1 Marks Question

Question 11 Mark

Find the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6).$

Answer

Let $(-1, 6)$ divides line segment joining the points $(-3, 10)$ and$(6, -8)$ in $k:1.$
Using Section formula, we get
$ - 1 = \frac{{( - 3) \times 1 + 6 \times k}}{{k + 1}}$
$4 \Rightarrow - k - 1 = ( - 3 + 6k)$
$\Rightarrow −7k = −2 \Rightarrow k=$ $\frac{2}{7}$
Therefore, the ratio is $\frac{2}{7}:1$ which is equivalent to $2:7$.
Therefore, $(-1, 6)$ divides line segment joining the points $(-3, 10)$ and $(6, -8)$ in $2:7$.

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Question 21 Mark

Find the area of the rhombus if its vertices are $(3, 0), (4, 5), (-1, 4)$ and $(-2, -1)$ taken in order.
[Hint: Area of a rhombus $=\frac{1}{2}$ (product of its diagonals)]

Answer

Let $A (3, 0), B (4, 5), C (-1, 4)$ and $D (-2, -1)$
$AC = \sqrt {{{( - 1 - 3)}^2} + {{(4 - 0)}^2}} = 4\sqrt 2 $
$BD = \sqrt {{{( - 2 - 4)}^2} + {{( - 1 - 5)}^2}} = \sqrt {36 + 36} = 6\sqrt 2 $
Area of rhombus = $\frac{1}{2}{d_1} \times {d_2}$
$ = \frac{1}{2}AC \times BD$
$ = \frac{1}{2} \times 4\sqrt 2 \times 6\sqrt 2 = 24$ Sq. unit.

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Question 31 Mark

Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$.

Answer

Let the coordinates of the required point be $(x, y)$. Then,
$x = \frac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}}$
$= \frac{{(2)(4) + (3)( - 1)}}{{2 + 3}}$
$= \frac{{8 - 3}}{5} = \frac{5}{5} = 1$
$y = \frac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}$
$ = \frac{{(2)( - 3) + (3)(7)}}{{2 + 3}}$
$ = \frac{{ - 6 + 21}}{5} = \frac{{15}}{5} = 3$
Hence, the required point is $(1, 3)$.

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Question 41 Mark

Find the distance between the pair of points $(a, b), (-a, -b)$

Answer

$(a, b), (-a, -b)$
Required distance
$= \sqrt {{{( - a - a)}^2} + {{( - b - b)}^2}}$
$= \sqrt {{{( - 2a)}^2} + {{( - 2b)}^2}}$
$= \sqrt {4{a^2} + 4{b^2}}$
$= \sqrt {4({a^2} + {b^2})}$
$= 2\sqrt {{a^2} + {b^2}}$

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Question 51 Mark

Find the distance between the pair of points $(-5, 7), (-1, 3)$

Answer

$(-5, 7), (-1, 3)$
Required distance
$= \sqrt {([- 1 - {{(-5)}^2]} + {{(3- 7)}^2}}$
$= \sqrt {16 + 16} = \sqrt {32}$
$= 4\sqrt 2$

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Question 61 Mark

Find the distance between the pair of points: $(2, 3), (4, 1)$

Answer

Applying Distance Formula to find distance between points $(2, 3)$ and $(4,1),$ we get
$d =$ $\sqrt {{{(4 - 2)}^2} + {{(1 - 3)}^2}} = \sqrt {{{(2)}^2} + {{( - 2)}^2}} = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2 \;units$

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Question 71 Mark

Find the ratio in which the $y$-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$. Also find the point of intersection.

Answer

Let the point on $y$-axis be $P(0, y)$ and $AP: PB = K: 1$
Therefore $\frac { 5 - k } { k + 1 } = 0$ gives $k = 5$
Hence required ratio is $5: 1$
$y = \frac { - 4 ( 5 ) - 6 } { 6 } = \frac { - 13 } { 3 }$
Hence point on y-axis is $\left( 0 , \frac { - 13 } { 3 } \right)$.

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Question 81 Mark

In what ratio does the point $(-4, 6)$ divide the line segment joining the points $A(-6, 10)$ and $B(3, -8)?$

Answer

Let $(-4, 6)$ divide $AB$ internally in the ratio $k:1$. Using the section formula, we get
$( - 4,6 ) = \left( \frac { 3 k - 6 } { k + 1 } , \frac { - 8 k + 10 } { k + 1 } \right)$
So, $- 4 = \frac { 3 k - 6 } { k + 1 }$
$i.e., -4k - 4 = 3k - 6$
$i.e., 7k = 2$
$i.e., k:1 = 2:7$
The same can be checked for the y-coordinate also.
Therefore, the ratio in which the point $(-4,6)$ divides the line segment $AB$ is $2: 7$.

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Question 91 Mark

If the points $A (6, 1), B (8, 2), C (9, 4)$ and $D (p, 3)$ are the vertices of a parallelogram, taken in order, find the value of $p$.

Answer

We know that the diagonals of a parallelogram bisect each other.
So, coordinates of the mid-point of diagonal $AC$ are same as the coordinates of the mid-point of diagonal $BD$.
$ \therefore \quad \left( \frac { 6 + 9 } { 2 } , \frac { 1 + 4 } { 2 } \right) = \left( \frac { 8 + p } { 2 } , \frac { 2 + 3 } { 2 } \right)$
$\Rightarrow \quad \left( \frac { 15 } { 2 } , \frac { 5 } { 2 } \right) = \left( \frac { 8 + p } { 2 } , \frac { 5 } { 2 } \right)$
$\Rightarrow \quad \frac { 15 } { 2 } = \frac { 8 + p } { 2 }$$\Rightarrow 15 = 8 + p \Rightarrow p = 7$

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