2 Marks Questions

Question 12 Marks

Find the value of a , if the distance between the points $A(-3, -14)$ and $B(a, -5)$ is $9$ units.

Answer

According to the question, Distance between $A (-3, -14)$ and $8 (a, -5), AB = 9$
$\Big[\because$ distance between two point $(x_1, y_1)$ and $(x_2, y_2)$ $=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\Rightarrow\sqrt{(\text{a}+3)^2+(-5+14)^2}=9$
$\Rightarrow\sqrt{(\text{a}+3)^2+(9)^2}=9$
On squaring both the sides, we get
$\Rightarrow (a + 3)^2 + 81 = 81$
$\Rightarrow (a + 3)^2 = 0$
$\Rightarrow a = -3$
Hence, the required of a is$ -3.$

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Question 22 Marks

Find the area of the triangle whose vertices are $(-8, 4), (-6, 6)$ and $(-3, 9).$

Answer

Given that, the vertices of triangles
Let $(x_1, y_1) → (-8, 4)$
$(x_2, y_2) → (-6, 6)$
and $(x_3, y_3) → (-3, 9)$
We know that, the area of triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$
$\triangle=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\triangle=\frac{1}{2}[-8(6-9)-6(9-4)+(-3)(4-9)]$
$\triangle=\frac{1}{2}[-8(-3)-6(5)-3(-2)]$
$\triangle=\frac{1}{2}(24-30+6)$
$\triangle=\frac{1}{2}(30-30)$
$\triangle=\frac{1}{2}(0)$
$\triangle=0$
Hence, the required area of triangle is 0.

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Question 32 Marks

The points $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$ are the vertices of $\triangle\text{ABC}.$
The median from A meets BC at D. Find the coordinates of the point D.

Answer

Median from A meets BC at D i. e., D is the mid-point of BC.
So, the coordinates of D are given by $\Big(\frac{\text{x}_2+\text{x}_3}{2},\frac{\text{y}_2+\text{y}_3}{2}\Big)$

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Question 42 Marks

The points $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$ are the vertices of $\triangle\text{ABC}.$
What are the coordinates of the centroid of the triangle ABC?

Answer

Coordinates of centroid G of $\triangle\text{ABC}$ are
$\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)$
It is observed that coordinates of P, Q, R and G are same.
Hence, the mediance intersect at the same point i. e., centroid which divides the medians in the ratio 2 : 1.

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