle
⇒ Orthocentre and circumcentre of ∆ABC coincide
⇒ Distance between orthocentre and circumcentre of ∆ABC is 0.

Page No 22.22:

Question 1:

The vertices of a triangle are O (0, 0), A (a, 0) and B (0, b). Write the coordinates of its circumcentre.

Answer:

The coordinates of circumcentre of a triangle are the intersection of perpendicular bisectors of any two sides of the triangle.

Thus, the coordinates of circumcentre of triangle OAB are $(\frac{a}{2}, \frac{b}{2})$ , as shown in the figure.

Page No 22.22:

Question 2:

In Q.No. 1, write the distance between the circumcentre and orthocentre of ∆OAB.

Answer:

The coordinates of circumcentre of a triangle are the point of intersection of perpendicular bisectors of any two sides of the triangle.

Thus, the coordinates of the circumcentre of triangle OAB is $(\frac{a}{2}, \frac{b}{2})$ ,as shown in the figure.
We know that the orthocentre of a triangle is the intersection of any two altitudes of the triangle.
So, the orthocentre of triangle OAB is the origin O(0, 0).

$∴$ Distance between the circumcentre and orthocentre of ∆OAB = OC

$\Rightarrow O C = \sqrt{{(\frac{a}{2} - 0)}^{2} + {(\frac{b}{2} - 0)}^{2}} = \frac{\sqrt{a^{2} + b^{2}}}{2}$

Page No 22.22:

Question 3:

Write the coordinates of the orthocentre of the triangle formed by points (8, 0), (4, 6) and (0, 0).

Answer:

The intersection point of three altitudes of a triangle is called orthocentre.

In the figure, two altitudes ON and BM of $∆$ OAB are shown.

Slope of AB = $\frac{6 - 0}{4 - 8} = - \frac{3}{2}$

$∴$ Slope of ON $= \frac{2}{3} (∵ Product of slopes = - 1)$
Equation of ON:

$(y - 0) = \frac{2}{3} (x - 0) y = \frac{2}{3} x$ ... (1)

Equation of BM:

x = 4 ... (2)

On solving equations (1) and (2), we get $(4, \frac{8}{3})$ as the coordinates of the orthocentre.

Page No 22.22:

Question 4:

Three vertices of a parallelogram, taken in order, are (−1, −6), (2, −5) and (7, 2). Write the coordinates of its fourth vertex.

Answer:

Let $A (- 1, - 6), B (2, - 5) and C (7, 2)$ be the vertices of the parallelogram ABCD.
Let the coordinates of D be (x, y).

Since, diagonals of a parallelogram bisect each other,

$\frac{- 1 + 7}{2} = \frac{2 + x}{2} and \frac{- 6 + 2}{2} = \frac{- 5 + y}{2} \Rightarrow x = 4 and y = 1$

Hence, the coordinates of the fourth vertex D are (4, 1).

Page No 22.22:

Question 5:

If the points (a, 0), (at₁², 2at₁) and (at₂², 2at₂) are collinear, write the value of t₁ t₂.

Answer:

For the points (a, 0), (at₁², 2at₁) and (at₂², 2at₂) to be collinear, the following condition has to be met:

$|\begin{matrix} a & 0 & 1 \\ a {t_{1}}^{2} & 2 a t_{1} & 1 \\ a {t_{2}}^{2} & 2 a t_{2} & 1 \end{matrix}| = 0 \Rightarrow a (2 a t_{1} - 2 a t_{2}) - 0 + 1 (2 a^{2} {t_{1}}^{2} t_{2} - 2 a^{2} t_{1} {t_{2}}^{2}) = 0 \Rightarrow 2 a^{2} (t_{1} - t_{2}) + 2 a^{2} t_{1} t_{2} (t_{1} - t_{2}) = 0 \Rightarrow 2 a^{2} (t_{1} - t_{2}) (1 + t_{1} t_{2}) = 0$

$\Rightarrow (t_{1} - t_{2}) = 0 or (1 + t_{1} t_{2}) = 0 (a \neq 0) \Rightarrow 1 + t_{1} t_{2} = 0 (∵ t_{1} \neq t_{2}) \Rightarrow t_{1} t_{2} = - 1$

Page No 22.22:

Question 6:

If the coordinates of the sides AB and AC of ∆ABC are (3, 5) and (−3, −3), respectively, then write the length of side BC.

Answer:

Disclaimer: In the question it should have been the coordinates of the mid points of AB and AC are (3, 5) and ( $-$ 3, $-$ 3)

Given: the coordinates of the midpoints of AB and AC are (3,5) and ( $-$ 3, $-$ 3).
Let, D and E be the midpoints of AB and AC, respectively.

$∴ D E = \sqrt{{[3 - (- 3)]}^{2} + {[5 - (- 3)]}^{2}} = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10 units$

Now, as D and E are midpoints of AB and AC respectively,
by the mid-points theorem,

$B C = 2 \times D E = 2 \times 10 units = 20 units$

Page No 22.22:

Question 7:

Write the coordinates of the circumcentre of a triangle whose centroid and orthocentre are at (3, 3) and (−3, 5), respectively.

Answer:

Let the coordinates of the circumcentre of the triangle be C(x, y).
Let the points O( $-$ 3, 5) and G(3, 3) represent the coordinates of the orthocentre and centroid, respectively.

We know that the centroid of a triangle divides the line joining the orthocentre and circumcentre in the ratio 2:1.

$∴ 3 = \frac{- 3 \times 1 + 2 x}{3} and 3 = \frac{5 \times 1 + 2 y}{3} \Rightarrow x = 6, y = 2$

Hence, the coordinates of the circumcentre is (6, 2).

Page No 22.22:

Question 8:

Write the coordinates of the in-centre of the triangle with vertices at (0, 0), (5, 0) and (0, 12).

Answer:

Let A(0,0), B(5, 0) and C(0, 12) be the vertices of the given triangle.
In-centre I of a triangle with vertices $A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3})$ is given by:

$I \equiv (\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c})$ , where a = BC, b = AC and c = AB.

Now,

$a = B C = \sqrt{{(5 - 0)}^{2} + {(0 - 12)}^{2}} = 13 b = A C = \sqrt{0 + 12^{2}} = 12 c = A B = \sqrt{0 + 5^{2}} = 5$

$∴ I \equiv (\frac{13 \times 0 + 12 \times 5 + 5 \times 0}{13 + 12 + 5}, \frac{13 \times 0 + 12 \times 0 + 5 \times 12}{13 + 12 + 5}) \Rightarrow I \equiv (\frac{60}{30}, \frac{60}{30}) = (2, 2)$

Hence, the coordinates of the in-centre of the triangle with vertices at (0, 0), (5, 0) and (0, 12) is (2, 2).

Page No 22.22:

Question 9:

If the points (1, −1), (2, −1) and (4, −3) are the mid-points of the sides of a triangle, then write the coordinates of its centroid.

Answer:

Let P(1, −1), Q(2, −1) and R(4, −3) be the mid-points of the sides AB, BC and CA, respectively, of $∆$ ABC.
Let $A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3})$ be the vertices of $∆$ ABC.
Since, P is the mid-point of AB,

$\frac{x_{1} + x_{2}}{2} = 1, \frac{y_{1} + y_{2}}{2} = - 1$               ... (1)

Q is the mid-point of BC.

$∴ \frac{x_{2} + x_{3}}{2} = 2, \frac{y_{2} + y_{3}}{2} = - 1$               ... (2)

R is the mid-point of AC.

$∴ \frac{x_{1} + x_{3}}{2} = 4, \frac{y_{1} + y_{3}}{2} = - 3$               ... (3)

Adding equations (1), (2) and (3), we get:

$x_{1} + x_{2} + x_{3} = 1 + 2 + 4 = 7 y_{1} + y_{2} + y_{3} = - 1 - 1 - 3 = - 5$

$∴ Centroid of ∆ A B C = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}) = (\frac{7}{3}, \frac{- 5}{3})$

Hence, the coordinates of the centroid of the triangle is $(\frac{7}{3}, \frac{- 5}{3})$ .

Page No 22.22:

Question 10:

Write the area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a + b).

Answer:

Let A(a, b + c), B(b, c + a) and C(c, a + b) be the vertices of the the given triangle.

$∴ Area of ∆ A B C = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| = \frac{1}{2} |a (c + a - a - b) + b (a + b - b - c) + c (b + c - c - a)| = \frac{1}{2} |a (c - b) + b (a - c) + c (b - a)| = \frac{1}{2} |a c - a b + a b - b c + b c - a c| = 0$

Hence, area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a + b) is 0.

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