(Each question 4 marks) - MATHS STD 11 Science Questions - Vidyadip

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16 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

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

Answer

We know that,$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{a}\text{c}}$
Where a = BC, b = CA and C = AB are the sides of the triangle ABC.
We have,
$\text{a}=\text{BC}=\sqrt{(9-2)^2+(2+1)^2}=\sqrt{49+9}=\sqrt{58}$
$\text{b}=\text{CA}=\sqrt{(0-9)^2+(0+2)^2}=\sqrt{81+4}=\sqrt{85}$
$\text{and}, \text{ c}=\text{AB}=\sqrt{(2-0)^2+(-1-0)^2}=\sqrt{4+1}=\sqrt{5}$
$\therefore\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{a}\text{c}}$
$=\frac{58+5-85}{2\times\sqrt{58}\times\sqrt{5}}$
$=\frac{63-85}{2\sqrt{290}}$
$=\frac{-22}{2\sqrt{290}}=\frac{-11}{\sqrt{290}}$
Hence, $\cos\text{B}=\frac{-11}{\sqrt{290}}$

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Question 24 Marks

Find the coording of the centre of the circle inscribed in a triangle whose vertics are (-36, 7), (20, 7) and (0, -8).

Answer

Let A(-36, 7), B(20, 7) and c(0, -8) be the vertices of the triangle ABC.Now,
$\text{a}=\text{BC}=\sqrt{(0-20)^2+(-8-7)^2}$
$=\sqrt{400+225}$
$=\sqrt{625}$
$=25$
$\text{and }\text{c }=48=\sqrt{(20+36)^2+(7-7)^2}$
$=\sqrt{(56)^2}$
$=56$
The coordinates of the center of the cricle are,
$\bigg(\frac{\text{a}\text{x}_1+\text{b}\text{x}_2+\text{c}\text{x}_3}{\text{a}+\text{b}+\text{c}},\frac{\text{a}\text{y}_1+\text{b}\text{y}_2+\text{c}\text{y}_3}{\text{a}+\text{b}+\text{c}}\bigg)$
$\text{or }, \bigg[\frac{25\times(-36)+39\times20+56\times0}{25+39+56},\frac{25\times7+39\times7+56\times(-8)}{25+39+56}\bigg]$
$\text{or},\bigg[\frac{-900+780}{120},\frac{175+273-448}{120}\bigg]$
$\text{or},\bigg[\frac{-120}{120},\frac{0}{120}\bigg]$
$\text{or},(-1,0)$
Hence, the coordinates of the centre of the crcle are (-1, 0).

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Question 34 Marks

Find the distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ when,

PQ is parallel to the y-axis.
PQ is parallel to the x-axis.

Answer

It is given that $P(x_1, y_1)$ and $Q(x_2, y_2)$ are two points.

PQ is parallel to the y-axis.

$\therefore\text{ x}_1=\text{x}_2 \ . . .(1)$
$\therefore \text{PQ}=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\bigg|$
$=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\bigg|$ $[\text{using}\text{ equaction }1]$
$=\bigg|\sqrt{(\text{y}_2-\text{y}_1)^2}\bigg|$
$=\big|\text{y}_2-\text{y}_1\big|$

PQ is parallel to the x-axis.

$\therefore\text{ y}_1=\text{y}_2 \ . . .(2)$
$\therefore \text{PQ}=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\bigg|$
$=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\bigg|$ $[\text{using}\text{ equaction }2]$
$=\bigg|\sqrt{(\text{x}_2-\text{x}_1)^2}\bigg|$
$\therefore \text{ PQ}=\big|\text{x}_2-\text{x}_1\big|$

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Question 44 Marks

At What point the origin be shifted so that the equation $x^2+ xy - 3x - y + 2 = 0$ cantain any first degree term and constand term?

Answer

We have, $x^2+x y-3 x-y+2=0$
Let the origin be shifted to $(h, k)$. Then $\mathrm{x}=\mathrm{X}$ and $\mathrm{y}=\mathrm{Y}+\mathrm{k}$.
Substituting $x=X+h, y=Y+k$ in the given equation, we get
$(X+h)^2+(X+h)(Y+k)-3(X+h)-(Y+k)+2=0$
$\Rightarrow X^2+h^2+2 X h+X Y+X k+Y h+h k-3 k-3 h-Y-k+2=0$
$\Rightarrow X^2+X Y+2 \times X k+Y h-Y 3-X+h^2+h k-3 h-k+2=0$
$\Rightarrow X^2+(2 X h+X k-3 X)+X Y+(Y h-Y)+\left(h^2+h k-3 h-k+2\right)=0$
$\Rightarrow X^2+(2 h+k-3) X+X Y+(h-1) Y+\left(h^2+h k-3 h-k+2\right)=0$
From this equation to be free from first degree and the constant term, we must have,
$2 \mathrm{~h}+\mathrm{k}-3=0 . \ldots . . \text { (ii) }$
$\mathrm{h}-1=0$
$\Rightarrow \mathrm{~h}=1 \ldots \ldots \text {. (iii) }$
And $\mathrm{h}^2+\mathrm{hk}-3 \mathrm{~h}-\mathrm{k}+2=0$.
Putting $h=1$ in equation (ii), we get
$2+k-3=0$
$\Rightarrow k=1$
Putting $\mathrm{h}=1$ and $\mathrm{k}=1$ in equation (iv), we get
$(1)^2+1-3-1+2=0 \text {. }$
Hence, the value of h and k satisfies the equation (iv)
$\therefore$ The origin is shifted at the point $(1,1)$.

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Question 54 Marks

A point moves as so that the difference its distances from (ae, 0) and (-ae, 0) is 2 a, prove that the equation to its locus is $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1 $, where $\text{b}^2=\text{a}^2(\text{e}^2-1).$

Answer

Let P(h, k) be any point on the locus and let A(ae, 0) and B(-ae, 0) be the given points. By the given condition$\text{PA}-\text{PB}=2\text{a}$
$\Rightarrow \text{PA}=2\text{a}+\text{PB}$
$\Rightarrow \sqrt{(\text{ae}-\text{h})^2+(0-\text{k})^2}=2\text{a}+\sqrt{(-\text{ae}-\text{h})^2+(0-\text{k})^2}$
$\Rightarrow (\text{ae}-\text{h})^2+\text{k}^2= \big(2\text{a}+\sqrt{(\text{ae}-\text{h}^2)+\text{k}^2}$ [Taking square on both sides]
$\Rightarrow \text{(ae)}^2+\text{h}^2-2\text{aeh}+\text{k}^2 =4\text{a}^2+(\text{a}\text{e}+\text{h})^2\\\ \ \ +\text{k}^2+2\times2\text{a}\times\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow \text{h}^2+\text{k}^2+\text{(ae)}^2-2\text{aeh} =4\text{a}^2+(\text{a}\text{e})^2+\text{h}^2\\\ \ \ +2\text{h}\text{a}\text{e}+\text{k}^2+4\text{a}\times\sqrt{(\text{a}\text{e}^2+\text{h}^2)+\text{k}^2}$
$\Rightarrow -4\text{a}-2\text{a}\text{e}\text{h}-2\text{a}\text{e}\text{h}=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -4\text{a}^2-4\text{a}\text{e}\text{h}=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -4\big[\text{a}^2+\text{a}\text{e}\text{h}\big]=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -\big[\text{a}^2+\text{a}\text{e}\text{h}\big]=\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -\text{a}[\text{a}+\text{e}\text{h}\big]=\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$ $\Rightarrow -[\text{a}+\text{e}\text{h}\big]=\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$ $\Rightarrow (\text{a}+\text{e}\text{h})^2=\bigg(\sqrt{(\text{a}\text{e}+\text{h}^2)+\text{k}^2}\bigg)^2$ [Taking square on both sides] $\Rightarrow \text{a}^2+(\text{e}\text{h})^22\text{h}\text{a}\text{e}=(\text{a}\text{e}+\text{h})^2+\text{k}^2$ $\Rightarrow \text{a}^2+(\text{e}\text{h})^2+2\text{h}\text{a}\text{e}=(\text{a}\text{e})+\text{h}^2+2\text{h}\text{a}\text{e}+\text{k}^2$ $\Rightarrow \text{a}^2+\text{e}^2\text{h}^2=\text{a}^2\text{e}^2+\text{h}^2+\text{k}^2$ $\Rightarrow \text{e}^2\text{h}^2-\text{h}^2-\text{k}^2=\text{a}^2\text{e}^2-\text{a}^2$ $\Rightarrow \text{h}^2(\text{e}^2-1)-\text{k}^2=\text{a}^2(\text{e}^2-1)$ $\Rightarrow \frac{\text{h}^2(\text{e}^2-1)}{\text{a}^2(\text{e}^2-1)}-\frac{\text{k}^2}{\text{a}^2(\text{e}^2-1)}=1$ $\Rightarrow \frac{\text{h}^2}{\text{a}^2}-\frac{\text{k}^2}{\text{b}^2},\text{Where }\text{b}^2=\text{a}^2(\text{e}^2-1)$ $\therefore$ The locus of (h, k) is $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1 $ Hence proved.

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Question 64 Marks

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

Answer

It is given that Clie on the x-axis. Let coordinates of C be (x, 0). Now, C is equidistant from the point A(7, 6) and B(3, 4). $\therefore \text{ AB}=\text{BC}\ [\text{given}]$ $\Rightarrow \text{AC}^2=\text{BC}^2$ $\Rightarrow \bigg[\sqrt{(\text{x}-7)^2+(0-6)^2}\bigg]= \bigg[\sqrt{(\text{x}-3)^2+(0-4)^2}\bigg]$ $\Rightarrow (\text{x}-7)^2+(-6)^2=(\text{x}-3)^2+(-4)^2$ $\Rightarrow \text{x}^2+49-14\text{x}+36=\text{x}^2+9-6\text{x}+16$ $\Rightarrow 49+36-36-16-9=\text{x}^2-\text{x}^2-6\text{x}+14\text{x}$ $\Rightarrow 85-25=8\text{x}$ $\Rightarrow 60=8\text{x}$ $\Rightarrow 8\text{x}=60$ $\Rightarrow \text{x}=\frac{60}{8}=\frac{15}{2}$ Hence, coordinates of c are $\bigg(\frac{15}{2},0\bigg).$

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Question 74 Marks

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

Answer

It is given that ABC is an equilatral triangle. $\therefore $ AB = BC = AC = 2a Area of equilatral triangle $=\frac{\sqrt{3}}{4}\text{(side)}^2$ $=\frac{\sqrt{3}}{4}\times(2\text{a})^2$ $=\frac{\sqrt{3}}{4}\times4\times\text{a}^2$ $=\sqrt{3}\text{a}^2$ But, area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}.$ $\Rightarrow\frac{1}{2}\times\text{Base}\times\text{Height}=\sqrt{3}\text{a}^2$ $\Rightarrow\frac{1}{2}\times\text{BC}\times\text{OA}=\sqrt{3}\text{a}^2$ $\Rightarrow\frac{1}{2}\times2\text{a}\times\text{OA}=\sqrt{3}\text{a}^2$ $\therefore $ coordinates of A are $(\sqrt{3}\text{a},0)\text{ or }\text{OA}(-\sqrt{3}\text{a},0)$ Clearly, the coordinates of B and C are (0, -a) and (0, a) respectively.
Hence, the vertices of the triangle are $(0,\text{a}),(0,-\text{a})\text{ and }\big(-\sqrt{3}\text{a},0\big) \text{ or }(0, \text{a}),(0,-\text{a})\text{ and }\big(\sqrt{3}\text{a},0\big).$

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Question 84 Marks

Verify that the area of the triangle with verticies (2, 3), (-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. $\therefore \text{ of } \triangle\text{ ABC }=\frac{1}{2}\big|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big|$ $=\frac{1}{2}\big|2(7+1)+5(-1-3)-3(3-7)\big|$ $=\frac{1}{2}\big|16-20+12\big|$ $=4$ Since the origin is shifted to the point (-1, 3), the vertices of the $\triangle\text{ ABC }$ will be$\text{A}'(2+1,3-3), \text{B}'(5+1,7-3)\text{ and } \text{C}'(-3+1,-1-3)$
$\text{ or } \text{A}'(3,0), \text{B}'(6,4),\text{ and }\text{C}'(-2-4)$
$\text{Now }, \text{ area }\text{of } \triangle\text{A}'\text{B}'\text{C}':$
$= \frac{1}{2}\big|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big|$ $= \frac{1}{2}\big|3(4+4)+6(-4-0)-2(0-4)\big|$ $=4$ Hence, area of the triangle remains invariant.

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Question 94 Marks

Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first deree terms: $x^2+y^2-5 x+2 y-5=0$

Answer

Let the origin be shifted to $(h, k)$. Then, $x=X+h$ and $y=Y+k$. Substituting $x=X+h, y=Y+k$ in the equation $x^2+y^2-5 x+2 y-5=0$,
We get $(X+h)^2+(Y+k)^2-5(X-h)-2(Y+k)+3=0$
$\Rightarrow x^2+h^2+2 X h+Y^2+k^2+2 Y k-5 X-5 h$
$+2 Y+2 k-5=0$
$\Rightarrow x^2+Y^2+2 Y k+2 Y+2 X h-5 X+h^2+k^2-5 h+2 k-5=0$
$\Rightarrow x^2+Y^2+(2 k+2) Y+(2 h-5) X+h^2$
$+k^2-5 h+2 k-5=0$ For this equation to be free from the term of first degree, we must have $2 k+2=0$ and $2 h-5$
$=0 \Rightarrow \mathrm{k}=-1$ and $\mathrm{h}=\frac{5}{2}$ Hence, the origin is shifted at the point $\left(\frac{5}{2},-1\right)$.

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Question 104 Marks

Four point A(6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are given in such a way that $\frac{\triangle\text{DBC}}{\triangle\text{ABC}}=\frac{1}{2}$, find x.

Answer

$\text{A}(6,3), \text{B}(-3,5),\text{C}(4,-2),\text{D}(\text{x},\text{3x})$ $\text{or }(\triangle\text{D}\text{B}\text{C})=\frac{1}{2}\big[\text{x}_1(\text{y}_2-\text{y}_1)+\text{x}_2(\text{y}_3-\text{y}_2)\big]$ $=\frac{1}{2}\big[-3(-2-3\text{x})+4(3\text{x}-5)+\text{x}(5+2)\big]$ $=\frac{1}{2}\big[1+9\text{x}+12\text{x}-20+5\text{x}+2\text{x}\big]$ $=\frac{1}{2}\big[28\text{x}-14\big]$ $=7\big[2\text{x}-1\big]$ $\text{or }(\triangle\text{ABC})=\frac{1}{2}\big[6(5+2)-3(-2-3)+4(3-5)\big]$ $=\frac{1}{2}[42+15-8]$ $\frac{\text{or }(\triangle \text{DBC})}{\text{or } (\triangle\text{ABC})}=\frac{1}{2}$ $\frac{7(2\text{x}-1)}{\frac{49}{2}}=\frac{1}{2}$ $\frac{14(2\text{x}-1)}{49}=\frac{1}{2}$ $\frac{28\text{x}-14}{49}=\frac{1}{2}$ $56\text{x}-28=49$ $56\text{x}=28+49$ $56\text{x}=77$ $\text{x}=\frac{11}{8}$

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Question 114 Marks

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. Let the coordinates of A and B be (a, 0) and (0, b) respectively. As the rod AB slides, the values. Let P(h, k) be a point on the locus. Then,$\text{h}=\frac{2\times \text{a}+1\times 0}{2+1}$
$\Rightarrow \text{h}=\frac{2\text{a}}{3}$
$\Rightarrow \text{a}=\frac{3\text{h}}{2}$
$\text{and }\text{k}=\frac{2\times0+\text{b}\times1}{2+1}$
$\Rightarrow \text{k}=\frac{\text{b}}{3}$
$\Rightarrow \text{b}=3\text{k}$
From $\triangle\text{AOB},$ We have
$\text{AB}^2=\text{OA}^2 +\text{OB}^2$
$\Rightarrow \text{l}^2=\big[(\text{a}-0)^2+(0,0)^2\big]+\big[(0,0)^2+(\text{a}-0)^2,\big]$
$\Rightarrow \text{l}^2=\text{a}^2+\text{b}^2$
$\Rightarrow \text{a}^2+\text{b}^2=\text{l}^2$
$\Rightarrow \bigg(\frac{3\text{h}}{2}\bigg)^2+(3\text{k})^2=\text{l}^2$
$\Rightarrow \frac{9\text{h}^2}{4}+9\text{k}^2=\text{l}^2$
$\Rightarrow \frac{\text{h}^2}{4}+\text{k}^2=\frac{\text{l}^2}{9}$
Hence, the locus of (h, k) is $\frac{\text{x}^2}{4}+\text{y}^2=\frac{\text{l}^2}{9}$

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Question 124 Marks

If O is the origin and Q is a variable point on $y^2 = x$, find the locus of the mid-point of OQ.

Answer

Let P(h, k) be the point on the locus and the coordinates of a are (a, b). Then,$\text{h}=\frac{\text{a}+0}{2}\text{ and }\frac{\text{b}+0}{2}=\text{k}$ $[\because$ P is thr mid-point of Q the origino $]$
$\text{h}=\frac{\text{a}}{2}\text{ and }\text{b}=2\text{k}$
$\Rightarrow \text{a}=2\text{h}\text{ and }\text{b}=2\text{k}$
point Q lies on the $y^2 = x$. Then,
$\text{b}^2=\text{a}$ $[\because \text{Q}:\text{(a,b)}]$
$\Rightarrow (2\text{k})^2=2\text{h}$ $[\because\text{a}=2\text{h} \text{ and } \text{b}=2\text{k}]$
$\Rightarrow 4\text{k}^2=2\text{h}$
$\Rightarrow 2\text{k}^2=\text{h}$
Hence, the locus of (h, k) is $2\text{y}^2=\text{x}$

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Question 134 Marks

Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first deree terms: $x^2 - 12x + 4 = 0$

Answer

Let the origin be shifted to ( $h, k$ ). Then, $x=X+h$ and $y=Y+k$. Substituting $x=X+h, y=Y+k$ in the equation $x^2-12 x+4=0$, We get $(X+h)^2-12(X+h)^2+4 \Rightarrow X^2+h^2+2 \times h-12 X-12 h+4=0 \Rightarrow X^2+(2 h-12) X+h^2-12 h+4$ $=0$ For this equation to be free from the term of first degree, we must have $2 \mathrm{~h}-12=0 \Rightarrow \mathrm{~h}=\frac{12}{2} \Rightarrow \mathrm{~h}=6$ Hence, the origin is shifted at the point $(6, k) k \in R$

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Question 144 Marks

Verify that the area of the triangle with vertice (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 co-ordinate of the vertex be A(4, 6), B(7, 10) and C(1, -2). Now area of the $\triangle \text{ABC}$ is given by, $\triangle= \frac{1}{2}\Big|(\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2))\Big|$ $=\frac{1}{2}\big|(4(10+2)+7(-2-6)+1(6-10))\big|$ $=\frac{1}{2}\big|(48-56-4)\big|$ $=6$ After transfroming the origin to (-2, 1), the co-ordinate of the vertex will be A(2, 7), B(5, 11) and C(-1, -1). Now the area will be: $\triangle_1= \frac{1}{2}\Big|(\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2))\Big|$ $=\frac{1}{2}\big|(2(11+1)+5(-1-7)-1(7-11))\big|$ $=\frac{1}{2}\big|(24-40+4)\big|$ $=6$ $\text{Here }\triangle=\triangle_1$ Hence proved.

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Question 154 Marks

If A(-1, 1) and B(2, 3) are two fixed points, find the locus of a point P, so that the area of $\triangle\text{PAB} = 8$ sq. units.

Answer

Let P(h, k) be any point on the locus. Then,Area (PAB) = 8 sq units
$\Rightarrow \frac{1}{2}\big|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big|=8$
$\Rightarrow \frac{1}{2}\big|-1(3-\text{k})+2(\text{k}-1)+\text{h}(1-3)\big|=8$
$\Rightarrow \frac{1}{2}\big|-3+\text{k}+2\text{k}+2-2\text{h}\big|=8$
$\Rightarrow \frac{1}{2}\big|-2\text{h}+3\text{k}-5\big|=8$
$\Rightarrow \big|-2\text{h}+3\text{k}-5\big|=16$
$\Rightarrow -2\text{h}+3\text{k}-5=\pm16$
$\Rightarrow 2\text{h}-3\text{k}+5\pm16=0$
$\Rightarrow 2\text{h}-3\text{k}+21=0 \ \text{or, }\ 2\text{h}-3\text{k}-11=0$
Hence, the locus of(h, k) is $2\text{h}-3\text{k}+21=0 \ \text{or, }\ 2\text{h}-3\text{k}-11=0$

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Question 164 Marks

If the line segment joining the points $P(x_1, y_1)$ and $Q(x_2, y_2)$ subtends an angle $\alpha$ at the origin O, prove that: $\text{O}\text{P}.\text{O}\text{Q}\cos\alpha =\text{x}_1\text{x}_2+\text{y}_1\text{y}_2. $

Answer

It is given that O is the origin. Then, $\text{OQ}^2=\text{x}_2^2+\text{y}_2^2$ $\text{OP}^2=\text{x}_1^2+\text{y}_1^2$ and, $\text{OP}^2=(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2$ Using cosine fromula in Rectangle OPQ, we have $\text{OP}^2=\text{OP}^2+\text{OQ}^2-2.(\text{OP})(\text{OQ})\cos\alpha$ $\Rightarrow (\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2=\text{x}_2^2+\text{y}_2^2+\text{x}_1^2+\text{y}_1^2-2(\text{OP}).(\text{OQ})\cos\alpha$ $\Rightarrow -2\text{x}_1\text{x}_2-2\text{y}_1\text{y}_2=-2\text{OP}.\text{O}\text{Q}\cos\alpha$ $\Rightarrow \text{x}_1\text{x}_2+\text{y}_1\text{y}_2=\text{OP}.\text{OQ}\cos\alpha$ $\Rightarrow\text{OP}.\text{OQ}\cos\alpha =\text{x}_1\text{x}_2+\text{y}_1\text{y}_2$ Hence, Proved.

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