Question 1: If the line segment joining the points $\displaystyle P (x_1 , y_1) \text{ and } Q (x_2 , y_2)$ subtends an angle $\displaystyle \alpha$ at the origin $\displaystyle O$, prove that: $\displaystyle OP \cdot OQ \cos \alpha = x_1x_2 + y_1y_2$. Consider the figure given. $\displaystyle OP^2 = {x_1}^2 + {y_1}^2$ $\displaystyle OQ^2 = {x_2}^2 + {y_2}^2$ $\displaystyle \therefore PQ^2 = ( x_2-x_1)^2 + ( y_2-y_1)^2$

Using cosine formula in $\displaystyle \triangle OPQ$ we get $\displaystyle PQ^2 = OP^2 + OQ^2 - 2 \cdot OP \cdot OQ. \cos \alpha$ $\displaystyle \Rightarrow ( x_2-x_1)^2 + ( y_2-y_1)^2 = ( {x_1}^2+{y_1}^2) + ( {x_2}^2+{y_2}^2) - 2 \cdot OP \cdot OQ. \cos \alpha$ $\displaystyle \Rightarrow {x_2}^2 + {x_1}^2 - 2 x_1 x_2 + {y_2}^2 + {y_1}^2 - 2y_1y_2 = {x_2}^2 + {x_1}^2 + {y_2}^2 + {y_1}^2 - 2 \cdot OP \cdot OQ. \cos \alpha$ $\displaystyle \Rightarrow 2 \cdot OP \cdot OQ. \cos \alpha = 2 x_1 x_2 + 2y_1y_2$ $\displaystyle \Rightarrow \cdot OP \cdot OQ. \cos \alpha = x_1 x_2 +y_1y_2$ $\displaystyle \\$

Question 2: The vertices of a $\displaystyle \triangle ABC \text{ and } A (0,0), B (2, -1) \text{ and } C (9,2)$. Find $\displaystyle \cos B$ $\displaystyle \text{Given } \triangle ABC, A ( 0,0), B( 2, -1) , C( 9, 2)$ $\displaystyle \text{We know } \cos B = \frac{a^2 + c^2 - b^2}{2ac} \text{ where } a = BC, b = CA, c = AB$ $\displaystyle a = BC = \sqrt{(2-9)^2 + ( -1-2)^2} = \sqrt{49 + 9} = \sqrt{58}$ $\displaystyle b = CA= \sqrt{(0-9)^2 + ( 2-0)^2} = \sqrt{81 + 4} = \sqrt{85}$ $\displaystyle c = AB= \sqrt{(0-2)^2 + ( 0+1)^2} = \sqrt{4 + 1} = \sqrt{5}$ $\displaystyle \therefore \cos B = \frac{58+5-85}{2 \cdot \sqrt{58} \cdot \sqrt{5}} = \frac{-11}{\sqrt{290}}$ $\displaystyle \\$

Question 3: Four points $\displaystyle A(6,3), B(-3, 5) , C(4, -2$) and $\displaystyle D(x, 3x)$ are given in such a way that $\displaystyle \frac{\triangle DBC}{\triangle ABC}= \frac{1}{2}$ , find $\displaystyle x$. $\displaystyle \text{Given } A(6,3), B(-3, 5) , C(4, -2$) and $\displaystyle D(x, 3x)$

We know that the area of a Triangle with vertices $\displaystyle (x_1, y_1), (x_2, y_2),(x_3, y_3)$

Area $\displaystyle = \frac{1}{2} \Big[ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3( y_1-y_2) \Big]$ $\displaystyle \therefore \ Ar. \ \triangle DBA = \frac{1}{2} \Big[ -3( -2-3x) + 4 ( 3x-5) + x( 5+2) \Big] = 7(2x-1)$

Also Area of $\displaystyle \triangle ABC = \frac{1}{2} \Big[ 6( 5+2( - 3 ( -2-3) + 4( 3 - 5) \Big] = \frac{49}{2}$ $\displaystyle \text{Given } \frac{\triangle DBA}{\triangle ABC} = \frac{7(2x-1)}{\frac{49}{2}}$ $\displaystyle \Rightarrow 14( 2x-1) = \frac{49}{2}$ $\displaystyle \Rightarrow 8x - 4 = 7$ $\displaystyle \Rightarrow x = \frac{11}{8}$ $\displaystyle \\$

Question 4: The points $\displaystyle A (2, 0), B (9,1), C (11, 6) \text{ and } D (4, 4)$ are the vertices of a quadrilateral $\displaystyle ABCD$. Determine whether $\displaystyle ABCD$ is a rhombus or not. $\displaystyle \text{Given } A (2, 0), B (9,1), C (11, 6) \text{ and } D (4, 4)$ $\displaystyle AB = \sqrt{(2-9)^2 + ( 0-1)^2} = \sqrt{49 + 1} = \sqrt{50}$ $\displaystyle BC= \sqrt{(9-11)^2 + ( 1-6)^2} = \sqrt{4 + 25} = \sqrt{29}$ $\displaystyle CD = \sqrt{(11-4)^2 + ( 6-4)^2} = \sqrt{49 + 4} = \sqrt{53}$ $\displaystyle AD = \sqrt{(2-4)^2 + ( 0-4)^2} = \sqrt{4 + 16} = \sqrt{20}$ $\displaystyle \text{Since } AB \neq BC \neq CD \neq AD , \text{ therefore } ABCD$ is not a rhombus. $\displaystyle \\$

Question 5: Find the coordinates of the center of the circle inscribed in a triangle whose vertices are $\displaystyle (-36, 7), (20, 7) \text{ and } (0, - 8)$.

Given the vertices of the triangle $\displaystyle (-36, 7), (20, 7) \text{ and } (0, - 8)$.

We know that if the vertices of a triangle are $\displaystyle (x_1, y_1), (x_2, y_2),(x_3, y_3)$, then the in-center of the circle is $\displaystyle \Big( \frac{ax_1+bx_2+cx_3}{a+b+c} , \frac{ay_1+by_2+cy_3}{a+b+c} \Big) \text{ where } a = BC, b = AC, c= AB$

Now $\displaystyle a = BC = \sqrt{(20-0)^2 + (7+8)^2} = \sqrt{400 + 225} = 25$ $\displaystyle b = CA= \sqrt{(0+36)^2 + ( =8-7)^2} = \sqrt{1296 + 225} = 39$ $\displaystyle c = AB= \sqrt{(20+36)^2 + ( 7-7)^2} = 56$

Therefore the coordinates of the in-center are $\displaystyle \Big( \frac{25(-36)+39(20)+56(0)}{25+39+56} , \frac{25(7)+39(7)+56(-8)}{25+39+56} \Big) = \Big( \frac{-120}{120} , 0 \Big) = (-1, 0)$ $\displaystyle \\$

Question 6: The base of an equilateral triangle with side $\displaystyle 2a$ lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle  $\displaystyle \text{Given } ABC$ is an equilateral triangle with side $\displaystyle 2a$. $\displaystyle \text{Let } A$ be $\displaystyle ( x, 0)$. $\displaystyle \therefore AB = BC = CA$ $\displaystyle \Rightarrow AB^2 = BC^2 = CA^2$ $\displaystyle \Rightarrow a^2 + a^2 = (2a)^2$ $\displaystyle \Rightarrow x^2 = 3a^2$ $\displaystyle \Rightarrow x = \pm \sqrt{3} a$

Therefore the vertices are $\displaystyle (0, a), ( 0, -a) , ( \sqrt{3} a, 0) \text{ or } (0, a), ( 0, -a) , ( -\sqrt{3} a, 0)$ $\displaystyle \\$

Question 7: Find the distance between $\displaystyle P (x_1, y1) \text{ and } Q (x_2, y_2)$ when (i) $\displaystyle PQ$ is parallel to the y-axis (ii) $\displaystyle PQ$ is parallel to the x-axis. $\displaystyle \text{Given } P (x_1, y1) \text{ and } Q (x_2, y_2)$ $\displaystyle \text{We know } PQ = \sqrt{ (x_1-x_2)^2 + ( y_1-y_2)^2}$ $\displaystyle \text{i) When } PQ \text{ is parallel to x-axis, then } x_1 = x_ 2$ $\displaystyle \therefore PQ = \sqrt{( y_1-y_2)^2 } = | y_1-y_2|$ $\displaystyle \text{i) When } PQ \text{ is parallel to y-axis, then } y_1 = y_ 2$ $\displaystyle \therefore PQ = \sqrt{( x_1-x_2)^2 } = | x_1-x_2|$ $\displaystyle \\$

Question 8: Find a point on the x-axis, which is equidistant from the point $\displaystyle (7,6) \text{ and } (3,4)$. $\displaystyle \text{Given points } A(7,6) \text{ and } B(3,4)$ $\displaystyle \text{Let } C(x, 0)$ be the equidistant point from $\displaystyle A \text{ and } B$ $\displaystyle \therefore AC = BC \Rightarrow AC^2 = BC^2$ $\displaystyle \Rightarrow ( 7-x)^2 + ( 6-0)^2 = ( 3-x)^2 + ( 4-0)^2$ $\displaystyle \Rightarrow 49 + x^2 - 14x + 36 = 9 + x^2 - 6x + 16$ $\displaystyle \Rightarrow 85 - 14 x = 25 - 6x$ $\displaystyle \Rightarrow 60 = 8x$ $\displaystyle \Rightarrow x = \frac{15}{2}$ $\displaystyle \therefore C = \Big( \frac{15}{2} , 0 \Big)$