Chapter 7: Coordinate Geometry

Given points: $$A(3,2)$$, $$B(-2,-3)$$ and $$C(2,3)$$.

1. Test for collinearity (area method)

The area of the triangle formed by three points $$ (x_1,y_1),(x_2,y_2),(x_3,y_3) $$ is

$$ \text{Area}=\tfrac12\, | x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |. $$

Substituting $$A(3,2),\;B(-2,-3),\;C(2,3):$$

$$ \begin{aligned} \text{Area} & = \tfrac12 |\, 3((-3)-3)+(-2)(3-2)+2(2-(-3)) \,| \\[2pt] & = \tfrac12 |\, 3(-6)+(-2)(1)+2(5) \,| \\[2pt] & = \tfrac12 |\,-18-2+10\,| \\[2pt] & = \tfrac12 |\,-10| \\[2pt] & = \tfrac12 \times 10 = 5.\end{aligned} $$

Since $$\text{Area}=5\neq0$$, the three points are not collinear; they do form a triangle.

2. Find the lengths of the sides

Thus
$$AB^2=50,\; BC^2=52,\; CA^2=2.$$

3. Classify the triangle

The largest square of a side is $$BC^2=52$$. Check the Pythagoras relation:

$$ AB^2 + CA^2 = 50 + 2 = 52 = BC^2. $$

Since the sum of the squares of the two shorter sides equals the square of the longest side, the triangle satisfies the converse of the Pythagoras theorem. Therefore, $$\triangle ABC$$ is a right-angled triangle.

All three side lengths are different (50, 52, 2), so it is specifically a right-angled scalene triangle.

Conclusion: The points $$ (3,2),\;(-2,-3),\;(2,3) $$ form a right-angled triangle, right-angled at $$A(3,2)$$.

Let us denote the four given points by

$$A(1,7), \; B(4,2), \; C(-1,-1), \; D(-4,4).$$

To prove that $$ABCD$$ is a square we shall show

1. all four sides are equal, and
2. one interior angle is a right angle.

These two facts together are sufficient because a quadrilateral with all sides equal is a rhombus, and a rhombus that has one right angle is automatically a square.

1. Length of the four sides

The distance formula between two points $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ is

$$PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.$$

Rather than carrying the square roots, we calculate the squares of the side–lengths.

Thus

$$AB = BC = CD = DA = \sqrt{34}.$$

Hence all four sides are equal.

2. Verification of a right angle

We now examine angle $$ABC$$ by computing the slopes of $$AB$$ and $$BC$$.

Slope of $$AB$$:

$$m_{AB} = \dfrac{2-7}{4-1} = -\dfrac{5}{3}.$$

Slope of $$BC$$:

$$m_{BC} = \dfrac{-1-2}{-1-4} = \dfrac{-3}{-5} = \dfrac{3}{5}.$$

The product of the two slopes is

$$m_{AB}\,m_{BC} = \Bigl(-\dfrac{5}{3}\Bigr)\Bigl(\dfrac{3}{5}\Bigr) = -1.$$

Since the product of the slopes of two lines is $$-1$$, the lines are perpendicular. Therefore $$\angle ABC = 90^{\circ}$$.

Conclusion

Quadrilateral $$ABCD$$ has all its sides equal and one right angle; hence it is a square.

Therefore the four given points are indeed the vertices of a square.

Given points: $$A(3,1)$$, $$B(6,4)$$ and $$C(8,6)$$.

To know whether the three students are sitting in a straight line, we must check if the three points are collinear. We do this by comparing the slopes of any two of the three possible line segments.

Step 1 – Slope of $$AB$$

$$m_{AB}=\frac{y_B-y_A}{x_B-x_A}=\frac{4-1}{6-3}=\frac{3}{3}=1$$

Step 2 – Slope of $$BC$$

$$m_{BC}=\frac{y_C-y_B}{x_C-x_B}=\frac{6-4}{8-6}=\frac{2}{2}=1$$

Step 3 – Comparison

The two slopes are equal: $$m_{AB}=m_{BC}=1$$. If two different pairs of points among three share the same slope, the three points lie on the same straight line; that is, they are collinear.

(Optional cross-check)  Area of $$\triangle ABC$$:

$$\text{Area}=\tfrac12\bigl|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigr|$$

$$=\tfrac12\bigl|3(4-6)+6(6-1)+8(1-4)\bigr|=\tfrac12\bigl|-6+30-24\bigr|=\tfrac12\times0=0$$

Zero area again confirms collinearity.

Conclusion: Ashima, Bharti and Camella are seated in a single straight line.

Let the required point be $$(x, y)$$.

Its distance from $$(7,1)$$ must be the same as its distance from $$(3,5)$$.

Using the distance formula, equate the two distances (it is easier to square them so that square‐roots disappear):

$$\bigl[(x-7)^2+(y-1)^2\bigr]=\bigl[(x-3)^2+(y-5)^2\bigr]$$

Expand both sides completely.

Left side:

$$ (x-7)^2 = x^2-14x+49, \qquad (y-1)^2 = y^2-2y+1 $$

So the left side becomes $$x^2-14x+49+y^2-2y+1$$.

Right side:

$$ (x-3)^2 = x^2-6x+9, \qquad (y-5)^2 = y^2-10y+25 $$

So the right side becomes $$x^2-6x+9+y^2-10y+25$$.

Set them equal:

$$x^2-14x+49+y^2-2y+1 = x^2-6x+9+y^2-10y+25$$

Cancel the identical terms $$x^2$$ and $$y^2$$ on both sides:

$$-14x+49-2y+1 = -6x+9-10y+25$$

Bring all terms to the left (subtract the right side from the left side so that the result equals 0):

$$\bigl(-14x+49-2y+1\bigr)-\bigl(-6x+9-10y+25\bigr)=0$$

Simplify term by term:

• x–terms: $$-14x+6x = -8x$$
• y–terms: $$-2y+10y = 8y$$
• constants: $$49+1-9-25 = 16$$

Hence

$$-8x+8y+16 = 0$$

Divide through by $$8$$ (or multiply by $$-1$$) to write it neatly:

$$x-y-2 = 0$$

or, equivalently,

$$y = x-2$$

Thus every point on the line $$x-y-2=0$$ is equidistant from $$(7,1)$$ and $$(3,5)$$.

Let the required point be $$P(0,y)$$ because any point on the $$y$$-axis has an $$x$$-coordinate $$0$$.

Using the distance formula,

$$PA=\sqrt{(0-6)^{2}+(y-5)^{2}}=\sqrt{36+(y-5)^{2}}$$

$$PB=\sqrt{(0+4)^{2}+(y-3)^{2}}=\sqrt{16+(y-3)^{2}}$$

Because $$P$$ is equidistant from $$A$$ and $$B$$, we set $$PA=PB$$ and square both sides:

$$36+(y-5)^{2}=16+(y-3)^{2}$$

Expand the squares:

$$36+y^{2}-10y+25=16+y^{2}-6y+9$$

Simplify:

$$y^{2}-10y+61=y^{2}-6y+25$$

Subtract $$y^{2}$$ from both sides:

$$-10y+61=-6y+25$$

Bring like terms together:

$$-4y=-36$$

$$y=9$$

Therefore the required point on the $$y$$-axis is $$P(0,9)$$.

To find the length of the straight line joining two points we use the Distance Formula.

If $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ are any two points in the Cartesian plane, then

$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

1. Distance between $$(0,0)$$ and $$(36,15)$$

Here, $$x_1 = 0,\; y_1 = 0,\; x_2 = 36,\; y_2 = 15.$$ Substituting in the formula,

$$PQ = \sqrt{(36-0)^2 + (15-0)^2} = \sqrt{36^2 + 15^2}.$$

Calculate the squares:

$$36^2 = 1296,\; 15^2 = 225.$$

Add the squares:

$$1296 + 225 = 1521.$$

Take the square root:

$$PQ = \sqrt{1521} = 39.$$

Therefore, the required distance is $$39$$ units.

2. Distance between the two towns A and B (Section 7.2)

In Section 7.2, town B is described as being $$40\text{ km}$$ east and $$30\text{ km}$$ north of town A. If we choose town A to be the origin $$A(0,0)$$, then town B is at $$B(40,30).$$

Using the same formula,

$$AB = \sqrt{(40-0)^2 + (30-0)^2} = \sqrt{40^2 + 30^2}.$$

Compute the squares and add:

$$40^2 = 1600,\; 30^2 = 900,\; 1600 + 900 = 2500.$$

$$AB = \sqrt{2500} = 50.$$

So, the two towns are $$50\text{ km}$$ apart.

Step 1 – Label the points

Let $$A(1,5), B(2,3), C(-2,-11)$$.

Step 2 – Write the slope formula

If two points $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ are distinct, the slope of the line $$PQ$$ is

$$m_{PQ}=\frac{y_2-y_1}{x_2-x_1}$$.

Step 3 – Find the slope of $$AB$$

$$m_{AB}=\frac{3-5}{2-1}=\frac{-2}{1}=-2$$.

Step 4 – Find the slope of $$BC$$

$$m_{BC}=\frac{-11-3}{-2-2}=\frac{-14}{-4}=\frac{7}{2}=3.5$$.

Step 5 – Compare the slopes

$$m_{AB}=-2\neq3.5=m_{BC}$$.

Since the slopes are different, the three points do not lie on the same straight line; hence they are not collinear.

Let us denote the three points as follows:

• $$A(5,-2)$$
• $$B(6,4)$$
• $$C(7,-2)$$

To know whether $$\triangle ABC$$ is isosceles, we compare the lengths of its sides using the distance formula.

Distance formula
For two points $$\bigl(x_1,y_1\bigr)$$ and $$\bigl(x_2,y_2\bigr)$$ the distance is $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$.

1. Side $$AB$$

   $$\begin{aligned} AB&=\sqrt{(6-5)^2+(4-(-2))^2}\\[2pt] &=\sqrt{1^2+6^2}\\[2pt] &=\sqrt{1+36}\\[2pt] &=\sqrt{37} \end{aligned}$$

2. Side $$BC$$

   $$\begin{aligned} BC&=\sqrt{(7-6)^2+((-2)-4)^2}\\[2pt] &=\sqrt{1^2+(-6)^2}\\[2pt] &=\sqrt{1+36}\\[2pt] &=\sqrt{37} \end{aligned}$$

3. Side $$AC$$

   $$\begin{aligned} AC&=\sqrt{(7-5)^2+((-2)-(-2))^2}\\[2pt] &=\sqrt{2^2+0^2}\\[2pt] &=\sqrt{4}\\[2pt] &=2 \end{aligned}$$

We have

$$AB = BC = \sqrt{37},\quad AC = 2.$$

Since two sides (here $$AB$$ and $$BC$$) are equal, $$\triangle ABC$$ is an isosceles triangle.

From Fig. 7.8 the coordinates of the four friends are

$$A(1,1),\;B(1,4),\;C(5,4),\;D(5,1).$$

The distance formula is

$$ ext{Distance}= oot{}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$

Calculate all four sides.

• Side $$AB$$:
  $$AB= oot{}\sqrt{(1-1)^2+(4-1)^2}= oot{}\sqrt{0+9}=3.$$
• Side $$BC$$:
  $$BC= oot{}\sqrt{(5-1)^2+(4-4)^2}= oot{}\sqrt{16+0}=4.$$
• Side $$CD$$:
  $$CD= oot{}\sqrt{(5-5)^2+(1-4)^2}= oot{}\sqrt{0+9}=3.$$
• Side $$DA$$:
  $$DA= oot{}\sqrt{(1-5)^2+(1-1)^2}= oot{}\sqrt{16+0}=4.$$

So $$AB=CD=3\text{ units}$$ and $$BC=DA=4\text{ units}$$. All four sides are not equal, therefore $$ABCD$$ cannot be a square.

For completeness, check the diagonals.

• Diagonal $$AC$$:
  $$AC= oot{}\sqrt{(5-1)^2+(4-1)^2}= oot{}\sqrt{16+9}=5.$$
• Diagonal $$BD$$:
  $$BD= oot{}\sqrt{(5-1)^2+(1-4)^2}= oot{}\sqrt{16+9}=5.$$

Although the two diagonals are equal, unequal sides show the quadrilateral is only a rectangle, not a square.

Conclusion : Chameli’s observation is correct; $$ABCD$$ is not a square, so Champa’s guess is wrong.

Step 1 – Introduce the unknown point
Because the point lies on the $$x$$-axis, its $$y$$-coordinate is zero. Write it as $$P(a,0)$$, where $$a$$ is a real number we have to find.

Step 2 – Write the two required distances
Using the distance formula between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$, the distances $$PA$$ and $$PB$$ are:

$$PA = \sqrt{(a-2)^2 + (0-(-5))^2} = \sqrt{(a-2)^2 + 25}$$

$$PB = \sqrt{(a-(-2))^2 + (0-9)^2} = \sqrt{(a+2)^2 + 81}$$

Step 3 – Use the condition “equidistant”
The point $$P$$ is equidistant from $$A(2,-5)$$ and $$B(-2,9)$$, so

$$\sqrt{(a-2)^2 + 25} = \sqrt{(a+2)^2 + 81}$$

Step 4 – Remove the square roots
Square both sides:

$$(a-2)^2 + 25 = (a+2)^2 + 81$$

Step 5 – Expand and simplify

$$(a^2 - 4a + 4) + 25 = (a^2 + 4a + 4) + 81$$

$$a^2 - 4a + 29 = a^2 + 4a + 85$$

Subtract $$a^2$$ from each side:

$$-4a + 29 = 4a + 85$$

Collect like terms:

$$-4a - 4a = 85 - 29$$

$$-8a = 56$$

Step 6 – Solve for $$a$$

$$a = \frac{56}{-8} = -7$$

Step 7 – State the coordinates
The required point on the $$x$$-axis is therefore $$P(-7,0)$$.

We know the distance formula between two points $$P(x_1, y_1)$$ and $$Q(x_2, y_2)$$: $$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

For the given points, we have

• $$x_1 = 2,\;y_1 = -3$$
• $$x_2 = 10,\;y_2 = y$$

The distance $$PQ$$ is given to be $$10$$ units, so

$$\sqrt{(10 - 2)^2 + (y - (-3))^2} = 10$$

Simplify step by step:

1. Compute the difference in $$x$$-coordinates:
   $$(10 - 2) = 8$$
2. Substitute this into the expression:
   $$\sqrt{8^2 + (y + 3)^2} = 10$$
3. Square both sides to remove the square root:
   $$(8^2) + (y + 3)^2 = 10^2$$
4. Simplify the squares:
   $$64 + (y + 3)^2 = 100$$
5. Isolate the squared term:
   $$(y + 3)^2 = 100 - 64 = 36$$
6. Take square roots of both sides:
   $$y + 3 = \pm 6$$
7. Solve the two linear equations:

• For the positive root, $$y + 3 = 6 \;\Rightarrow\; y = 6 - 3 = 3$$
• For the negative root, $$y + 3 = -6 \;\Rightarrow\; y = -6 - 3 = -9$$

Hence, the required values of $$y$$ are $$3$$ and $$-9$$.

Given points
P(5, -3), Q(0, 1), R(x, 6)

Since Q is equidistant from P and R, we must have $$\mathrm{QP}=\mathrm{QR}$$.

Step 1  Compute $$\mathrm{QP}$$

Using the distance formula,
$$\mathrm{QP}=\sqrt{(0-5)^2+(1-(-3))^2} =\sqrt{(-5)^2+(4)^2} =\sqrt{25+16} =\sqrt{41}.$$

Step 2  Write an expression for $$\mathrm{QR}$$

$$\mathrm{QR}=\sqrt{(0-x)^2+(1-6)^2} =\sqrt{x^2+(-5)^2} =\sqrt{x^2+25}.$$

Step 3  Equate the two distances

$$\sqrt{41}=\sqrt{x^2+25}.$$

Square both sides to remove the square roots:

$$41=x^2+25 \;\;\Rightarrow\;\; x^2=16 \;\;\Rightarrow\;\; x=4 \text{ or } x=-4.$$

Step 4  Find $$\mathrm{QR}$$

For either value of $$x$$ we have already seen that $$\mathrm{QR}=\sqrt{41}\;(\approx6.40)\text{ units}.$$

Step 5  Find $$\mathrm{PR}$$ for each value of $$x$$

General expression:
$$\mathrm{PR}=\sqrt{(5-x)^2+(-3-6)^2} =\sqrt{(5-x)^2+(-9)^2} =\sqrt{(5-x)^2+81}.$$

• When $$x=4$$:
  $$\mathrm{PR}=\sqrt{(5-4)^2+81} = \sqrt{1+81} = \sqrt{82}\;(\approx9.06)\text{ units}.$$
• When $$x=-4$$:
  $$\mathrm{PR}=\sqrt{(5+4)^2+81} = \sqrt{81+81} = \sqrt{162} = 9\sqrt{2}\;(\approx12.73)\text{ units}.$$

Result

$$x=4\text{ or }x=-4;\quad \mathrm{QR}=\sqrt{41}\text{ units};\quad \mathrm{PR}=\sqrt{82}\text{ units (if }x=4\text{)}\text{ or }9\sqrt{2}\text{ units (if }x=-4\text{).}$$

The point $$(x,y)$$ is said to be equidistant from $$(3,6)$$ and $$(-3,4)$$ when the two distance-formula values are equal.

Step 1 · Write the two distances

Distance from $$(x,y)$$ to $$(3,6):$$
$$\sqrt{(x-3)^2+(y-6)^2}$$

Distance from $$(x,y)$$ to $$(-3,4):$$
$$\sqrt{(x+3)^2+(y-4)^2}$$

Step 2 · Set the distances equal

$$\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}$$

Step 3 · Eliminate the square roots by squaring

$$(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2$$

Step 4 · Expand each binomial square

$$x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16$$

Step 5 · Cancel the common terms on both sides

• $$x^2$$ appears on both sides → cancels.
• $$y^2$$ appears on both sides → cancels.

That leaves

$$-6x-12y+45=6x-8y+25$$

Step 6 · Gather all terms on one side

$$-6x-12y+45-6x+8y-25=0$$

Step 7 · Combine like terms

$$-12x-4y+20=0$$

Step 8 · Divide by −4 to simplify

$$3x+y-5=0$$

Step 9 · Write the relation

Therefore, the required relation between $$x$$ and $$y$$ is

$$3x+y-5=0$$  or  $$y=5-3x$$.

Let the given points be
A$$\,(x_1,\,y_1)=(4,-3)$$ and B$$\,(x_2,\,y_2)=(8,5)$$.

The point P$$\,(x,\,y)$$ divides AB internally in the ratio $$3:1$$, that is

$$\;AP:PB = 3:1\;.$$

For an internal division in the ratio $$m:n$$, the section formula gives

$$\;P\,(x,\,y)=\Bigl(\dfrac{m\,x_2+n\,x_1}{m+n},\;\dfrac{m\,y_2+n\,y_1}{m+n}\Bigr).$$

Here $$m=3$$ and $$n=1$$. Substituting the known coordinates:

x–coordinate

$$x = \dfrac{3\,(8)+1\,(4)}{3+1} = \dfrac{24+4}{4} = \dfrac{28}{4} = 7.$$

y–coordinate

$$y = \dfrac{3\,(5)+1\,(-3)}{3+1} = \dfrac{15-3}{4} = \dfrac{12}{4} = 3.$$

Therefore the required point is $$\,(7,\,3)$$.

Let the point $$P(-4,6)$$ divide the line segment joining $$A(-6,10)$$ and $$B(3,-8)$$ internally in the ratio $$m:n$$, where $$m:n = AP:PB$$.

By the section formula, the coordinates of $$P$$ are given by
$$x_P = \dfrac{m\,x_B + n\,x_A}{m+n}, \qquad y_P = \dfrac{m\,y_B + n\,y_A}{m+n}.$$

Substituting the given coordinates:

For the abscissa ($$x$$-coordinate):
$$-4 = \dfrac{m\,(3) + n\,(-6)}{m+n}.$$

Cross-multiplying:
$$-4(m+n) = 3m - 6n \;\Longrightarrow\; -4m - 4n = 3m - 6n.$$

Rearranging terms:
$$-4m - 4n - 3m + 6n = 0 \;\Longrightarrow\; -7m + 2n = 0.$$
Hence, $$2n = 7m \;\Longrightarrow\; n = \dfrac{7}{2}m.$$

For the ordinate ($$y$$-coordinate):
$$6 = \dfrac{m\,(-8) + n\,(10)}{m+n}.$$

Cross-multiplying:
$$6(m+n) = -8m + 10n \;\Longrightarrow\; 6m + 6n = -8m + 10n.$$

Simplifying:
$$6m + 6n + 8m - 10n = 0 \;\Longrightarrow\; 14m - 4n = 0.$$
So, $$14m = 4n \;\Longrightarrow\; n = \dfrac{14}{4}m = \dfrac{7}{2}m.$$

The same relation $$n = \dfrac{7}{2}m$$ is obtained from both coordinates, confirming consistency.

Therefore,

$$\dfrac{AP}{PB} = \dfrac{m}{n} = \dfrac{m}{\tfrac{7}{2}m} = \dfrac{2}{7}.$$

Hence the point $$(-4,6)$$ divides the line segment $$AB$$ in the ratio $$2:7$$.

Given end-points:
$$A(2,-2), \; B(-7,4).$$

Let $$P$$ and $$Q$$ be the points of trisection, counted from $$A$$ towards $$B$$.

Because $$A P : P B = 1 : 2$$, the internal section formula gives

$$P\bigl(x_P,y_P\bigr)=\left(\dfrac{1\,x_B+2\,x_A}{1+2},\;\dfrac{1\,y_B+2\,y_A}{1+2}\right).$$

Substitute $$x_A=2,\;y_A=-2,\;x_B=-7,\;y_B=4$$:

$$x_P = \dfrac{1(-7)+2(2)}{3}=\dfrac{-7+4}{3}=-1,\quad y_P = \dfrac{1(4)+2(-2)}{3}=\dfrac{4-4}{3}=0.$$

Thus $$P(-1,0).$$

The second point $$Q$$ satisfies $$A Q : Q B = 2 : 1$$, so

$$Q\bigl(x_Q,y_Q\bigr)=\left(\dfrac{2\,x_B+1\,x_A}{2+1},\;\dfrac{2\,y_B+1\,y_A}{2+1}\right).$$

Compute:

$$x_Q = \dfrac{2(-7)+1(2)}{3}=\dfrac{-14+2}{3}=-4,\quad y_Q = \dfrac{2(4)+1(-2)}{3}=\dfrac{8-2}{3}=2.$$

Therefore $$Q(-4,2).$$

Hence the points which divide $$\overline{AB}$$ into three equal parts are
$$(-1,0)\;\text{and}\;(-4,2).$$

Let $$A(5,-6)$$ and $$B(-1,-4)$$ be the given end-points.
Because the required point $$P$$ lies on the $$y$$-axis, its coordinates are of the form $$P(0, y)$$.

Suppose $$P$$ divides $$AB$$ internally in the ratio $$m:n$$, taken as
$$AP:PB = m:n$$.

Section formula (internal division):
If $$P(x,y)$$ divides $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ in the ratio $$m:n$$, then

$$x = \dfrac{m x_2 + n x_1}{m+n}, \qquad y = \dfrac{m y_2 + n y_1}{m+n}. $$

Here $$x_1 = 5,\; y_1 = -6,\; x_2 = -1,\; y_2 = -4.$$ Since $$x = 0$$ for every point on the $$y$$-axis, we have

$$0 = \dfrac{m(-1) + n(5)}{m+n}.$$

Equating the numerator to zero (the denominator is non-zero):
$$ -m + 5n = 0 \;\Longrightarrow\; m = 5n. $$

Thus the required ratio is
$$ m:n = 5n:n = 5:1. $$

Substitute $$m = 5,\; n = 1$$ in the formula for the $$y$$-coordinate:

$$ y = \dfrac{5(-4) + 1(-6)}{5 + 1} = \dfrac{-20 - 6}{6} = \dfrac{-26}{6} = -\dfrac{13}{3}. $$

Therefore

• the $$y$$-axis divides the segment internally in the ratio 5 : 1,
• the point of intersection is $$P\bigl(0, -\dfrac{13}{3}\bigr).$$

The vertices of the parallelogram are taken in order:

$$\mathrm{A}(6,1),\;\mathrm{B}(8,2),\;\mathrm{C}(9,4),\;\mathrm{D}(p,3).$$

In a parallelogram, each pair of opposite sides is parallel. Therefore:

• $$\overline{AB}\;\parallel\;\overline{CD}$$
• $$\overline{BC}\;\parallel\;\overline{AD}$$

Using the pair $$\overline{AB}\parallel\overline{CD}$$

1. Find the slope of $$\overline{AB}$$:
   $$m_{AB}=\frac{2-1}{8-6}=\frac{1}{2}.$$
2. Find the slope of $$\overline{CD}$$:
   $$m_{CD}=\frac{3-4}{p-9}=\frac{-1}{p-9}.$$
3. Since the lines are parallel, their slopes are equal:
   $$m_{AB}=m_{CD}\;\Longrightarrow\;\frac{-1}{p-9}=\frac{1}{2}.$$
4. Solve for $$p$$ by cross-multiplying:
   $$-1\times2=1\times(p-9)\;\Longrightarrow\;-2=p-9.$$
   Add $9$ to both sides:
   $$p=7.$$

Verification with the other pair $$\overline{BC}\parallel\overline{AD}$$ (optional)

Slope of $$\overline{BC}$$: $$m_{BC}=\frac{4-2}{9-8}=2.$$

Slope of $$\overline{AD}$$: $$m_{AD}=\frac{3-1}{p-6}=\frac{2}{p-6}.$$

Set $$m_{BC}=m_{AD}$$:
$$\frac{2}{p-6}=2\;\Longrightarrow\;p-6=1\;\Longrightarrow\;p=7.$$
The value is consistent.

Hence, the required value of $$p$$ is $$7$$.

The required point P divides the line segment joining
$$A(-1,7)$$ and $$B(4,-3)$$ internally in the ratio $$2:3$$, i.e. $$AP:PB = 2:3$$.

For internal division in the ratio $$m_1:m_2$$, the coordinates of the point are given by the section formula

$$\bigl(\,x,\,y\bigr) = \left( \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2},\; \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \right).$$

Here
$$m_1 = 2, \; m_2 = 3,$$
$$x_1 = -1, \; y_1 = 7,$$
$$x_2 = 4, \; y_2 = -3.$$

Finding the x-coordinate

$$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} = \dfrac{2\times4 + 3\times(-1)}{2+3} = \dfrac{8 - 3}{5} = \dfrac{5}{5} = 1.$$

Finding the y-coordinate

$$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} = \dfrac{2\times(-3) + 3\times7}{2+3} = \dfrac{-6 + 21}{5} = \dfrac{15}{5} = 3.$$

Thus the point which divides $$AB$$ in the ratio $$2:3$$ is $$\bigl(1, 3\bigr).$$

Given end-points of the line segment:
A $$\bigl(x_1,\,y_1\bigr)=(4,-1)$$,  B $$\bigl(x_2,\,y_2\bigr)=(-2,-3)$$.

Let P and Q be the required points such that the segment AB is divided into three equal parts, i.e.
$$AP = PQ = QB$$.

That means

• P divides AB internally in the ratio $$1:2$$ (one part from A, two parts from B).
• Q divides AB internally in the ratio $$2:1$$ (two parts from A, one part from B).

Section formula
If a point R divides AB in the ratio $$m:n$$ (AR : RB), its coordinates are
$$\Bigl( \dfrac{nx_1+mx_2}{m+n},\; \dfrac{ny_1+my_2}{m+n} \Bigr).$$

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1. Coordinates of P  (ratio $$1:2$$)

Here $$m=1,\;n=2$$.

$$\begin{aligned} P_x &= \dfrac{2\,x_1 + 1\,x_2}{1+2} = \dfrac{2\times4 + 1\times(-2)}{3} = \dfrac{8-2}{3} = 2,\\[4pt] P_y &= \dfrac{2\,y_1 + 1\,y_2}{1+2} = \dfrac{2\times(-1) + 1\times(-3)}{3} = \dfrac{-2-3}{3} = -\dfrac{5}{3}. \end{aligned}$$

Therefore $$P\,(2,-\tfrac{5}{3}).$$

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2. Coordinates of Q  (ratio $$2:1$$)

Now $$m=2,\;n=1$$.

$$\begin{aligned} Q_x &= \dfrac{1\,x_1 + 2\,x_2}{2+1} = \dfrac{1\times4 + 2\times(-2)}{3} = \dfrac{4-4}{3} = 0,\\[4pt] Q_y &= \dfrac{1\,y_1 + 2\,y_2}{2+1} = \dfrac{1\times(-1) + 2\times(-3)}{3} = \dfrac{-1-6}{3} = -\dfrac{7}{3}. \end{aligned}$$

Hence $$Q\,(0,-\tfrac{7}{3}).$$

Conclusion
The points that trisect the line segment joining $$A(4,-1)$$ and $$B(-2,-3)$$ are
$$\bigl(2,-\tfrac{5}{3}\bigr)\;\text{and}\;\bigl(0,-\tfrac{7}{3}\bigr).$$

Step 1 : Choice of coordinate axes

Place the rectangle ABCD in the Cartesian plane with

• vertex $$A(0,0)$$,
• side $$AB$$ on the x-axis,
• side $$AD$$ on the y-axis.

Because 100 flower-pots, each 1 m apart, are arranged on $$AD$$, we take

$$AD = 100\;\text{m},\;\text{ so }\;D(0,100).$$

Chalk lines parallel to $$AD$$ are drawn at every 1 m, i.e. at $$x = 0,1,2,3,\dots$$ m.
Thus

• the 1^st line is $$x = 1$$,
• the 2^nd line is $$x = 2$$,
• the 8^th line is $$x = 8$$.

Step 2 : Coordinates of the two flags

Niharika runs $$\dfrac14$$ of $$AD$$ on the 2^nd line.

Vertical distance covered $$= \dfrac14 \times 100 = 25\;\text{m}$$.
Hence the green flag is at $$N\bigl(2,25\bigr).$$

Preet runs $$\dfrac15$$ of $$AD$$ on the 8^th line.

Vertical distance covered $$= \dfrac15 \times 100 = 20\;\text{m}$$.
Hence the red flag is at $$P\bigl(8,20\bigr).$$

Step 3 : Distance between the flags

Using the distance formula,

$$\begin{aligned} NP &= \sqrt{(8-2)^2 + (20-25)^2} \\ &= \sqrt{6^2 + (-5)^2} \\ &= \sqrt{36 + 25} \\ &= \sqrt{61}\;\text{m} \approx 7.81 \text{ m.} \end{aligned}$$

Step 4 : Mid-point of the segment joining the flags

For points $$N(x_1,y_1)$$ and $$P(x_2,y_2)$$, the midpoint is

$$M\left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right).$$

Therefore

$$\begin{aligned} M &= \left(\dfrac{2+8}{2},\;\dfrac{25+20}{2}\right) \\ &= (5,22.5). \end{aligned}$$

Interpretation

• Rashmi must stand on the 5^th chalk line (because $$x = 5$$ m from $$AD$$).
• She should run $$22.5$$ m up from $$AB$$ along that line.

Hence her blue flag will be exactly midway between the green and red flags.

Let the given points be

• $$A(-3, 10)\;\,(x_1, y_1)$$
• $$B(6, -8)\;\,(x_2, y_2)$$
• $$P(-1, 6)\;\,(x, y)$$ – the point which divides AB.

Suppose $$P$$ divides the line segment $$AB$$ internally in the ratio $$m : n$$, where

• $$m$$ corresponds to the part $$AP$$,
• $$n$$ corresponds to the part $$PB$$.

By the section formula for internal division, the coordinates of $$P$$ are

$$x = \dfrac{m x_2 + n x_1}{m + n}, \qquad y = \dfrac{m y_2 + n y_1}{m + n}.$$

Substituting all the known values:

For the x-coordinate

$$-1 = \dfrac{m(6) + n(-3)}{m + n}.$$

Multiplying both sides by $$(m + n)$$ gives

$$-m - n = 6m - 3n \;\;\Longrightarrow\;\; 0 = 7m - 2n.$$

Hence

$$\boxed{\;2n = 7m\;} \;\;\Longrightarrow\;\; n = \dfrac{7}{2} m. \quad(1)$$

For the y-coordinate

$$6 = \dfrac{m(-8) + n(10)}{m + n}.$$

Multiplying by $$(m + n)$$:

$$6m + 6n = -8m + 10n \;\;\Longrightarrow\;\; 14m - 4n = 0.$$

This simplifies to

$$\boxed{\;4n = 14m\;} \;\;\Longrightarrow\;\; n = \dfrac{7}{2} m. \quad(2)$$

Equations (1) and (2) are identical, so the value is consistent.
Choosing the smallest integral values that satisfy $$n = \dfrac{7}{2}m$$ gives

$$m = 2, \; n = 7.$$

Therefore

$$AP : PB = m : n = 2 : 7.$$

Because both $$m$$ and $$n$$ are positive, the point $$(-1, 6)$$ lies between $$A$$ and $$B$$ – the division is internal.

Given: $$\mathrm{A}(1,-5)$$ and $$\mathrm{B}(-4,5)$$.

Let $$\mathrm{P}(x,0)$$ be the required point on the $$x$$-axis. Suppose $$\mathrm{P}$$ divides $$\overline{\mathrm{AB}}$$ internally in the ratio $$k:1$$ (that is, $$\mathrm{AP}:\mathrm{PB}=k:1$$).

Section formula

• $$x=\dfrac{k x_2+x_1}{k+1}$$
• $$y=\dfrac{k y_2+y_1}{k+1}$$

Here $$x_1=1,\;y_1=-5,\;x_2=-4,\;y_2=5$$.

Because $$\mathrm{P}$$ lies on the $$x$$-axis, its $$y$$-coordinate is zero:

$$0=\dfrac{k(5)+(-5)}{k+1}=\dfrac{5k-5}{k+1}$$

The denominator $$k+1\neq0$$, so

$$5k-5=0\;\Rightarrow\;k=1$$

Hence $$\mathrm{AP}:\mathrm{PB}=1:1$$.

Coordinates of \(\mathrm{P}\)

With $$k=1$$,

$$x=\dfrac{1\cdot(-4)+1}{1+1}=\dfrac{-4+1}{2}=\dfrac{-3}{2}=-1.5$$

Therefore $$\mathrm{P}\Bigl(-\dfrac{3}{2},\,0\Bigr)$$.

Result
Ratio: $$1:1$$
Point of division: $$\left(-\dfrac{3}{2},\,0\right)$$

Let the vertices of the parallelogram in order be

• $$A(1, 2)$$
• $$B(4, y)$$
• $$C(x, 6)$$
• $$D(3, 5)$$

In a parallelogram the diagonals bisect each other, so the midpoint of $$AC$$ equals the midpoint of $$BD$$.

Midpoint of $$AC$$:

$$\left(\frac{1 + x}{2},\;\frac{2 + 6}{2}\right)=\left(\frac{1+x}{2},\;4\right)$$

Midpoint of $$BD$$:

$$\left(\frac{4 + 3}{2},\;\frac{y + 5}{2}\right)=\left(\frac{7}{2},\;\frac{y+5}{2}\right)$$

Equating the corresponding coordinates:

$$\frac{1 + x}{2}=\frac{7}{2}\;\Rightarrow\;1 + x = 7\;\Rightarrow\;x = 6$$

$$4 = \frac{y + 5}{2}\;\Rightarrow\;8 = y + 5\;\Rightarrow\;y = 3$$

Hence, $$x = 6$$ and $$y = 3$$.

The problem involves the mid-point formula.

If a point $$M(x_m, y_m)$$ is the mid-point of the segment joining $$A(x_1, y_1)$$ and $$B(x_2, y_2)$$, then

$$x_m = \frac{x_1 + x_2}{2}, \;\; y_m = \frac{y_1 + y_2}{2}.$$

Here, the centre of the circle is the mid-point of the diameter $$AB$$.

• Centre (mid-point) $$M = (2,\,-3).$$
• Given end-point $$B = (1,\,4).$$
• Let the unknown end-point be $$A = (x,\,y).$$

Apply the mid-point formulas separately for the x– and y–coordinates.

1. x-coordinate
   $$\frac{x + 1}{2} = 2 \;\;\Longrightarrow\; x + 1 = 4 \;\;\Longrightarrow\; x = 3.$$
2. y-coordinate
   $$\frac{y + 4}{2} = -3 \;\;\Longrightarrow\; y + 4 = -6 \;\;\Longrightarrow\; y = -10.$$

Hence $$A = (3,\,-10).$$

Let the given points be

$$A(-2,-2), \; B(2,-4).$$

P lies on the line segment $$\overline{AB}$$ and the length condition is

$$\mathrm{AP}=\dfrac{3}{7}\,\mathrm{AB}.$$

This means that P divides AB internally in the ratio

$$\mathrm{AP}:\mathrm{PB}=3:4,$$

because $$\mathrm{PB}=\mathrm{AB}-\mathrm{AP}=\dfrac{4}{7}\,\mathrm{AB}.$$

Denote the ratio by $$m:n=3:4$$ (with $$m=3$$ toward B and $$n=4$$ toward A).

By the Section Formula, if P divides the segment joining

$$A(x_1,y_1) \text{ and } B(x_2,y_2)$$

internally in the ratio $$m:n$$, then

$$ P\left(\dfrac{mx_2+nx_1}{m+n},\;\dfrac{my_2+ny_1}{m+n}\right). $$

Substitute $$x_1=-2,\;y_1=-2,\;x_2=2,\;y_2=-4,\;m=3,\;n=4$$:

\[\begin{aligned} x_P &= \dfrac{3\,(2)+4\,(-2)}{3+4} = \dfrac{6-8}{7} = -\dfrac{2}{7},\\[4pt] y_P &= \dfrac{3\,(-4)+4\,(-2)}{3+4} = \dfrac{-12-8}{7} = -\dfrac{20}{7}. \end{aligned}\]

Therefore, the required point is

$$P\left(-\dfrac{2}{7},\,-\dfrac{20}{7}\right).$$

Given: End-points of the line segment are $$A(-2,2)$$ and $$B(2,8)$$.

To divide $$\overline{AB}$$ into four equal parts we need the three interior points which cut the segment in the ratios

• $$1:3$$ (nearest to A)
• $$1:1$$ (mid-point)
• $$3:1$$ (nearest to B)

Let those points be $$P_1,\;P_2,\;P_3$$ respectively.

1. Coordinates of $$P_1$$ (ratio $$1:3$$)

Section formula for a point dividing $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ internally in the ratio $$m:n$$ is

$$\bigl(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\bigr).$$

Here $$m=1,\;n=3,$$ $$x_1=-2,\;y_1=2,\;x_2=2,\;y_2=8$$.

$$\displaystyle x_{P_1}=\frac{1\times2+3\times(-2)}{1+3}=\frac{2-6}{4}=-1,\;\;y_{P_1}=\frac{1\times8+3\times2}{4}=\frac{8+6}{4}=\frac{14}{4}=\tfrac72.$$

Thus $$P_1(-1,\tfrac72).$$

2. Coordinates of $$P_2$$ (ratio $$1:1$$: the mid-point)

Mid-point formula:

$$\displaystyle x_{P_2}=\frac{-2+2}{2}=0,\;\;y_{P_2}=\frac{2+8}{2}=5.$$

Hence $$P_2(0,5).$$

3. Coordinates of $$P_3$$ (ratio $$3:1$$)

Now $$m=3,\;n=1$$.

$$\displaystyle x_{P_3}=\frac{3\times2+1\times(-2)}{3+1}=\frac{6-2}{4}=1,\;\;y_{P_3}=\frac{3\times8+1\times2}{4}=\frac{24+2}{4}=\frac{26}{4}=\tfrac{13}{2}.$$

Therefore $$P_3(1,\tfrac{13}{2}).$$

Result

The required points dividing $$AB$$ into four equal parts are

$$P_1(-1,\tfrac72),\;P_2(0,5),\;P_3(1,\tfrac{13}{2}).$$

Label the vertices in the order given:

• $$A(3,0)$$
• $$B(4,5)$$
• $$C(-1,4)$$
• $$D(-2,-1)$$

Step 1  (optional check): All four sides are equal, so the quadrilateral is indeed a rhombus.

Since all four sides are equal, the given quadrilateral is a rhombus, so we can use the diagonal formula for its area.

Step 2 : Find the lengths of the two diagonals.

• Diagonal $$AC$$
  $$AC=\sqrt{\,( -1-3 )^2 + ( 4-0 )^2}\;=\;\sqrt{(-4)^2+4^2}\;=\;\sqrt{16+16}\;=\;\sqrt{32}=4\sqrt{2}$$
• Diagonal $$BD$$
  $$BD=\sqrt{\,( -2-4 )^2 + ( -1-5 )^2}\;=\;\sqrt{(-6)^2+(-6)^2}\;=\;\sqrt{36+36}\;=\;\sqrt{72}=6\sqrt{2}$$

Step 3 : Apply the rhombus area formula.

The area of a rhombus is one-half the product of its diagonals:

$$\text{Area}=\tfrac12\times AC \times BD =\tfrac12\times\bigl(4\sqrt{2}\bigr)\times\bigl(6\sqrt{2}\bigr) =\tfrac12\times24\times2 =24$$

Therefore, the area of the rhombus is $$24$$ square units.