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#### If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then(A) a = b(B) a = 2b(C) 2a = b(D) a = –b

If points A(1, 2), O(0, 0),C (a, b) are collinear then area of triangle

formed by these points must be zero.

$\\A(1, 2)=A(x\textsubscript{1} ,y\textsubscript{1})\\ O(0, 0)=O(x\textsubscript{2} ,y\textsubscript{2})\\ C (a, b)=C(x\textsubscript{3} ,y\textsubscript{3})\\ area of triangle= \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})]\\ \Rightarrow \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})] = 0\\ \Rightarrow [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})] = 0\\ \Rightarrow [1(0 - b) + 0(b - 2) + a(2 - 0)] = 0\\ \Rightarrow [-b + 0 + 2a] = 0\\ \Rightarrow -b + 2a = 0\\ \Rightarrow 2a = b\\$

Hence, option C is correct.

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#### If the distance between the points (4, p) and (1, 0) is 5, then the value of p is(A) 4 only(B) ± 4(C) – 4 only(D) 0

Points are A(4, p) and B(1, 0)

Distance

$\ (AB)\ = 5 (given )\\$

Here

$\\{{x}_{1}} =4\ \ {{x}_{2}} =1\\ {{y}_{1}} =p \: \: \: \: \: {{y}_{2}} =0\\$

Using distance formula

$\\=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ AB =\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ Squaring both sides\\ (1 - 4)\textsuperscript{2} + (0 - p)\textsuperscript{2} = (5)\textsuperscript{2}\\ (-3)\textsuperscript{2} + (p)\textsuperscript{2} = 25\\ 9 + p\textsuperscript{2} = 25\\ p\textsuperscript{2} = 25 - 9\\ p\textsuperscript{2} = 16\\ p = \pm 4\\$

Hence, option B is correct.

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#### The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is(A) (a+b+c)2(B) 0(C)( a + b + c)(D) abc

Vertices are (a, b + c), (b, c + a), (c, a + b)

Here $A(x\textsubscript{1}, y\textsubscript{1}) = (a , b + c)\\$

$\\B\left( {{x}_{2}},{{y}_{2}} \right)= (b , c + a)\\ C\left( {{x}_{3}},{{y}_{3}} \right)= (c , a + b)\\$

We know that

$Area of triangle = \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})]\\$
$\\ = \frac{1}{2} [a(c + a - a - b) + b(a + b - b - c) + c(b + c - c -a)]\\ = \frac{1}{2} [a(c - b) + b(a - c) + c(b - a)]\\ = \frac{1}{2} [ac - ab + ab - bc + bc - ac] = 0\\$

Hence, option B is correct.

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#### A line intersects the y-axis and x-axis at the points P and Q, respectively. If(2, –5) is the mid-point of PQ, then the coordinates of P and Q are respectively(A) (0, – 5) and (2, 0)(B) (0, 10) and (– 4, 0)(C) (0, 4) and (– 10, 0)(D) (0, – 10) and (4, 0)

Point

$P=(0, y)\{ at\ y -axis x=0\}$
$\\\mathrm{Q}=(\mathrm{x}, 0) \quad\{ at \mathrm{x} axis \mathrm{y}=0\} \\\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0, \mathrm{y}) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(\mathrm{x}, 0)$

$Mid-point =(2,-5)$

$\\\left[\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}\right),\left(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)\right]=(2,-5)\\\left(\frac{0+x}{2}, \frac{y+0}{2}\right)=2,-5 \\\frac{x}{2}=2 \quad \frac{y}{2}=-5$

$\\ x=4 \: \: \: \: \: \: \: \: y=-10 \\ \text { Point } P(0,-10), \ and\ Q(4,0)$

Hence, option D is correct.

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#### A circle drawn with origin as the Centre passes through(13/2, 0) . The point which does not lie in the interior of the circle is :$\\(A) \frac{-3}{4}, 1 \\(B) 2, \frac{7}{3} \\(C) 5, \frac{-1}{2} \\(D) \left(-6, \frac{5}{2}\right)$

A) Distance of the point (-3/4, 1) from (0,0) is

$=\sqrt{\left(\frac{-3}{4}-0\right)^{2}+(1-0)^{2}}=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}=1.25 \mathrm{units}$

The distance  1.25<6.5 .  so the point (-3/4, 1)  is lie
B) Distance point (2, 7/3) from (0,0) is

$=\sqrt{\left(\frac{7}{3}-0\right)^{2}+(2-0)^{2}}=\sqrt{\frac{49}{9}+4}=\sqrt{\frac{85}{9}}=\frac{9.2195}{3}=3.0731<6.25$

So the point is lie
C) Distance point (5,-1/2) from (0,0) is

$\\=\sqrt{\left(-\frac{1}{2}-0\right)^{2}+(5-0)^{2}}=\sqrt{\frac{1}{4}+25}=\sqrt{\frac{101}{4}}=\frac{10.0498}{2}=5.0249<6.25$

So the point is lie in the interior of the circle
D)
The circle passes through (13/2, 0) having a centre (0,0)

$\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(13 / 2,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,0)$

$\text{Apply distance formula}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$

$\text{Radius}=\sqrt{(6.5-0)^{2}+(0-0)^{2}}$
$\\=\sqrt{(6.5)^{2}}=6.5 \\=5^{2}+0.5^{2}-6.5^{2} \quad (Negative) \\\left(-6, \frac{5}{2}\right)=(-6)^{2}+(2.5)^{2}-6.5 \\=6^{2}+2.5^{2}-6.5 \quad(positive)$

Hence, point D is the correct answers.

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#### The coordinates of the point whichis equidistant from the three verticesof the ΔAOB as shown in the figure is$(A) (x, y)$$(B) (y, x)$$(C) \frac{x}{2},\frac{y}{2}$$(D)\frac{y}{2},\frac{x}{2}$

In the given figure, it is clear that DAOB is a right-angle triangle.

And in a right-angle triangle, the mid-point of the hypotenuse is equidistant from the three vertices. Thus, co-ordinates must be mid-point of AB

$\\ A (0, 2y), B(2x, 0)\\ (x\textsubscript{1}, y\textsubscript{1}) = (0, 2y) ,(x\textsubscript{2}, y\textsubscript{2}) = (2x, 0)\\$

Now find mid-point of AB using mid-point formula

$\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\$

$\left( \frac{0+2x}{2},\frac{{2y}+{0}}{2} \right)= \left ( x,y \right ) \\$

Hence, option A is correct.

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#### The perpendicular bisector of the line segment joining the points A (1, 5) andB (4, 6) cuts the y-axis at(A) (0, 13)(B) (0, –13)(C) (0, 12)(D) (13, 0)

$\\At y-axis, x = 0 \therefore point P is (0, y).\\ A (1, 5) and B (4, 6)\\ \therefore AP = BP\\$

Squaring both sides we get

$\\AP\textsuperscript{2} = BP\textsuperscript{2}\\ (x\textsubscript{1} - 0)\textsuperscript{2} + (y\textsubscript{1} + y)\textsuperscript{2} = (x\textsubscript{2}- 0)\textsuperscript{2} + (x\textsubscript{2} - y)\textsuperscript{2 }\\ \textsuperscript{ }(Because distance formula =\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} )\\ (1 - 0)\textsuperscript{2} + (5 - y)\textsuperscript{2} = (4 - 0)\textsuperscript{2} + (6 - y)\textsuperscript{2}\\ (1)\textsuperscript{2} + (5)\textsuperscript{2} + (y)\textsuperscript{2} - 2 \times 5 \times y = (4)\textsuperscript{2} + (6)\textsuperscript{2} + (y)\textsuperscript{2} - 2 \times 6 \times y\\ \{ using (a - b)\textsuperscript{2} = a\textsuperscript{2} + b\textsuperscript{2} - 2ab \} \\ 1 + 25 + y\textsuperscript{2} - 10y = 16 + 36 + y\textsuperscript{2} - 12y\\ 26 - 10y + 12y = 52\\ 2y = 52 - 26\\ 2y = 26\\ y = 13\\$

Hence, point P is (0, 13)

Therefore, option A is correct.

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#### If P(a/3 , 4) is the mid-point of the line segment joining the points Q (– 6, 5) and R (– 2, 3), then the value of a is(A) – 4(B) – 12(C) 12(D) – 6

$\\ \mathrm{Q}(-6,5) \quad \mathrm{R}(-2,3)\\\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$

By using mid-point formula

$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$
$P=\left(\frac{-6-2}{2}, \frac{5+3}{2}\right)$
$\\ \left(\frac{a}{3}, 4\right)=\left(\frac{-8}{2}, \frac{8}{2}\right) \\\left(\frac{a}{3}, 4\right)=(-4,4)$
Compare x co-ordinate
$\\\frac{a}{3}=-4 \\a=-4 \times 3 \\a=-12$

Hence, option B is correct.

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#### If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4),then $\\(A) \mathrm{AP}=\frac{1}{3} \mathrm{AB} \\(B) \mathrm{AP}=1 / 3 \mathrm{~PB} \\(C) \mathrm{PB}=\frac{1}{3} \mathrm{AB} \\(D) \mathrm{AP}=\frac{1}{2} \mathrm{AB}$

Let P divide  A B  in ratio  k: 1
$\\\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(4,2) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(8,4)\\P=\left(\frac{k(8)+1(4)}{k+1}, \frac{k(4)+1(2)}{k+1}\right)$
By using section formula
Here,

$\mathrm{P}=(2,1)$
$(2,1)=\left(\frac{8 k+4}{k+1}, \frac{4 k+12}{k+1}\right)$
Compare x co-ordinate
$\\\frac{8 k+4}{k+1}=2 \\8 k+4=2 k+2 \\6 k=-2 \\k=\frac{-2}{6}=\frac{-1}{3} \\Here, k is negative. Hence P divides A B in ratio 1: 3 externally. \\\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{1}{3} \\\mathrm{AP}=\frac{1}{3} \mathrm{~PB}$

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#### The fourth vertex D of a parallelogram ABCD whose three vertices areA (–2, 3), B (6, 7) and C (8, 3) is(A) (0, 1)(B) (0, –1)(C) (–1, 0)(D) (- 2 , 0)

Let D(x, y)

A(–2, 3), B(6, 7), C(8, 3) (given)

We know that in parallelogram diagonals are equal

mid-point of AC = mid-point of BD

$\\ \left(\frac{-2+8}{2}, \frac{3+3}{2}\right)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ \left(\frac{6}{2}, \frac{6}{2}\right)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ (3,3)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ \frac{6+x}{2}=3 \quad \frac{7+y}{2}=3 \\ 6+x=6 \quad 7+y=6 \\ x=0 \quad y=-1 \\ \text { Hence, } D=(0,-1)$

option B is correct.

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