## A median of a triangle divides it into two triangles of equal areas

[question] Question. A median of a triangle divides it into two triangles of equal areas. Verify this result for ABC whose vertices are A(4, – 6), B(3,–2) and C(5,2) [/question] [solution] Solution: Here, the vertices of the triangles are A(4, –6), B(3, –2) and C(5, 2). Let D be the midpoint of BC. $\therefore \quad$ The coordinates of the mid point D are $\therefore \quad$ The coordinates of the mid point $D$ are $\left\{\frac{\mathbf{3}+\mathbf{5}}{\mathbf{2}}, \frac{-\mathbf{2}+\mathbf{2}}{\...

## Find the area of the quadrilateral whose vertices taken in order are

[question] Question. Find the area of the quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2) and (2, 3). [/question] [solution] Solution: Join A and C. The given points are $\mathrm{A}(-4,-2), \mathrm{B}(-3,-5), \mathrm{C}(3,-2)$ and $\mathrm{D}(2,3)$ Area of $\triangle \mathrm{ABC}$ $=\frac{\mathbf{1}}{\mathbf{2}}[(-4)(-5+2)-3(-2+2)+3(-2+5)]$ $=\frac{\mathbf{1}}{\mathbf{2}}[12+0+9]=\frac{\mathbf{2 1}}{\mathbf{2}}=10.5$ sq. units Area of $\Delta \mathrm{ACD}$ $=\frac{1}{...

## Find the area of the triangle formed by joining

[question] Question. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, – 1), (2, 1) and (0, 3). Find the ratio of this area of the area of the given triangle. [/question] [solution] Solution: Let the vertices of the triangle be A(0, –1), B(2, 1) and C(0, 3). Let D, E and F be the mid-points of the sides BC, CA and AB respectively. Then : Coordinates of D are $\left(\frac{\mathbf{2}+\mathbf{0}}{\mathbf{2}}, \frac{\mathbf{1}+\mathbf{...

## In each of the following find the value of 'k',

[question] Question. In each of the following find the value of 'k', for which the points are collinear. (i) (7, – 2), (5, 1), (3, k) (ii) (8,1), (k – 4), (2,–5). [/question] [solution] Solution: The given three points will be collinear if the $\Delta$ formed by them has equal to zero area. (i) Let $\mathrm{A}(7,-2), \mathrm{B}(5,1)$ and $\mathrm{C}(3, \mathrm{k})$ be the vertices of a triangle. $\therefore$ The given points will be collinear, if $\operatorname{ar}(\Delta \mathrm{ABC})=0$ or $7(...

## Find the area of the triangle whose vertices are :

[question] Question. Find the area of the triangle whose vertices are : (i) (2,3), (–1, 0), (2, –4) (ii) (– 5, – 1), (3,–5), (5,2) [/question] [solution] Solution: (i) Let the vertices of the triangles be A(2, 3), B (–1, 0) and C(2, –4) Here $x_{1}=2, y_{1}=3$ $x_{2}=-1, y_{2}=0$ $x_{3}=2, y_{3}=-4$ $\because \quad$ Area of a $\Delta$ $=\frac{1}{\boldsymbol{2}}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\le...

## Find the area of a rhombus if its vertices are

[question] Question. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, – 1) taken in order. [/question] [solution] Solution: Diagonals AC and BD bisect each other at right angle to each other at O. $A C=\sqrt{(-1-3)^{2}+(4-0)^{2}}$ $=\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}$ $B D=\sqrt{(4+2)^{2}+(5+1)^{2}}=\sqrt{36+36}=6 \sqrt{2}$ Then $\mathrm{OA}=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \times 4 \sqrt{2}=2 \sqrt{2}$ $\mathrm{OB}=\frac{1}{2} \mathrm{BD}=\frac{1}{2} \times \mat...

## Find the coordinates of the points which divide the line segment joining

[question] Question. Find the coordinates of the points which divide the line segment joining A (– 2, 2) and B (2,8) into four equal parts. [/question] [solution] Solution: Here, the given points are A(–2, 2) and B(2, 8) Let $P_{1}, P_{2}$ and $P_{3}$ divide $A B$ in four equal parts. $\because \quad \mathrm{AP}_{1}=\mathrm{P}_{1} \mathrm{P}_{2}=\mathrm{P}_{2} \mathrm{P}_{3}=\mathrm{P}_{3} \mathrm{~B}$ Obviously, $\mathrm{P}_{2}$ is the mid-point of $\mathrm{AB}$ $\therefore \quad$ Coordinates o...

## If A and B are (– 2, – 2) and (2, – 4),

[question] Question. If $\mathrm{A}$ and $\mathrm{B}$ are $(-2,-2)$ and $(2,-4)$, respectively, find the coordinates of $\mathrm{P}$ such that $\mathrm{AP}=$ $\frac{\mathbf{3}}{\mathbf{7}} \mathrm{AB}$ and $\mathrm{P}$ lies on the line segment $\mathrm{AB}$. [/question] [solution] Solution: $\mathrm{AP}=\frac{3}{7} \mathrm{AB}$ $\mathrm{BP}=\mathrm{AB}-\mathrm{AP}=\mathrm{AB}-\frac{\mathbf{3}}{7} \mathrm{AB}=\frac{4}{7} \mathrm{AB}$ $\frac{A P}{B P}=\frac{\frac{3}{7} A B}{\frac{4}{7} A B}=\frac{...

## Find the coordinates of a point A,

[question] Question. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4). [/question] [solution] Solution: Here, centre of the circle is O(2, –3) Let the end points of the diameter be A(x, y) and B(1, 4) The centre of a circle bisects the diameter. $\therefore \quad 2=\frac{\mathbf{x}+\mathbf{1}}{\mathbf{2}} \Rightarrow \mathbf{x}+1=4$ or $\mathbf{x}=3$ And $-3=\frac{\mathbf{y}+\mathbf{4}}{\mathbf{2}} \Rightarrow \mathrm{y}+4=-6$ or $\m...

## If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order,

[question] Question. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. [/question] [solution] Solution: Mid-point of the diagonal AC has x-coordinate $=\frac{\mathbf{x}+\mathbf{1}}{\mathbf{2}}$ and $y$-coordinate $=\frac{\mathbf{6}+\mathbf{2}}{\mathbf{2}}=4$ i.e., $\left(\frac{\mathbf{x}+\mathbf{1}}{\mathbf{2}}, \mathbf{4}\right)$ is the mid-point of $\mathrm{AC}$. Similarly, mid-point of the diagonal BD is $\left(\frac{\mathbf{4}+\mathbf{3}}{...

## Find the ratio in which the line segment joining

[question] Question. Find the ratio in which the line segment joining A(1, – 5) and B( – 4, 5) is divided by the x-axis. Also find the coordinates of the point of division. [/question] [solution] Solution: The given points are : A(1, – 5) and B(–4, 5). Let the required ratio = k : 1 and the required point be P(x, y) Part-I : To find the ratio Since, the point P lies on x-axis, $\therefore$ Its $y$-coordinate is 0 . $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$ and $0=\frac{m_{1} y_{2}+m_{2} y_...

## Find the ratio in which the line segment joining the points

[question] Question. Find the ratio in which the line segment joining the points (– 3, 10) and (6, –8) is divided by (–1, 6). [/question] [solution] Solution: Let the required ratio be K : 1 [/solution]...

## Find the coordinates of the points of trisection of the line segment joining

[question] Question. Find the coordinates of the points of trisection of the line segment joining $(4,-1)$ and $(-2,-3)$. [/question] [solution] Solution: Points P and Q trisect the line segment joining the points A(4, – 1) and B(2, – 3), i.e., AP = PQ = QB. Here, P divides AB in the ratio 1 : 2 and Q divides AB in the ratio 2 : 1. $x$-coordinate of $P=\frac{1 \times(-2)+2 \times(4)}{1+2}=\frac{6}{3}=2$; $y-$ coordinate of $P=\frac{1 \times(-3)+2 \times(-1)}{1+2}=\frac{-5}{3}$ Thus, the coordina...

## Find the co-ordinates of the point which divides the line joining

[question] Question. Find the co-ordinates of the point which divides the line joining of (–1, 7) and (4, –3) in the ratio 2 : 3. [/question] [solution] Solution: Let the required point be P(x, y). Here the end points are (–1, 7) and (4, –3) $\because \quad$ Ratio $=2: 3=\mathrm{m}_{1}: \mathrm{m}_{2}$ $\therefore \quad x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{(2 \times 4)+3(-1)}{2+3}$ $=\frac{8-3}{5}=\frac{5}{5}=1$ And $y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}$ $=\frac{2 \times(...

## Find a relation between x and y such that the point

[question] Question. Find a relation between x and y such that the point (x,y) is equidistant from the point (3, 6) and (– 3, 4). [/question] [solution] Solution: A(3,6) and B(–3, 4) are the given points. Point P (x, y) is equidistant from the points A and B. $\Rightarrow \mathrm{PA}=\mathrm{PB}$ $\Rightarrow \sqrt{(x-3)^{2}+(y-6)^{2}}=\sqrt{(x+3)^{2}+(y-4)^{2}}$ $\Rightarrow(x-3)^{2}+(y-6)^{2}=(x+3)^{2}+(y-4)^{2}$ $\Rightarrow\left(x^{2}-6 x+9\right)+\left(y^{2}-12 y+36\right)$ $=\left(x^{2}+6 ...

## If Q (0,1) is equidistant from P(5, – 3) and R(x, 6),

[question] Question. If Q (0,1) is equidistant from P(5, – 3) and R(x, 6), find the values of x. Also find the distances QR and PR. [/question] [solution] Solution: Here, $Q P=\sqrt{(5-0)^{2}+\mid(-3)-11^{2}}=\sqrt{5^{2}+(-4)^{2}}$ $=\sqrt{25+16}=\sqrt{41}$ $Q R=\sqrt{(x-0)^{2}+(6-1)^{2}}=\sqrt{x^{2}+5^{2}}=\sqrt{x^{2}+25}$ $\because \quad \mathrm{QP}=\mathrm{QR}$ $\therefore \quad \sqrt{41}=\sqrt{x^{2}+25}$ Squaring both sides, we have $x^{2}+25=41$ $\Rightarrow x^{2}+25-41=0$ $\Rightarrow x^{2...

## Find the values of y for which the distance between the points P(2, –3)

[question] Question. Find the values of y for which the distance between the points P(2, –3) and Q(10, y) is 10 units [/question] [solution] Solution: Distance between P(2, – 3) and Q(10, y) = 10 units $\Rightarrow \sqrt{(10-2)^{2}+(y+3)^{2}}=10$ $\Rightarrow 64+(y+3)^{2}=100$ $\Rightarrow(y+3)^{2}=36$ $\Rightarrow y^{2}+6 y+9=36$ $y^{2}+6 y-27=0$ $\Rightarrow y^{2}+9 y-3 y-27=0$ $\Rightarrow y(y+9)-3(y+9)=0$ $\Rightarrow(\mathrm{y}+9)(\mathrm{y}-3)=0$ $\Rightarrow y+9=0$ or $y-3=0$ $\Rightarrow...

## Find the point on the x-axis which is equidistant from

[question] Question. Find the point on the x-axis which is equidistant from (2, – 5) and (– 2, 9). [/question] [solution] Solution: We know that any point on x-axis has its ordinate = 0 Let the required point be P(x, 0). Let the given points be A(2, –5) and B(–2, 9) $\therefore \quad \mathrm{AP}=\sqrt{(x-2)^{2}+5^{2}}=\sqrt{x^{2}-4 x+4+25}$ $=\sqrt{x^{2}-4 x+29}$ $\mathrm{BP}=\sqrt{[\mathbf{x}-(-2)]^{2}+(-\mathbf{9})^{2}}=\sqrt{(\mathbf{x}+2)^{2}+(-\mathbf{9})^{2}}$ $=\sqrt{x^{2}+4 x+4+81}=\sqrt...

## Name the quadrilateral formed,

[question] Question. Name the quadrilateral formed, if any, by the following points, and give reasons for your answer. (i) (–1, –2), (1, 0), (–1, 2), (– 3, 0) (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) (iii) (4, 5), (7, 6), (4, 3), (1, 2) [/question] [solution] Solution: (i) A(–1, –2), B(1, 0), C(–1, 2), D(–3, 0) Determine distances : AB, BC, CD, DA, AC and BD. $\mathrm{AB}=\sqrt{(1+1)^{2}+(0+2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$ $\mathrm{BC}=\sqrt{(-1-1)^{2}+(2-0)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqr...

## Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

[question] Question. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. [/question] [solution] Solution: Let the points be A(5, –2), B(6, 4) and C(7, –2). $\therefore \quad \mathrm{AB}=\sqrt{(\mathbf{6}-\mathbf{5})^{2}+\mathbf{4}-(\mathbf{- 2})^{2}}$ $=\sqrt{(1)^{2}+(B)^{2}}=\sqrt{1+36}=\sqrt{37}$ $B C=\sqrt{(7-6)^{2}+(-2-4)^{2}}$ $=\sqrt{(1)^{2}+(-6)^{2}}=\sqrt{1+36}=\sqrt{37}$ $A C=\sqrt{(7-5)^{2}+(-2-(-2))^{2}}$ $=\sqrt{(+2)^{2}+(0)^{2}}=\sqrt{4+0}=2$ We ha...

## Find the distance between the points (0,0) and (36,15).

[question] Question. Find the distance between the points (0,0) and (36,15). [/question] [solution] Solution: Part-I Let the points be A(0, 0) and B(36, 15) $\therefore \quad \mathrm{AB}=\sqrt{(36-0)^{2}+(15-0)^{2}}$ $=\sqrt{(36)^{2}+(15)^{2}}=\sqrt{1296+225}$ $=\sqrt{\mathbf{1 5 2 1}}=\sqrt{\mathbf{3 9}^{\mathbf{2}}}=39$ Part-II We have A(0, 0) and B(36, 15) as the positions of two towns Here $x_{1}=0, x_{2}=36$ and $y_{1}=0, y_{2}=15$ $\therefore \quad \mathrm{AB}=\sqrt{(\mathbf{3 6}-\mathbf{0...

## Find the distance between the following pairs of points :

[question] Question. Find the distance between the following pairs of points : (a) (2,3), (4, 1) (b) (–5, 7), (–1,3) (c) (a, b), (– a, – b) [/question] [solution] Solution: (a) The given points are : A (2, 3), B (4, 1). Required distance $=\mathrm{AB}=\mathrm{BA}=\sqrt{\left(\mathbf{x}_{2}-\mathbf{x}_{1}\right)^{2}+\left(\mathbf{y}_{2}-\mathbf{y}_{1}\right)^{2}}$ $A B=\sqrt{(4-2)^{2}+(1-3)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}$ $=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$ units (b) Here $x_{1}=-5, y_{1}=7$ and $x_{...