2 Marks Questions

Question 12 Marks

Find the distance between the points:
$A(1, -3)$ and $B(4, -6)$

Answer

The given points are $A(1, -3)$ and $B(4, -6)$
Then, $\left(x_1=1, y_1=-3\right)$ and $\left(x_2=4\right.$ and $\left.y_2=-6\right)$
$\therefore\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{4}-\text{1})^2+(-\text{6}+\text{3})^2}$
$=\sqrt{(\text{3})^2+(-\text{3})^2}$
$=\sqrt{\text{9}+\text{9}}$
$=\sqrt{\text{18}}=3\sqrt{2}\text{ units}.$

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Question 22 Marks

If the coordinates of points $A$ and $B$ are $(-2, -2)$ and $(2, -4)$ respectively, find the coordinates of the points $P$ such that $\text{AP}=\frac{3}{7}\text{AB},$ where $P$ lies on the line segment $AB$.

Answer

Here, $P(x, y)$ divides line segment $AB$, such that
$\text{AP}=\frac{3}{7}\text{AB}$
$\Rightarrow\frac{\text{AP}}{\text{AB}}=\frac{3}{7}$
$\Rightarrow\frac{\text{AB}}{\text{AP}}=\frac{7}{3}$
$\Rightarrow\frac{\text{AB}}{\text{AP}}-1=\frac{7}{3}-1$
$\Rightarrow\frac{\text{AB}-\text{AP}}{\text{AP}}=\frac{7-3}{3}$
$\Rightarrow\frac{\text{BP}}{\text{AP}}=\frac{4}{3}$
$\Rightarrow\frac{\text{AP}}{\text{BP}}=\frac{3}{4}$
$\therefore$ $P$ divides $AB$ in the ratio $3 : 4$
$\text{x}=\frac{3\times2+4(-2)}{3+4};\text{y}=\frac{3\times(-4)+4(2)}{3+4}$
$\text{x}=\frac{6-8}{7};\text{y}=\frac{-12-8}{7}$
$\text{x}=\frac{-2}{7};\text{y}=\frac{-20}{7}$
$\therefore$ The coordinate of $P$ are $\Big(\frac{-2}{7},\frac{-20}{7}\Big).$

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Question 32 Marks

Find the coordinate of the points of trisection of the line segment joining the points $A(7, -2)$ and $B(1, -5)$.

Answer

We have given that the coordinates of the Points of trisection of the line segment joining the points $A(7, -2)$ and $B(1, -5)$
Therefore, the line $AB$ divides sectionally into two parts $C$ and $D$ into $1 : 2$ and $2 : 1$
Now, we have use the Section Formula,
For $1 : 2$ $= \text{ma}+\frac{\text{nb}}{\text{a+b}}, \text{ma}+\frac{\text{nb}}{\text{a+b}} $
$\Rightarrow1\times1+2\times\frac{7}{1+2}, 1\times(-5)+2\times(-2)\frac{7}{1+2})$
$ = \frac{15}{3}, \frac{-9}{3} = (5, -3)$
Similarly, For the Ratio $2 : 1$
For $1 : 2$ $\Rightarrow\frac{\text{ma+nb}}{\text{a+b}},\frac{\text{ma+nb}}{\text{a+b}}$$=2\times1+1\times\frac{7}{2}+1, 2\times(-5)+1\times\frac{(-2)}{2+1} -4)$
$= \frac{9}{3}, \frac{-12}{3} = (3, -4)$
Hence, the coordinates of the point of the trisection of the line segments are $C(5, -3)$ and $D(3, -4)$

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Question 42 Marks

Find the area of $\triangle\text{ABC}$ whose vertices are:
$A(3, 8), B(-4, 2)$ and $C(5, -1)$

Answer

$A(3, 8), B(-4, 2)$ and $C(5, -1)$ are the vertices of $\triangle\text{ABC}.$
Then, $\left(x_1=3, y_1=8\right),\left(x_2=-4, y_2=2\right)$ and $\left(x_3=5, y_3=-1\right)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{3(2-(-1))+(-4)(-1-8)+5(8-2)\big\}$
$=\frac{1}{2}\big\{3(2+1)-4(-9)+5(6)\big\}$
$=\frac{1}{2}\big\{9+36+30\big\}$
$=\frac{1}{2}(75)$
$=37.5\ \text{sq}.\text{units}$

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Question 52 Marks

Find the distance between the points:
$A(-6, -4)$ and $B(9, -12)$.

Answer

The given points are $A(-6, -4)$ and $B(9, -12)$
Then,$\left(x_1=-6, y_1=-4\right)$ and $\left(x_2=9\right.$ and $\left.y_2=-12\right)$
$\therefore\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{6}+\text{9})^2+(-\text{12}+\text{4})^2}$
$=\sqrt{(\text{15})^2+(-\text{8})^2}$
$=\sqrt{\text{225}+\text{64}}$
$=\sqrt{\text{289}}=17\text{ units}.$

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Question 62 Marks

Find the distance between the points:
$A(9, 3)$ and $B(15, 11)$.

Answer

The given points are $A(9, 3)$ and $B(15, 11)$
Then, $\left(x_1=9, y_1=3\right)$ and $\left(x_2=15\right.$ and $\left.y_2=11\right)$
$\therefore\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{15}-\text{9})^2+(\text{11}-\text{3})^2}$
$=\sqrt{\text{6}^2+\text{8}^2}$
$=\sqrt{\text{36}+\text{64}}$
$=\sqrt{\text{100}}=10\text{ units}.$

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Question 72 Marks

Find the distance between the points:
$A(7, -4)$ and $B(-5, 1)$

Answer

The given points are $A(7, 4)$ and $B(-5, 1)$
Then, $\left(x_1=7, y_1=-4\right)$ and $\left(x_2=-5\right.$ and $\left.y_2=1\right)$
$\therefore\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(-\text{5}-\text{7})^2+(\text{1}+\text{4})^2}$
$=\sqrt{(-\text{12})^2+(\text{5})^2}$
$=\sqrt{\text{144}+\text{25}}$
$=\sqrt{\text{169}}=13\text{ units}.$

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Maths 2 Marks Questions

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