MCQ 11 Mark
If the line segment joining the points $(3, -4)$ and $(1, 2)$ is trisected at points $P(a, -2)$ and $\text{Q}\Big(\frac{5}{3},\text{b}\Big).$ Then,
- A
$\text{a}=\frac{8}{3},\text{b}=\frac{2}{3}$
- ✓
$\text{a}=\frac{7}{3},\text{b}=0$
- C
$\text{a}=\frac{1}{3},\text{b}=1$
- D
$\text{a}=\frac{2}{3},\text{b}=\frac{1}{3}$
AnswerCorrect option: B. $\text{a}=\frac{7}{3},\text{b}=0$
We have two points $A(3, -4)$ and $B(1, 2)$. There are two points $P(a, -2)$ and $\text{Q}\Big(\frac{5}{3},\text{b}\Big)$ which trisect the line segment joining $A$ and $B.$
Now according to the section formula if any point $P$ divides a line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m: n$ internally than,
$\text{P(x, y)}=\Big(\frac{\text{nx}_1+\text{mx}_2}{\text{m}+\text{n}},\frac{\text{ny}_1+\text{my}_2}{\text{m}+\text{n}}\Big)$
The point $P$ is the point of trisection of the line segment $AB$.
So, $P$ divides $AB$ in the ratio $1: 2$.
Now we will use section formula to find the co$-$ordinates of unknown point A as,
$\text{P(a,}-2)=\Big(\frac{2(3)+1(1)}{1+2},\frac{2(-4)+1(2)}{1+2}\Big)$
$=\Big(\frac{7}{3},-2\Big)$
Equate the individual terms on both the sides. We get,
$\text{a}=\frac{7}{3}$
Similarly, the point $Q$ is the point of trisection of the line segment $AB$.
So, $Q$ divides $AB$ in the ratio $2: 1$
Now we will use section formula to find the co$-$ordinates of unknown point $A$ as,
$\text{Q}\Big(\frac{5}{3},\text{b}\Big)=\Big(\frac{2(1)+1(3)}{1+2},\frac{2(2)+1(-4)}{1+2}\Big)$
$=\Big(\frac{5}{3},0\Big)$
Equate the individual terms on both the sides. We get,
$b = 0$
View full question & answer→MCQ 21 Mark
If points $(t, 2t), (-2, 6)$ and $(3, 1)$ are collinear, then $t =$
- A
$\frac{3}{4}$
- ✓
$\frac{4}{3}$
- C
$\frac{5}{3}$
- D
$\frac{3}{5}$
AnswerCorrect option: B. $\frac{4}{3}$
We have three collinear points $A(t, 2t), B(-2, 6), C(3, 1).$
In general if $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ are collinear then,
$\frac{1}{2}\Big[\text{x}_1(\text{y}_2 - \text{y}_3) + \text{x}_2(\text{y}_3 - \text{y}_1) + \text{x}_3\text{(y}_1 - \text{y}_2)\Big] = 0$
So, $t(6 - 1) - 2(1 - 2t) + 3(2t - 6) = 0$
So, $5t + 4t + 6t - 2 - 18 = 0$
So, $15t = 20$
Therefore, $\text{t}=\frac{4}{3}$
View full question & answer→MCQ 31 Mark
The distance between the points $(\text{a}\cos25^\circ,0)$ and $(0,\text{a}\cos65^\circ)$ is:
AnswerDistance between $(\text{a}\cos25^\circ,0)$ and $(0,\text{a}\cos65^\circ)$
$=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(0-\text{a}\cos25^\circ)^2+(\text{a}\cos65^\circ-0)^2}$
$=\sqrt{\text{a}^2\cos^225^\circ+\text{a}^2\cos^265^\circ}$
$=\sqrt{\text{a}^2[\cos^225^\circ+\cos^265^\circ]}$
$=\text{a}\sqrt{\cos^2(90^\circ-65^\circ)+\cos^265^\circ}$
$=\text{a}\sqrt{\sin^265^\circ+\cos^265^\circ}$
$=\text{a}(\sqrt{1})$
$=\text{a}$
View full question & answer→MCQ 41 Mark
The coordinates of the fourth vertex of the rectangle formed by the points $(0, 0), (2, 0), (0, 3)$ are,
- A
$(3, 0)$
- B
$(0, 2)$
- ✓
$(-2, 3)$
- D
$(3, 2)$
AnswerCorrect option: C. $(-2, 3)$
Three vertices of a rectangle are $A(0, 0), B(2, 0), C(0, 3).$
Let fourth vertex be $D(x, y).$
The diagonals of a rectangle bisect eachother at $O.$
O is the mid$-$point of $AC,$ then
Coordinates of $O$ will be $\Big(\frac{0+0}{2},\frac{0+3}{2}\Big)$
or $\Big(0,\frac{3}{2}\Big)$
$\because O$ is also the mid$-$point of $BD$
$0=\frac{2+\text{x}}{2}$
$\Rightarrow\ 2+\text{x}=0$
$\Rightarrow\ \text{x}=-2$
and $\frac{3}{2}=\frac{0+\text{y}}{2}$
$\Rightarrow\ \text{y}=3$
$\therefore$ Co$-$ordinates of $D$ are $(-2, 3).$
View full question & answer→MCQ 51 Mark
A line segment is of length $10$ units. If the coordinates of its one end are $(2, -3)$ and the abscissa of the other end is $10$, then its ordinate is:
- A
$9, 6$
- ✓
$3, -9$
- C
$-3, 9$
- D
$9, -6$
AnswerCorrect option: B. $3, -9$
Let the ordinate of other end =$ y$
Then distance between $(2, -3)$ and $(10, y) = 10$ units
$\Rightarrow\ \sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}=10$
$\Rightarrow\ \sqrt{(10-2)^2+(\text{y}+3)^2}=10$
$\Rightarrow\ \sqrt{(8)^2+(\text{y}+3)^2}=10$
Squaring both sides
$\Rightarrow (8)^2 + (y + 3)^2 = (10)^2$
$\Rightarrow 64 + (y + 3)^2 = 100$
$\Rightarrow (y + 3)^2 = 100 - 64 = 36 = (6)^2$
$\Rightarrow (y + 3)^2 - (6)^2 = 0$
$\Rightarrow (y + 3 + 6)(y + 3 - 6) = 0$
$\{\because$ $a^2 - b^2 = (a + b)(a - b)$$\}$
$\Rightarrow (y + 9)(y - 3) = 0$
Either$ y + 9 = 0$, then $y = -9$
or$ y - 3,$ then $y = 3$
$\therefore$ $y = 3, -9$
View full question & answer→MCQ 61 Mark
If $x$ is a positive integer such that the distance between points $P(x, 2)$ and $Q(3, -6)$ is $10$ units, then $x =$
AnswerDistance between $P(x, 2)$ and $Q(3, -6) = 10$ units
$\Rightarrow\ \sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+(-6-2)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+(-8)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+64}=10$
Squaring both sides,
$(3 - x)^2 + 64 = 100$
$\Rightarrow 9 + x^2 - 6x + 64 - 100 = 0$
$\Rightarrow x^2 - 6x - 27 = 0$
$\Rightarrow x^2 - 9x + 3x - 27 = 0$
$\begin{Bmatrix}\because\ 27=-9\times3\\\ \ -6=-9+3\end{Bmatrix}$
$\Rightarrow x(x - 9) + 3(x - 9) = 0$
$\Rightarrow (x - 9)(x - 3) = 0$
Either $x - 9 = 0$, then $x = 9$ or $x + 3 = 0$, then $x = -3$
$x$ is positive integer.
Hence $x = 9.$
View full question & answer→MCQ 71 Mark
If three points $(0, 0),(3,\sqrt{3})$ and $(3,\lambda)$ form an equilateral triangle, then $\lambda=$
AnswerLet the points $(0, 0)$, $(3,\sqrt{3})$ and $(3,\lambda)$ form an equilateral triangle
$AB = BC = CA$
$\Rightarrow AB^2 = BC^2 = CA^2$
Now, $AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
$=(3-0)^2+(\sqrt{3}-0)^2$
$=(3)^2+(\sqrt{3})^2$
$=9+3=12$
$\text{BC}^2=(3-3)^2+(\lambda-\sqrt{3})^2$
$=(0)^2+(\lambda-\sqrt{3})^2=(\lambda-\sqrt{3})^2$
and $\text{CA}^2=(0-3)^2+(0-\lambda)^2$
$=(-3)^2+(-\lambda)^2$
$=9+\lambda^2$
$\text{AB}^2=\text{CA}^2\Rightarrow\ 12=9+\lambda^2$
$\Rightarrow\ \lambda^2=12-9=3$
$\therefore\ \lambda=\pm\sqrt{3}$
View full question & answer→MCQ 81 Mark
The ratio in which the line segment joining $P(x_1, y_1)$ and $Q(x_2, y_2)$ is divided by $x-$axis is:
- A
$y_1 : y_2$
- ✓
$-y_1 : y_2$
- C
$x_1 : x_2$
- D
$-x_1 : x_2$
AnswerCorrect option: B. $-y_1 : y_2$
Let a point $A$ on $x-$axis divides the line segment joining the points $P(x_1, y_1), Q(x_2, y_2)$ in the ratio $m_1 : m_2$
and let co$-$ordinates of $A$ be $(x, 0)$
$\therefore\ 0=\frac{\text{m}_1\text{y}_2+\text{m}_2\text{y}_1}{\text{m}_1+\text{m}_2}$
$\Rightarrow\ 0=\text{m}_1\text{y}_2+\text{m}_2\text{y}_1$
$\Rightarrow\ \text{m}_1\text{y}_2=-\text{m}_2\text{y}_1$
$\Rightarrow\ \frac{\text{m}_1}{\text{m}_2}=\frac{-\text{y}_1}{\text{y}_2}$
$\therefore$ Ratio is $-y_1 : y_2$
View full question & answer→MCQ 91 Mark
If points $(1, 2), (-5, 6)$ and $(a, -2)$ are collinear, then $a =$
AnswerThe area of a triangle whose vertices are $(1, 2), (-5, 6)$ and $(a, -2)$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[1(6+2)+(-5)(-2-2)+\text{a}(2-6)]$
$=\frac{1}{2}[1\times8+(-5)(-4)+\text{a}(-4)]$
$=\frac{1}{2}[8+20-4\text{a}]$
$=\frac{1}{2}(28-4\text{a})$
$=(14-2\text{a})\text{ sq.units}$
$\because$ The points are collinear.
$\therefore$ Area $= 0$
$\Rightarrow\ 14-2\text{a}=0$
$\Rightarrow\ 2\text{a}=14$
$\Rightarrow\ \text{a}=\frac{14}{2}=7$
Hence, $a = 7$
View full question & answer→MCQ 101 Mark
In the figure, the area of $\triangle ABC ($in square units$)$ is:
Answer
The coordinates of $A$ are $(1, 3).$
$\therefore$ Distance of $A$ from the $x-$axis, $AD = y-$coordinate of $A = 3$ units.
The number of units between $B$ and $C$ on the $x-$axis are $5.$
$\therefore BC = 5$ units
Now,
Area of $\triangle\text{ABC}=\frac{1}{2}\times\text{BC}\times\text{AD}$
$=\frac{1}{2}\times5\times3$
$=\frac{15}{2}$
$=7.5\text{ sq.units}$
Thus, the area of $\triangle ABC$ is $7.5$sq. units. View full question & answer→MCQ 111 Mark
If $P(2, 4), Q(0, 3), R(3, 6)$ and $S(5, y)$ are the vertices of a paralelogram $\text{PQRS,}$ then the value of $y$ is:
AnswerIt is given that $P(2, 4), Q(0, 3), R(3, 6)$ and $S(5, y)$ are the vertices of a parallelogram $\text{PQRS.}$
Join $PR$ and $QS,$ intersecting each other at $O.$
We know that the diagonals of the parallelogram bisect each other.
So, $O$ is the mid$-$point of $PR$ and $QS.$
Coordinates of mid$-$point of $\text{PR}=\Big(\frac{2+3}{2},\frac{4+6}{2}\Big)$
$=\Big(\frac{5}{2},\frac{10}{2}\Big)=\Big(\frac{5}{2},5\Big)$
Coordinates of mid$-$point of $\text{QS}=\Big(\frac{0+5}{2},\frac{3+\text{y}}{2}\Big)$
$=\Big(\frac{5}{2},\frac{3+\text{y}}{2}\Big)$
Now, these points coincides at the point $O.$
$\therefore\ \Big(\frac{5}{2},\frac{3+\text{y}}{2}\Big)=\Big(\frac{5}{2},5\Big)$
$\Rightarrow\ \frac{3+\text{y}}{2}=5$
$\Rightarrow 3 + y = 10$
$\Rightarrow y = 7$
Thus, the value of $y$ is $7.$ View full question & answer→MCQ 121 Mark
The coordinates of the point on $x$-axis which are equidistant from the points $(-3, 4)$ and $(2, 5)$ are:
Answer$\because$ The point is on $x$-axis.
$\therefore$ Its ordinate will be $=0$
Let the points be $P(x, 0)$ which is equidistant from $A(-3,4)$ and $B(2,5)$
$PA=PB \Rightarrow PA^2=PB^2$
Now, $PA ^2=(-3- x )^2+(4-0)^2$
$=(-3-x)^2+(4)^2$
$=9+x^2+6 x+16=x^2+6 x+25$
$\text { and } P B^2=(2-x)^2+(5-0)^2=(2-x)^2+(5)^2$
$=4-4 x+x^2+25$
$=x^2-4 x+29$
$\therefore x^2+6 x+25=x^2-4 x+29$
$\Rightarrow x^2+6 x-x^2+4 x=29-25$
$\Rightarrow 10 x=4$
$\Rightarrow x=\frac{4}{10}=\frac{2}{5}$
Point will be $\left(\frac{2}{5}, 0\right)$.
View full question & answer→MCQ 131 Mark
The line segment joining points $(-3, -4)$ and $(1, -2)$ is divided by $y-$axis in the ratio:
- A
$1 : 3$
- B
$2 : 3$
- ✓
$3 : 1$
- D
$2 : 3$
AnswerCorrect option: C. $3 : 1$
The point lies on $y-$axis.
Its abscissa will be zero.
Let the point divides the line segment joining the points $(-3, -4)$ and $(1, -2)$ in the ratio $m : n$
$\therefore\ 0=\frac{\text{mx}_2+\text{nx}_1}{\text{m}+\text{n}}$
$\Rightarrow\ 0=\frac{\text{m}\times1+\text{n}\times(-3)}{\text{m}+\text{n}}$
$\Rightarrow\ \frac{\text{m}-3\text{n}}{\text{m}+\text{n}}=0$
$\Rightarrow\ \text{m}-3\text{n}=0$
$\Rightarrow\ \text{m}=3\text{n}$
$\Rightarrow\ \frac{\text{m}}{\text{n}}=\frac{3}{1}$
$\therefore$ Ratio $= 3 : 1$
View full question & answer→MCQ 141 Mark
If the point $P(2, 1)$ lies on the line segment joining points $A(4, 2)$ and $B(8, 4),$ then:
- A
$\text{AP}=\frac{1}{3}\text{AB}$
- B
$\text{AP}=\text{BP}$
- C
$\text{BP}=\frac{1}{3}\text{AB}$
- ✓
$\text{AP}=\frac{1}{2}\text{AB}$
AnswerCorrect option: D. $\text{AP}=\frac{1}{2}\text{AB}$
Use section formula for finding out the ratio in which $P$ divided the line segment $AB.$
$(2,1)=\Big(\frac{\text{m}_1(8)+\text{m}_2(4)}{\text{m}_1+\text{m}_2},\frac{\text{m}_1(4)+\text{m}_2(2)}{\text{m}_1+\text{m}_2}\Big)$
$\Rightarrow\ 2=\frac{8\text{m}_1+4\text{m}_2}{\text{m}_1+\text{m}_2};\ 1=\frac{4\text{m}_1+2\text{m}_2}{\text{m}_1+\text{m}_2}$
$\Rightarrow\ 3\text{m}_1+\text{m}_2=0$
$\Rightarrow\ \frac{\text{m}_1}{\text{m}_2}=-\frac{1}{3}$
$\therefore$ The point $P$ divided $AB$ in ratio $1 : 3$ externally.
$AP : PB = 1 : 3$
$\Rightarrow AP : AB = 1 : 2$
$\Rightarrow\ \text{AP}=\frac{1}{2}\text{AB}$
View full question & answer→MCQ 151 Mark
If $(x, 2), (-3, -4)$ and $(7, -5)$ are coliinear, then $x =$
AnswerArea of triangle whose vertices are $(x, 2), (-3, -4)$ and $(7, -5)$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{x}(-4+5)+(-3)(-5-2)+7(2+4)]$
$=\frac{1}{2}[\text{x}\times1+(-3)(-7)+7\times6]$
$=\frac{1}{2}(\text{x}+21+42)$
$=\frac{1}{2}(\text{x}+63)$
$\because$ The points are collinear
$\therefore$ Area of the $\triangle=0$
$\Rightarrow\ \frac{1}{2}(\text{x}+63)=0$
$\Rightarrow\ \text{x}+63=0$
$\Rightarrow\ \text{x}=-63$
View full question & answer→MCQ 161 Mark
The ratio in which $(4, 5)$ divides the join of $(2, 3)$ and $(7, 8)$ is:
- A
$-2 : 3$
- B
$-3 : 2$
- C
$3 : 2$
- ✓
$2 : 3$
AnswerCorrect option: D. $2 : 3$
Let the point $(4, 5)$ divides the line segment joining the points $(2, 3)$ and $(7, 8)$ in the ratio $m : n$
$\therefore\ 4=\frac{\text{mx}_2+\text{nx}_1}{\text{m}+\text{n}}=\frac{\text{m}\times7+\text{n}\times2}{\text{m}+\text{n}}$
$\Rightarrow 4(m + n) = 7m + 2n$
$\Rightarrow 4m + 4n = 7m + 2n$
$\Rightarrow 4n - 2n = 7m - 4m$
$\Rightarrow 2n = 3m$
$\Rightarrow\ \frac{\text{m}}{\text{n}}=\frac{2}{3}$
$\therefore m : n = 2 : 3$
View full question & answer→MCQ 171 Mark
If points $(a, 0), (0, b)$ and $(1, 1)$ are collinear, then $\frac{1}{\text{a}}+\frac{1}{\text{b}}=$
AnswerThe area of triangle whose vertices are $(a, 0), (0, b)$ and $(1, 1).$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{a}(\text{b}-1)+0(1-0)+1(0-\text{b})]$
$=\frac{1}{2}[\text{ab}-\text{a}+0-\text{b}]$
$=\frac{1}{2}(\text{ab}-\text{a}-\text{b})$
$\because$ The points are collinear
$\therefore\ \frac{1}{2}(\text{ab}-\text{a}-\text{b})=0$
$\Rightarrow\ \text{ab}-\text{a}-\text{b}=0$
$\Rightarrow\ \text{ab}=\text{a}+\text{b}$
$\Rightarrow\ \frac{\text{a}+\text{b}}{\text{ab}}=1$
$\Rightarrow\ \frac{\text{a}}{\text{ab}}+\frac{\text{b}}{\text{ab}}=1$
$\Rightarrow\ \frac{1}{\text{b}}+\frac{1}{\text{a}}=1$
$\Rightarrow\ \frac{1}{\text{a}}+\frac{1}{\text{b}}=1$
View full question & answer→MCQ 181 Mark
The perimeter of the triangle formed by the points $(0, 0), (0, 1)$ and $(0, 1)$ is:
- A
$1\pm\sqrt{2}$
- B
$\sqrt{2}+1$
- C
$3$
- ✓
$2+\sqrt{2}$
AnswerCorrect option: D. $2+\sqrt{2}$
We have a triangle $\triangle\text{ABC}$ whose co$-$ordinates are $A(0, 0), B(1, 0), C(0, 1).$
So clearly the triangle is right angled triangle, right angled at $A$.
So,
$AB = 1$ units
$AC = 1$ units
Now apply Pythagoras theorem to get the hypotenuse,
$\text{BC}=\sqrt{\text{AB}^2+\text{AC}^2}$
$=\sqrt{2}$
So the perimeter of the triangle is,
$=\text{AB}+\text{BC}+\text{AC}$
$=1+1+\sqrt{2}$
$=2+\sqrt{2}$
View full question & answer→MCQ 191 Mark
If the centroid of the triangle formed by $(7, x), (y, -6)$ and $(9, 10)$ is at $(6, 3)$, then $(x, y) =$
- A
$(4, 5)$
- B
$(5, 4)$
- C
$(-5, -2)$
- ✓
$(5, 2)$
AnswerCorrect option: D. $(5, 2)$
We have to find the unknown co-ordinates.
The co-ordinates of vertices are $A(7, x), B(y, -6), C(9, 10).$
The co-ordinate of the centroid is $(6, 3).$
We know that the co-ordinates of the centroid of a triangle whose vertices are $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is,
$\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)$
So,
$(6,3)=\Big(\frac{\text{y}+7+9}{3},\frac{\text{x}-6+10}{3}\Big)$
Compare individual terms on both the sides,
$\frac{\text{x}+4}{3}=3$
So,
$x = 5$
Similarly,
$\frac{\text{y}+16}{3}=6$
So,
$y = 2$
View full question & answer→MCQ 201 Mark
If the centroid of the triangle formed by the points $(3, -5), (-7, 4), (10, -k)$ is at the point $(k, -1)$, then $k =$
AnswerWe have to find the unknown co$-$ordinates.
The co$-$ordinates of vertices are $A(3, -5), B(-7, 4), C(10, -k)$
The co$-$ordinate of the centroid is $(k, -1)$
We know that the co$-$ordinates of the centroid of a triangle whose vertices are $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:
$\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)$
So, $(\text{k}, -1)=\Big(\frac{3-7+10}{3},\frac{-5+4-\text{k}}{3}\Big)$
Compare individual terms on both the sides,
$\text {k} = 2$
View full question & answer→MCQ 211 Mark
A line intersects the y-axis and x-axis at $P$ 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 $(-10, 0)$
- ✓
$(0, -10)$ and $(4, 0)$
AnswerCorrect option: D. $(0, -10)$ and $(4, 0)$
Let the coordinates of $P(0, y)$ and $Q(x, 0)$, respectively.
So, the mid-point of $P(0, y)$ and $Q(x, 0)$ is
$\text{M}\Big(\frac{0+\text{x}}{2},\frac{\text{y}+0}{2}\Big)$
$\bigg[\because$ mid-point of a line segment having points $(x_1, y_1)$ and $(x_2, y_2)$ $=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\bigg]$
But it is given that, mid-point of $PQ$ is $(2, -5)$
$2=\frac{\text{x}+0}{2}$ and $-5=\frac{\text{y}+0}{2}$
$\Rightarrow 4 = x$ and $-10 = y$
$\Rightarrow x = 4$ and $y = -10$
So, the coordinates of $P$ and $Q$ are $(0, -10)$ and $(4, 0)$. View full question & answer→MCQ 221 Mark
The perimeter of a triangle with vertices $(0, 4)$ and $(0, 0)$ and $(3, 0)$ is:
- A
$7+\sqrt{5}$
- B
$5$
- C
$10$
- ✓
$12$
Answer$A(0, 4)$ and $B(0, 0)$ and $C(3, 0)$ are the vertices of $\triangle ABC$
$\therefore\ \text{AB}=\sqrt{(0-0)^2+(4-0)^2}=\sqrt{0+4^2}$
$=\sqrt{0+16}=\sqrt{16}=4$
$\text{BC}=\sqrt{(0-3)^2+(0-0)^2}=\sqrt{(-3)^2+0^2}$
$=\sqrt{9+0}=\sqrt{9}=3$
$\text{CA}=\sqrt{(3-0)^2+(0-4)^2}=\sqrt{3^2+4^2}$
$=\sqrt{9+16}=\sqrt{25}=5$
$\therefore$ Perimeter $= AB + BC + CA$
$= 4 + 3 + 5$
$= 12$ units
View full question & answer→MCQ 231 Mark
If the point $P(x, y)$ is equidistant from $A(5, 1)$ and $B(-1, 5)$, then
- A
$5x = y$
- B
$x = 5y$
- ✓
$3x = 2y$
- D
$2x = 3y$
AnswerCorrect option: C. $3x = 2y$
Points $P(x, y)$ is equidistant from $A(5,1), B(-1,5)$ then $A P=B P $
$\Rightarrow A P^2=B P^2$
$\Rightarrow(5-x)^2+(1-y)^2=(-1-x)^2+(5-y)^2$
$\Rightarrow 25+x^2-10 x+1+y^2-2 y=1+x^2+2 x+25+y^2-10 y$
$\Rightarrow-10 x-2 y+26=2 x-10 y+26$
$\Rightarrow 2 x+10 x=-2 y+10 y$
$\Rightarrow 12 x=8 y$
$\Rightarrow 3 x=2 y$
View full question & answer→MCQ 241 Mark
If $A(5, 3)$, $B(11, -5)$ and $P(12, y)$ are the vertices of a right triangle right angled at $P$, then $y =$
- A
$-2, 4$
- B
$-2, 4$
- ✓
$2, -4$
- D
$2, 4$
AnswerCorrect option: C. $2, -4$
We have a right angled triangle $\triangle\text{APC}$ whose co$-$ordinates are $A (5, 3), B (11, -5), P(12, y).$
So clearly the triangle is, right angled at $A.$ So,
$AP^2 = (12 - 5)^2 + (y - 3)^2$
$BP^2 = (12 - 11)^2 + (y + 5)^2$
$AB^2 = (11 - 5)^2 + (-5 - 3)^2$
Now apply Pythagoras theorem to get,
$AB^2 = AP^2 + BP^2$
So, $100 = 50 + 2y^2 + 34 + 4y$
On further simplification we get the quadratic equation as,
$2y^2 + 4y - 16 = 0$
$y^2 + 2y - 8 = 0$
Now solve this equation using factorization method to get,
$(y + 4)(y - 2) = 0$
Therefore, $y = (2, -4)$
View full question & answer→MCQ 251 Mark
If the centroid of a triangle is $(1, 4)$ and two of its vertices are $(4, -3)$ and $(-9, 7)$, then the area of the triangle is:
AnswerCorrect option: B. $\frac{183}{2}\text{ sq.units}$
Centroid of a triangle $= (1, 4)$
and two vertices of the triangle are $(4, -3)$ and $(-9, 7)$
Let the third vertex be $(x, y),$ then
$\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3}=1$
$\Rightarrow\ \frac{4-9+\text{x}}{3}=1$
$\Rightarrow\ -5+\text{x}=3$
$\Rightarrow\ \text{x}=3+5=8$
and $\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}=4$
$\Rightarrow\ \frac{-3+7+\text{y}}{3}=4$
$\Rightarrow\ 4+\text{y}=12$
$\Rightarrow\ \text{y}=12-4=8$
$\therefore$ Third vertex $= (8, 8)$
Now area of the triangle,
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[4(7-8)+(-9)(8+3)+8(-3-7)]$
$=\frac{1}{2}[4\times(-1)+(-9)\times11+8\times(-10)]$
$=\frac{1}{2}[-4-99-80]$
$=\frac{1}{2}\times(-183)$
$=\frac{183}{2}\text{ sq.units}$
View full question & answer→MCQ 261 Mark
If $P$ is a point on $x-$axis such that its distance from the origin is $3$ units, then the coordinates of a point $Q$ on $\text{OY}$ such that $\text{OP = OQ,}$ are
- ✓
$(0, 3)$
- B
$(3, 0)$
- C
$(0, 0)$
- D
$(0, -3)$
AnswerCorrect option: A. $(0, 3)$
Given: If $P$ is a point on $x$ axis such that its distance from the origin is $3$ units.
To find: The coordinates of a point $Q$ on $\text{OY}$ such that $\text{OP = OQ.}$
On $x$ axis $y$ coordinates is $0.$
Hence the coordinates of point $P$ will be $(3, 0)$ as it is given that the distance from origin is $3$ units.
Now then the coordinates of $Q$ on $\text{OY}$ such that $\text{OP = OQ}$
On $y$ axis $x$ coordinates is $0.$
Hence the coordinates of point $Q$ will be $(0, 3).$
View full question & answer→MCQ 271 Mark
If the coordinates of one end of a diameter of a circle are $(2, 3)$ and the coordinates of its centre are $(-2, 5),$ then the coordinates of the other end of the diameter are:
- ✓
$(-6, 7)$
- B
$(6, -7)$
- C
$(6, 7)$
- D
$(-6, -7)$
AnswerCorrect option: A. $(-6, 7)$
Let $AB$ be the diameter of a circle with centre $O$.
Coordinates of $A(2, 3)$ and $O(-2, 5)$
Let coordinates of $B$ be $(x, y)$
$\because\ -2=\frac{2+\text{x}}{2}$
$\Rightarrow\ -4=2+\text{x}$
$\Rightarrow\ \text{x}=-4-2=-6$
and $5=\frac{3+\text{y}}{2}$
$\Rightarrow\ 3+\text{y}=10$
$\Rightarrow\ \text{y}=10-3=7$
Coordinates of other end will be $(-6, 7).$ View full question & answer→MCQ 281 Mark
If $(-1, 2), (2, -1)$ and $(3, 1)$ are any three vertices of a parallelogram, then:
- A
$a = 2, b = 0$
- B
$a = -2, b = 0$
- C
$a = -2, b = 6$
- ✓
$a = 0, b = 4$
AnswerCorrect option: D. $a = 0, b = 4$
In $\ce{\| gm ABCD,}$ diagonals $AC$ and $AD$ bisect each other at $O.$
$O$ is mid$-$point of $AC$
$\therefore$ Co$-$ordinates of $O$ will be
$\Big(\frac{-1+3}{2},\frac{2+1}{2}\Big)$ or $\Big(\frac{2}{2},\frac{3}{2}\Big)$ or $\Big(1,\frac{3}{2}\Big)$
$\because O$ is mid$-$point of $BD$
$\therefore\ \frac{2+\text{a}}{2}=1$ and $\frac{-1+\text{b}}{2}=\frac{3}{2}$
$\Rightarrow\ 2+\text{a}=2$
$\Rightarrow a = 2 - 2 = 0$
and $-1 + b = 3$
$\Rightarrow b = 3 + 1 = 4$
$\therefore a = 0, b = 4$ View full question & answer→MCQ 291 Mark
If $(-2, 1)$ is the centroid of the triangle having its vertices at $(x, 0), (5, -2), (-8, y),$ then $x, y$ satisfy the relation:
- A
$3x + 8y = 0$
- B
$3x – 8y = 0$
- ✓
$-3x + 5y = 0$
- D
$8x = 3y$
AnswerCorrect option: C. $-3x + 5y = 0$
$(-2, 1)$ is the centroid of triangle whose vertices are $(x, 0), (5, -2), (-8, y)$
$\therefore$ Then $\text{x}=\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3}$
$\Rightarrow\ -2=\frac{\text{x}+5-8}{3}$
$\Rightarrow\ -6=\text{x}-3$
$\Rightarrow\ \text{x}=-6+3=-3$
and $1=\frac{0-2+\text{y}}{3}$
$\Rightarrow\ -2+\text{y}=3$
$\Rightarrow\ \text{y}=3+2=5$
$\therefore\ \text{x}=-3,\text{y}=5$
Value of $x$ and $y$ do not satisfy any choice.
View full question & answer→MCQ 301 Mark
If the distance between the points $(4, p)$ and $(1, 0)$ is $5,$ then $p =$
AnswerCorrect option: A. $\pm4$
Distance between $(4, p)$ and $(1, 0) = 5$
$\Rightarrow\ \sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}=5$
$\Rightarrow\ \sqrt{(1-4)^2+(0-\text{p})^2}=5$
$\Rightarrow\ \sqrt{(-3)^2+(-\text{p})^2}=5$
Squaring, both sides
$\Rightarrow\ (-3)^2+(-\text{p})^2=(5)^2$
$\Rightarrow\ 9+\text{p}^2=25$
$\Rightarrow\ \text{p}^2=25-9=16$
$\therefore\ \text{p}=\pm\sqrt{16}=\pm4$
View full question & answer→MCQ 311 Mark
The coordinates of the circumcentre of the triangle formed by the points $O(0, 0), A(a, 0)$ and $B(0, b)$ are,
- A
$(\text{a, b})$
- ✓
$\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$
- C
$\Big(\frac{\text{b}}{2},\frac{\text{a}}{2}\Big)$
- D
$(\text{b, a})$
AnswerCorrect option: B. $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$
Let co$-$ordinates of $C$ be $(x, y)$ which is the centre of the circumcircle of $\triangle OAB.$
Radii of a circle are equal
$\therefore$ $OC = CA = CB $
$\Rightarrow OC^2 = CA^2 = CB^2$
$\therefore$ $(x - 0)^2 + (y - 0)^2 = (x - a)^2 + (y - 0)^2$
$\Rightarrow x^2 + y^2 = (x - a)2 + y^2$
$\Rightarrow x^2 = (x - a)^2$
$\Rightarrow x^2 = x^2 + a^2 - 2ax$
$\Rightarrow a^2 - 2ax = 0$
$\Rightarrow a(a - 2x) = 0$
$\Rightarrow a = 2x$
$\Rightarrow\ \text{x}=\frac{\text{a}}{2}$
and $(x - 0)^2 + (y - 0)^2 = (x - 0)^2 + (y - b)^2$
$x^2 + y^2 = x^2 + y^2 - 2by + b^2$
$\Rightarrow 2by = b^2$
$\Rightarrow\ \text{y}=\frac{\text{b}}{2}$
$\therefore$ Co$-$ordinates of circumcentre are $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big).$ View full question & answer→MCQ 321 Mark
If $A(x, 2), B(-3, -4)$ and $C(7, -5)$ are collinear, then the value of $x$ is:
Answer$A(x, 2), B(-3, -4)$ and $C(7, -5)$ are collinear, then area $\triangle ABC = 0$
Now area of $\triangle\text{ABC}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{x}(-4+5)+(-3)(-5-2)+7(2+4)]$
$=\frac{1}{2}[\text{x}+21+42]$
$=\frac{1}{2}(63+\text{x})$
$\therefore$ Points are collinear
$\therefore$ area $\triangle \text{ABC} = 0$
$=\frac{1}{2}(63+\text{x})=0$
$\Rightarrow\ 63+\text{x}=0$
$\Rightarrow\ \text{x}=-63$
View full question & answer→MCQ 331 Mark
If $A(2, 2), B(-4, -4)$ and $C(5, -8)$ are the vertices of a triangle, then the length of the median through vertices $C$ is:
- A
$\sqrt{65}$
- B
$\sqrt{117}$
- ✓
$\sqrt{85}$
- D
$\sqrt{113}$
AnswerCorrect option: C. $\sqrt{85}$
Let midpoint of $A(2, 2), B(-4, -4)$ be $D$ whose coordinates will be
$=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)=\Big(\frac{2-4}{2},\frac{2-4}{2}\Big)$
or $\Big(\frac{-2}{2},\frac{-2}{2}\Big)$ or $(-1, -1)$
$\therefore$ Length of median $CD$
$=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(5+1)^2+(-8+1)^2}$
$=\sqrt{(6)^2+(-7)^2}$
$=\sqrt{36+49}$
$=\sqrt{85}\text{ units}$
View full question & answer→MCQ 341 Mark
The ratio in which the x-axis divides the segment joining $(3, 6)$ and $(12, -3)$ is:
- ✓
$2 : 1$
- B
$1 : 2$
- C
$-2 : 1$
- D
$1 : -2$
AnswerCorrect option: A. $2 : 1$
Let $P(x, 0)$ be the point of intersection of x-axis with the line segment joining $A(3, 6)$ and $B(12, -3)$ which divides the line segment AB in the ratio $\lambda:1.$
Now, according to the section formula if point a point P divides a line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m : n$ internally than,
$\text{P(x, y)}=\Big(\frac{\text{nx}_1+\text{mx}_2}{\text{m}+\text{n}},\frac{\text{ny}_1+\text{my}_2}{\text{m}+\text{n}}\Big)$
Now we will use section formula as,
$(\text{x},0)=\Big(\frac{12\lambda+3}{\lambda+1},\frac{-3\lambda+6}{\lambda+1}\Big)$
Now equate the y component on both the sides,
$\frac{-3\lambda+6}{\lambda+1}=0$
On further simplification,
$\lambda=\frac{2}{1}$
So, $x$-axis divides AB in the ratio $\frac{2}{1}.$
View full question & answer→MCQ 351 Mark
The coordinates of a point on $x-$axis which lies on the perpendicular bisector of the line segment joining the points $(7, 6)$ and $(-3, 4)$ are,
- A
$(0, 2)$
- ✓
$(3, 0)$
- C
$(0, 3)$
- D
$(2, 0)$
AnswerCorrect option: B. $(3, 0)$
The given point $P$ lies on $x-$axis.
Let the co$-$ordinates of $P$ be $(x, 0).$
The point $P$ lies on the perpendicular bisector of of the line segment joining the points $A(7, 6), B(-3, 4)$
$\therefore$ $PA = PB $
$\Rightarrow PA^2 = PB^2$
$\Rightarrow (x - 7)^2 + (0 - 6)^2 = (x + 3)^2 + (0 - 4)^2$
$\Rightarrow x^2 - 14x + 49 + 36 = x^2 + 6x + 9 + 16$
$\Rightarrow -14x + 85 = 6x + 25$
$\Rightarrow 6x + 14x = 85 - 25$
$\Rightarrow 20x = 60$
$\Rightarrow\ \text{x}=\frac{60}{20}=3$
$\therefore$ Co$-$ordinates of $P$ will be $(3, 0).$
View full question & answer→MCQ 361 Mark
The distance of the point $(4, 7)$ from the $x-$axis is:
- A
$4$
- ✓
$7$
- C
$11$
- D
$\sqrt{65}$
AnswerThe distance of the point $A(4, 7)$ from $x-$axis is $B(x, 0)$
where $x = 4$
$\text{AB}=\sqrt{(4-4)^2+(0-7)^2}$
$=\sqrt{0^2+49}$
$=7$
View full question & answer→MCQ 371 Mark
If points $A(5, p)$,$ B(1, 5)$, $C(2, 1)$ and$ D(6, 2)$ form a square $\text {ABCD}$, then $\text {p} =$
AnswerIn a square all the sides are equal to each other.
Here the four points are $A(5, p), B(1,5), C(2,1)$ and $D(6,2)$.
The vertex $'A'$ should be equidistant from $' B \ '$ as well as $' D\ '.$
Let us now find out the distance $'AB\ '$ and $'AD\ '.$
$AB=\sqrt{(5-1)^2+(p-5)^2}$
$AB=\sqrt{(4)^2+(p-5)^2}$
$AD=\sqrt{(5-6)^2+(p-2)^2}$
$AD=\sqrt{(-1)^2+(p-2)^2}$
These two need to be equal.
Equating the above two equations we have,
$AB=AD$
$\sqrt{(4)^2+(p-5)^2}=\sqrt{(-1)^2+(p-2)^2}$
Squaring on both sides we have,
$(4)^2+(p-5)^2=(-1)^2+(p-2)^2$
$16+p^2+25-10 p=1+p^2+4-4 p$
$6 p=36$
$p=6$
View full question & answer→MCQ 381 Mark
The distance of the point $(4, 7)$ from the $y-$axis is:
- ✓
$4$
- B
$7$
- C
$11$
- D
$\sqrt{65}$
AnswerThe distance of the point $(4, 7)$ from $y-$axis means $B(0, y)$ where $y = 7$
$\text{AB}=\sqrt{(0,-4)^2+(7-7)^2}$
$=\sqrt{(-4)^2+(0)^2}$
$=\sqrt{16+0}$
$=\sqrt{16}$
$=4$
View full question & answer→MCQ 391 Mark
If the centroid of the triangle formed by the points $(a, b), (b, c)$ and $(c, a)$ is at the origin, then $a^3 + b^3 + c^3 =$
- A
$abc$
- B
$0$
- C
$a + b + c$
- ✓
$3abc$
AnswerCorrect option: D. $3abc$
Centroid of the triangle formed by the points $(a, b), (b, c)$ and $(c, a)$ is origin $(0, 0).$
$\therefore\ \frac{\text{a}+\text{b}+\text{c}}{3}=0$
$\Rightarrow a + b + c = 0$
$\therefore$ $a^3 + b^3 + c^3 = 3abc$
$\because$ $a + b + c = 0$
Hence $a^3 + b^3 + c^3 = 3abc$
View full question & answer→MCQ 401 Mark
The area of the triangle formed by $(a, b + c), (b, c + a)$ and $(c, a + b)$ is:
- A
$a + b + c$
- B
$abc$
- C
$(a + b + c)^2$
- ✓
$0$
AnswerVertices of a triangle are $(a, b + c), (b, c + a)$ and $(c, a + b)$
Area of $\triangle=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{a}(\text{c}+\text{a}-\text{a}-\text{b})+\text{b}(\text{a}+\text{b}-\text{b}-\text{c})+\text{c}(\text{b}+\text{c}-\text{c}-\text{a})]$
$=\frac{1}{2}[\text{a}(\text{c}-\text{b})+\text{b}(\text{a}-\text{c})+\text{c}(\text{b}-\text{a})]$
$=\frac{1}{2}[\text{ac}-\text{ab}+\text{ab}-\text{bc}+\text{bc}-\text{ac}]$
$=\frac{1}{2}\times0$
$=0$
View full question & answer→MCQ 411 Mark
The distance between the points $(\cos\theta,\sin\theta)$ and $(\sin\theta,-\cos\theta)$ is:
- A
$\sqrt{3}$
- ✓
$\sqrt{2}$
- C
$2$
- D
$1$
AnswerCorrect option: B. $\sqrt{2}$
We have to find the distance between $\text{A}(\cos\theta,\sin\theta)$ and $\text{B}(\sin\theta,-\cos\theta).$
In general, the distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by,
$\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
So, $\text{AB}=\sqrt{(\sin\theta-\cos\theta)^2+(-\cos\theta-\sin\theta)^2}$
$=\sqrt{2(\sin^2\theta+\cos^2\theta)}$
But according to the trigonometric identity,
$\sin^2\theta+\cos^2\theta=1$
Therefore, $\text{AB}=\sqrt{2}$
View full question & answer→MCQ 421 Mark
The distance between the points $(\text{a}\cos\theta+\text{b}\sin\theta,0)$ and $(0,\text{a}\sin\theta-\text{b}\cos\theta)$ is:
AnswerCorrect option: D. $\sqrt{\text{a}^2+\text{b}^2}$
We have to find the distance between $\text{A}(\text{a}\cos\theta+\text{b}\sin\theta,0)$ and $\text{B}(0,\text{a}\sin\theta-\text{b}\cos\theta).$
In general, the distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by,
$\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
So, $=\sqrt{(\text{a}\cos\theta+\text{b}\sin\theta)^2+(-\text{a}\sin\theta+\text{b}\cos\theta)^2}$
$=\sqrt{\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta}$
$\text{AB}=\sqrt{(\text{a}\cos\theta+\text{b}\sin\theta-0)^2+(0-\text{a}\sin\theta+\text{b}\cos\theta)^2}$
$=\sqrt{\text{a}^2(\sin^2\theta+\cos^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)}$
But according to the trigonometric identity,
$\sin^2\theta+\cos^2\theta=1$
Therefore, $\text{AB}=\sqrt{\text{a}^2+\text{b}^2}$
View full question & answer→MCQ 431 Mark
If the area of the triangle formed by the points $(x, 2x), (-2, 6)$ and $(3, 1)$ is $5$ square units, then $x =$
- A
$\frac{2}{3}$
- B
$\frac{3}{5}$
- ✓
$2$
- D
$5$
AnswerArea of triangle whose vertices are $(x, 2x), (-2, 6)$ and $(3, 1)$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{x}(6-1)+(-2)(1-2\text{x})+3(2\text{x}-6)]$
$=\frac{1}{2}[5\text{x}-2+4\text{x}+6\text{x}-18]$
$=\frac{1}{2}[15\text{x}-20]$
$\because$ Area $=$ 5sq. units
$\therefore\ \frac{1}{2}(15\text{x}-20)=5$
$\Rightarrow\ 15\text{x}-20=10$
$\Rightarrow\ 15\text{x}=10+20=30$
$\Rightarrow\ \text{x}=\frac{30}{15}=2$
$\therefore\ \text{x}=2$
View full question & answer→MCQ 441 Mark
If the points $(k, 2k), (3k, 3k)$ and $(3, 1)$ are collinear, then $k$:
- A
$\frac{1}{3}$
- ✓
$\frac{-1}{3}$
- C
$\frac{2}{3}$
- D
$\frac{-2}{3}$
AnswerCorrect option: B. $\frac{-1}{3}$
We have three collinear points $A(k, 2k), B(3k, 3k)$ and $C(3, 1).$
In general if $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ are collinear then, area of the triangle is $0.$
$\frac{1}{2}\Big[\text{x}_1(\text{y}_2 - \text{y}_3) + \text{x}_2(\text{y}_3 - \text{y}_1) + \text{x}_3(\text{y}_1 - \text{y}_2)\Big] = 0$
So, $k(3k - 1) + 3k(1 - 2k) + 3(2k - 3k) = 0$
So, $-3k^2 - k = 0$
Take out the common terms,
$-k(3k + 1) = 0$
Therefore, $\text{k}=-\frac{1}{3}$
View full question & answer→MCQ 451 Mark
The length of a line segment joining $A(2, -3)$ and $B$ is $10$ units. If the abscissa of $B$ is $10$ units, then its ordinates can be,
- ✓
$3$ or $-9$
- B
$-3$ or $9$
- C
$6$ or $27$
- D
$-6$ or $-27$
AnswerCorrect option: A. $3$ or $-9$
It is given that distance between $P(2, -3)$ and $Q(10, y)$ is $10$.
In general, the distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by,
$AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
So, $10^2 = (10 - 2)^2 + (y + 3)^2$
On further simplification,
$(y + 3)^2 = 36$
$\text{y}=-3\pm6$
$= -9, 3$
$y = -9, 3$
View full question & answer→MCQ 461 Mark
The coordinates of the point $P$ dividing the line segment joining the points $A(1, 3)$ and $B(4, 6)$ in the ratio $2 : 1$ are:
- A
$(2, 4)$
- ✓
$(3, 5)$
- C
$(4, 2)$
- D
$(5, 3)$
AnswerCorrect option: B. $(3, 5)$
Point $P$ divides the line segment joining the points $A(1, 3)$ and $B(4, 6)$ in the ratio $2 : 1$
Let coordinates of $P$ be $(x, y),$ then
$\text{x}=\frac{\text{m}_1\text{x}_2+\text{m}_2\text{x}_1}{\text{m}_1+\text{m}_2}=\frac{2\times4+1\times1}{2+1}$
$=\frac{8+1}{3}=\frac{9}{3}=3$
$\text{y}=\frac{\text{m}_1\text{y}_2+\text{m}_2\text{y}_1}{\text{m}_1+\text{m}_2}=\frac{2\times6+1\times3}{2+1}$
$=\frac{12+3}{3}$
$=\frac{15}{3}$
$=5$
$\therefore$ Coordinates of $P$ are $(3, 5).$
View full question & answer→MCQ 471 Mark
The point on the $x-$axis which is equidistant from points $(-1, 0)$ and $(5, 0)$ is:
- A
$(0, 2)$
- ✓
$(2, 0)$
- C
$(3, 0)$
- D
$(0, 3)$
AnswerCorrect option: B. $(2, 0)$
$\text{PA}=\sqrt{(\text{x}+1)^2+(0-0)^2}$
$\text{PA}=\sqrt{(\text{x}-1)^2}\ \ ...(\text{i})$
Similarly,
$\text{PB}=\sqrt{(\text{x}-5)^2+(0-0)^2}$
$\text{PB}=\sqrt{(\text{x}-5)^2}\ \ ...\text{(ii)}$
From equation $(i)$ and $(ii)$
$\text{PA}=\text{PB}$
$\sqrt{(\text{x}+1)^2}=\sqrt{(\text{x}-5)^2}$
squaring both the side
$(\text{x}+1)^2=(\text{x}-5)^2$
$\text{x}^2+2\text{x}+1=\text{x}^2+25-10\text{x}$
$10\text{x}+2\text{x}=25-1$
$12\text{x}=24$
$\text{x}=2$
Correct option $(b) (2, 0).$ View full question & answer→MCQ 481 Mark
If $A(4, 9), B(2, 3)$ and $C(6, 5)$ are the vertices of $\triangle ABC,$ then the length of median through $C$ is:
- A
$5$ untis
- ✓
$\sqrt{10}\text{ units}$
- C
$25$ units
- D
$10$ units
AnswerCorrect option: B. $\sqrt{10}\text{ units}$
$A(4, 9), B(2, 3)$ and $C(6, 5)$ are the vertices of $\triangle ABC$
Let median $CD$ has been drawn $C(6, 5)$
$\therefore D$ is mid$-$point of $AB$
$\text{D}=\Big(\frac{4+2}{2},\frac{9+3}{2}\Big)$
$\therefore D(3, 6)$
$\therefore$ Length of $\text{CD}=\sqrt{(6-3)^2+(5-6)^2}$
$=\sqrt{3^2+(-1)^2}$
$=\sqrt{9+1}$
$=\sqrt{10}\text{ units}$ View full question & answer→MCQ 491 Mark
If the point $(x, 4)$ lies on a circle whose centre is at the origin and radius is $5$, then $x =$
- A
$\pm5$
- ✓
$\pm3$
- C
$0$
- D
$\pm4$
AnswerCorrect option: B. $\pm3$
Point $A(x, 4)$ is on a circle with centre $O(0, 0)$ and radius $= 5$
$\therefore\ \text{OA}=\sqrt{(\text{x}-0)^2+(4-0)^2}=\sqrt{\text{x}^2+16}$
$\therefore\ \sqrt{\text{x}^2+16}=5$
Squaring both sides,
$\Rightarrow\ \text{x}^2+16=25$
$\Rightarrow\ \text{x}^2=25-16=9=(\pm3)^2$
$\therefore\ \text{x}=\pm3$
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