3 Marks Question

Question 13 Marks

qussion The points $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$ are the vertices of $\triangle A B C$. Find the coordinates of the point $P$ on $A D$ such that $A P: P D=2: 1$

Answer

The coordinates of the point P on AD such that AP : PD = 2 : 1 are given by
$\text{x}=\frac{2\Big(\frac{\text{x}_2+\text{x}_3}{2}\Big)+1(\text{x}_1)}{2+1},$ $\text{y}=\frac{2\Big(\frac{\text{y}_2+\text{y}_3}{2}\Big)+1(\text{y}_1)}{2+1}$
$\Rightarrow\text{x}=\frac{\text{x}_2+\text{x}_3+\text{x}_1}{3},$ $\text{y}=\frac{\text{y}_2+\text{y}_3+\text{y}_1}{3}$
$\therefore\text{P}\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)$ is the required point.

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Question 23 Marks

Find the value of m if the points (5, 1), (-2, -3) and (8, 2m ) are collinear.

Answer

Let $\text{A}\equiv(\text{x}_1,\text{ y}_1)\equiv(5, 1),\text{ B}\equiv(\text{x}_2,\text{ y}_2)\equiv(-2,-3),$
and$\text{ C}\equiv(\text{x}_3,\text{ y}_3)\equiv(8,2\text{m})$
Since, the points $\text{A}\equiv(5,1),\text{B}\equiv(-2,-3)$
and $\text{C}\equiv(8,2\text{m})$ are collinear.
$\therefore\text{ Area of }\triangle\text{ABC}=0$
$\Rightarrow\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]=0$
$\Rightarrow\frac{1}{2}[5(-3-2\text{m})+(-2)(2\text{m}-1)+8\{1-(-3)\}]=0$
$\Rightarrow\frac{1}{2}(-15-10\text{m}-4\text{m}+2+32)=0$
$\Rightarrow\frac{1}{2}(-14\text{m}+19)=0$
$\Rightarrow\text{m}=\frac{19}{14}$
Hence, the required value of m is $\frac{19}{14}.$

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Question 33 Marks

Find the points on the x–axis which are at a distance of $2\sqrt{5}$ from the point $(7, -4)$. How many such points are there?

Answer

We know that, every point on the X-axis in the form (x, 0). Let P(x, 0) the point on the X-axis have $2\sqrt{5}$ distance from the point Q (7, -4).
$\Big[\because$ distance formula $=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
By given condition, $\text{PQ}=2\sqrt{5}$
$(PQ)^2= 4 \times 5$
$\Rightarrow (x - 7)^2 + (0 + 4)^2 = 20$
$\Rightarrow X^2 + 49 - 14x + 16 = 20$
$\Rightarrow x^2 - 14x + 65 - 20 = 0$
$\Rightarrow x^2 - 14x + 45 = 0$
$\Rightarrow x^2- 9x - 5x + 45 = 0$ [by factorisation method]
$\Rightarrow x(x - 9) - 5(x - 9) = 0$
$\Rightarrow (x - 9)(x - 5) = 0$
$x = 5, 9$
Hence, there are two points lies on the axis, which are $(5, 0)$ and $(9, 0)$, have $2\sqrt{5}$ distance from the point $(7, – 4).$

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Question 43 Marks

Name the type of triangle formed by the points $A(-5, 6), B(-4, -2)$ and $C(7, 5).$

Answer

To find the type of triangle, first we determine the length of all three sides and see whatever condition of triangle is satisfy by these sides. Now, using distance formula between two points,
$\Big[\because\text{d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\Rightarrow\text{AB}=\sqrt{(-4+5)^2+(-2-6)^2}$
$\Rightarrow\text{AB}=\sqrt{(1)^2+(-8)^2}$
$\Rightarrow\text{AB}=\sqrt{1+64}$
$\Rightarrow\text{AB}=\sqrt{65}$
$\Rightarrow\text{BC}=\sqrt{(7+4)^2+(5+2)^2}$
$\Rightarrow\text{BC}=\sqrt{(11)^2+(7)^2}$
$\Rightarrow\text{BC}=\sqrt{121+49}$
$\Rightarrow\text{BC}=\sqrt{170}$
$\Rightarrow\text{CA}=\sqrt{(-5-7)^2+(6-5)^2}$
$\Rightarrow\text{CA}=\sqrt{(-12)^2+(1)^2}$
$\Rightarrow\text{CA}=\sqrt{144+1}$
$\Rightarrow\text{CA}=\sqrt{145}$
we see that, $\text{AB}\neq\text{BC}\neq\text{CA}$
and not hold the condition of Pythagoras in a $\triangle\text{ABC}.$
i.e., (Hypotenuse)$^2$ = ( Base)$^2$ + (Perpendicular)$^2$
Hence, the required triangle is scalene because all of its sides are not equal i.e., different to each other.

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