Q1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Solution:

Let coordinate be (x, y) which divides the join of (-1,7) and (4,-3) in ratio 2:3 $$\begin{aligned}x=\dfrac{mx_{2}+nx_{1}}{m+n}\\ y=\dfrac{my_{2}+ny_{1}}{m+n}\end{aligned}$$ Substituting values $$\begin{aligned}x_{1}=-1&,\ x_{2}=4,\\ y_{1}=7&,\ y_2=-3\\ m=2&,\ n=3\\\\ x&=\dfrac{mx_{2}+nx_{1}}{m+n}\\ &=\dfrac{2\times 4+3\times \left( -1\right) }{2+3}\\ &=\dfrac{8-3}{5}\\ &=\dfrac{5}{5}\\ &=1\\\\ y&=\dfrac{my_{2}+ny_{1}}{m+n}\\ &=\dfrac{2\times \left( -3\right) +3\times \left( 7\right) }{2+3}\\ &=\dfrac{-6+21}{5}\\ &=\dfrac{15}{5}\\ &=3\\\\ \left( x,y\right) &=\left( 1,3\right) \end{aligned}$$ (1,3) will divide the line segment in 2:3

Q2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Solution:

Trisection implies that coordinate will divide the line segment in 1: 2,
Here m = 1, n = 2 and 2: 1 (that is ⅓ and ⅔ part of line segment) $$\begin{aligned}x_{1}=4&,\ x_{2}=-2\\ y_{1}=-1&,\ y_{2}=-3\\ m=1&,\ n=2\end{aligned}$$

Substituting values in Section Formula First coordinate to trisect the line segment (m=1, n = 2)

$$\begin{aligned}x&=\dfrac{mx_{2}+nx_{1}}{m+n}\\ &=\dfrac{1\times \left( -2\right) +2\times \left( 4\right) }{1+2}\\ &=\dfrac{-2+8}{3}\\ &=\dfrac{6}{3}\\ &=2\\\\ y&=\dfrac{my_{2}+ny_{1}}{m+n}\\ &=\dfrac{1\times \left( -3\right) +2\times \left( -1\right) }{1+2}\\ &=\dfrac{-3-2}{3}\\ &=-\dfrac{5}{3}\end{aligned}$$

2nd coordinate to trisect the line segment will be when m =2, n = 1

$$\begin{aligned}x&=\dfrac{mx_{2}+nx_{1}}{m+n}\\ &=\dfrac{2\times \left( -2\right) +1\times \left( 4\right) }{2+1}\\ &=\dfrac{-4+4}{3}\\ &=0\\\\ y&=\dfrac{my_{2}+ny_{1}}{m+n}\\ &=\dfrac{2\cdot \left( -3\right) +1\cdot \left( -1\right) }{2+1}\\ &=\dfrac{-6-1}{3}\\ &=\dfrac{-7}{3}\\ \left( x,y\right) &=\left( 0,\frac{-7}{3}\right) \end{aligned}$$

(2,-5/3) and (0,-7/3) will trisect the given line segment

Q3.To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in Fig. 7.12. Niharika runs \(frac{1}{4}\)th the distance AD on the 2nd line and posts a green flag. Preet runs \(\frac{1}{5}\)th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution:

Coordinates where of Niharika post the flag y= 100/4 = 25 and x = 2, hence = (2, 25)
Coordinates of Preet post the flag who runs 1/5, y = 100/5 = 20 and x = 8, hence co ordinates = (8,20)
Distance between the flags

$$\begin{aligned}d&=\sqrt{\left( 8-2\right) ^{2}+\left( 20-25\right) ^{2}}\\ &=\sqrt{\left( b\right) ^{2}+\left[ b\right] ^{2}}\\ &=\sqrt{36+25}\\ &=\sqrt{61}m\end{aligned}$$

Midpoint of the line joining the two flag

$$\begin{aligned}x&=\dfrac{x_{1}+x_{2}}{2}\\ &=\dfrac{2+8}{2}\\ &=\dfrac{10}{2}\\ &=5\\\\ y&=\dfrac{y_{1}+y_{2}}{2}\\ &=\dfrac{20+25}{2}\\ &=\dfrac{45}{2}\\ &=22.5\\\\ \left( x,y\right) &=\left( 5,22.5\right) \end{aligned}$$

Midpoint of the flags would be the fifth line at a distance of 22.5 m

Q4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Solution:

Coordinates of line segment are (-3,10) and (6,-8)
Let (-1, 6) divides the line segment in the ratio of m: 1

$$\begin{aligned}x&=\dfrac{mx_{2}+nx_{1}}{m+n}\\ -1&=\dfrac{m\cdot 6+1\cdot \left( -3\right) }{m+1}\\ -\left( m+1\right) &=6m-3\\ -m-6m&=-3+1\\ 7m&=-2\\ m&=\dfrac{2}{7}\end{aligned}$$

Coordinate (-1, 6) will divide the line segment in the ratio of 2: 7

Q5. 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.

Solution:

Coordinates of line segment A(-1,-5) and B (-4,5)
line segment is divided by x-axis, therefore y coordinate will be zero.
Let x-axis will deride the line segment in ratio of m: 1

$$\begin{aligned}y&=\dfrac{my_{2}+y_{1}}{m+1}\\ 0&=\dfrac{m\times 5+\left( -5\right) }{m+1}\\ 0&=5m-5\\ 5m&=5\\ m&=1\end{aligned}$$

Let coordinate at which line segment divides be (x, 0)

$$\begin{aligned}x&=\dfrac{mx_{2}+nx_{1}}{m+n}\\ &=-4+1\\ &=\dfrac{-3}{2}\\ y&=0\\ \left( x,y\right) &=\left( -\frac{3}{2},0\right) \end{aligned}$$

Q6 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution:

Lets coordinate be A (1,2),B(4, y), C (x, 6), D (3,5) where A, B,C,D are the vertex of parallelogram. Diagonals of parallelogram bisect each other. Let o be the point of intersection and then its coordinates will be midpoint of A and C and also mid point of B and D Let Midpoint of \(AC (x_1,y_1\)) and mid point of BD is \((x_2, y_2)\)

$$\begin{aligned}x_{1}&=\dfrac{1+x}{2}\\\\ &=\dfrac{2+6}{2}\\ &=\dfrac{8}{2}\\ &=4\\\\ x_{2}&=\dfrac{4+3}{2}\\ &=\dfrac{7}{2}\\\\ &=\dfrac{y+5}{2}\end{aligned}$$ But \(x_{1}=x_{2}\) and \(y_{1}=y_{2}\) as both points coincides each other on the point of intersection od diagonals $$\begin{aligned}\dfrac{y+5}{2}&=4\\ y+5&=8\\ y&=8-5\\ y&=3\\\\ \dfrac{7}{2}&=\dfrac{1+x}{2}\\ x&=6\\\\ x=6,&\ y=3\end{aligned}$$

Q7. 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).

Solution:

Centre of the circle is with coordinate at(2,-3) and coordinate of one end of the diameter is at (1,4)
Center of the circle divides the diameter in the ratio of 1: 1
Hence, coordinate of Center \((x,\; y)\)

$$\begin{aligned}x&=\dfrac{mx_{2}+nx_{1}}{m+n}\\ 2&=\dfrac{1+x}{2}\\ 4&=1+x\\ \Rightarrow x&=3\\\\ y&=\dfrac{y+4}{2}\\ -3&=\dfrac{y+4}{2}\\ -6&=y+4\\ \Rightarrow y=&-10\\ y&=-10\\\\ (x.\; y)& =\left( 3,-10\right) \end{aligned}$$

Q8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =\(\frac{3}{7}\)AB and P lies on the line segment AB.

Solution:

Coordinate of A (-2,-2)
coordinate of B (2,-4)

$$\begin{aligned}AP&=\dfrac{3}{7}AB\\ PB&=AB-AP\\ &=AB-\dfrac{3}{7}AB\\ &=\dfrac{7AB-3AB}{7}\\ &=\dfrac{4AB}{7}\\\\ \dfrac{AP}{PB}&=\dfrac{3AB\times 7}{4AB\times 7}\\\\ \dfrac{AP}{AB}&=\dfrac{3}{4}\end{aligned}$$

Point P divides AB in ratio of 3: 4

$$\begin{aligned}x&=\dfrac{mx_{2}+nx_{1}}{m+n}\\ &=\dfrac{3\times \left( 2\right) +4\times \left( -2\right) }{3+4}\\ &=\dfrac{-6+\left( -8\right) }{7}\\ &=\dfrac{-2}{7}\\\\ y&=\dfrac{my_{2}+ny_{1}}{m+n}\\ &=\dfrac{3\times \left( -4\right) +4\times \left( -2\right) }{3+4}\\ &=\dfrac{-12-8}{7}\\ &=\dfrac{-2}{7}\\\\ P\left( -\dfrac{2}{7},-\dfrac{20}{7}\right) \end{aligned}$$

Q9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Solution:

Coordinates of First end of line segmental A(-2, 2)
coordinates of Second end of line segment B (2,8)
To find Coordinates of the point which will divide line segment into 4 equal parts,
Let first find a point P which divides the line segment into two equal part, Hence coordinates of \(P (x, y)\) will be

$$\begin{aligned}P\left( x\right) &=\dfrac{x_{1}+x_{2}}{2}\\ &=\dfrac{-2+2}{2}\\ &=0\\\\ P\left( y\right)&=\dfrac{y_{1}+y_{2}}{2}\\ &=\dfrac{2+8}{2}\\ &=\dfrac{10}{2}\\ &=5\end{aligned}$$

Coordinates of point P which divides the line segmeth into two equal half is (0,5)
Now let us assume that Point R on the line segment AP such that it divides the segment in ratio of 1: 1 coordinates of \(R(x,y)\)

$$\begin{aligned}R\left( x\right) &=\dfrac{x_{1}+x_{2}}{2}\\ &=\dfrac{-2+0}{2}\\ &=-1\\\\ R\left( y\right)&=\dfrac{y_{1}+y_{2}}{2}\\ &=\dfrac{2+5}{2}\\ &=\dfrac{7}{2}\\ \end{aligned}$$

Similarly assume a point S on the line segment PB
such that it divides the line segment PB in ratio of 1: 1
Hence, coordinates of point \(P(x,\ y)\)

$$\begin{aligned}Q\left( x\right) &=\dfrac{x_{1}+x_{2}}{2}\\ &=\dfrac{0+2}{2}\\ &=1\\\\ Q\left( y\right)&=\dfrac{y_{1}+y_{2}}{2}\\ &=\dfrac{5+8}{2}\\ &=\dfrac{13}{2}\end{aligned}$$

Hence coordinates that divides the line segment AB equally in 4 parts is

$$\left( -1,\dfrac{7}{2}\right) ,\left( 0,5\right) ,\left( 1,\dfrac{13}{2}\right) $$

Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Solution:

Lets Vertices of Rhombus are A, B, C, D and its coordinates are

$$\begin{aligned}A\left( 3,0\right) \\ B\left( 4,5\right) \\ C\left( -1,4\right) \\ D\left( -2,-1\right) \end{aligned}$$

Points AC and BD will form the diagonals, therefore by distance formula, length of AC and BD is

$$\scriptsize\begin{aligned}AC&=\sqrt{\left[ 3-\left( -1\right) \right] ^{2}+\left( 0-4\right) ^{2}}\\ &=\sqrt{\left( 3+1\right) ^{2}+\left( -4\right) ^{2}}\\ &=\sqrt{4^{2}+4^{2}}\\ &=\sqrt{16+16}\\ &=4\times \sqrt{2}\\\\ BD&=\sqrt{\left[ 1-\left( -2\right) \right] ^{2}+\left[ -\left( -1\right) \right] ^{2}}\\ &=\sqrt{6^{2}+6^{2}}\\ &=\sqrt{36+36}\\ &=6\times \sqrt{2}\end{aligned}$$

Area of Rhombus = ½ (produts of its diagonals) therefore

$$\begin{aligned}A&=\dfrac{1}{2}\times 4\times \sqrt{2}\times 6\times \sqrt{2}\\ A&=24\end{aligned}$$

Area of Rhombus = 24 square units