Chapter 1: Orienting Yourself: The Use of Coordinates

Step 1 ‒ Understand the question
The question does not ask for any calculation; it simply wants the generally accepted ("standard") width for an ordinary room door. You are also invited to verify this by measuring a few doors at home or at school.

Step 2 ‒ Recall/consult the standard dimensions
In Indian building practice, the Bureau of Indian Standards (BIS) and most architectural hand-books list the nominal size for an internal (room-to-room) door as:

• Width = $$0.90\,\text{m} = 90\,\text{cm}$$
• Height = $$2.10\,\text{m} = 210\,\text{cm}$$ (height is not asked here, but is often quoted together).

Step 3 ‒ Verify with actual measurements
Take a measuring tape or a metre scale and check a few doors:

The small variations you will observe (e.g. 88 cm → 95 cm) are due to finishing tolerances or special design needs, but they all cluster around the nominal standard of $$0.90\,\text{m}$$.

Conclusion
Hence, the standard width specified for a normal room door is about $$0.90\,\text{metre}$$ (90 cm).

Step 1 : Understand the meaning of “suitable”
For a wheelchair user to move comfortably through a doorway, Building Codes in India (and in most other countries) recommend a clear opening of at least $$90\text{ cm}$$ (0.9 m). This allows a standard wheelchair of width $$75\text{ cm}$$ to pass with about $$7{-}8\text{ cm}$$ clearance on each side.

Thus a doorway is suitable if

Step 2 : Measure a representative door in your school

1. Take a steel measuring-tape.
   
2. Open the door fully so that it does not block the opening.
   
3. Measure the horizontal distance between the inner faces of the two door-jambs (ignore the thickness of the door itself). Call this measurement $$W$$.

Example measurement
Suppose you obtain
$$W = 83\text{ cm}$$

Step 3 : Compare with the requirement
Insert the measured value in inequality (1):
$$83\text{ cm}<90\text{ cm}$$

The condition $$W\ge90\text{ cm}$$ is not satisfied.

Step 4 : Draw the conclusion
Because the door-width is less than the recommended minimum, the door in this example is not suitable for wheelchair users.

Step 5 : Check other doors
Repeat Steps 2 – 4 for the main gate, library, washrooms, laboratories, etc. If any frequently used door has $$W<90\text{ cm}$$, that door needs widening, automatic opening, or an alternate accessible route.

Final statement
The suitability of your school’s doors depends solely on the measured width $$W$$. Use criterion (1):

• If $$W\ge90\text{ cm}$$ ⇒ the door is suitable.
• If $$W<90\text{ cm}$$ ⇒ it is not suitable.

Step 1 | Recall the definition of Cartesian coordinates
A point in the plane is written as an ordered pair $$\bigl(x , y\bigr)$$, where

• $$x$$ is called the x‑coordinate (or abscissa).
• $$y$$ is called the y‑coordinate (or ordinate).

Step 2 | Understand what the y‑axis is
The y‑axis is the vertical line that passes through the origin and is described by the equation $$x = 0$$. Every point lying on this line therefore satisfies $$x = 0$$.

Step 3 | State the form of any point on the y‑axis
If a point is on the y‑axis, its coordinates must be of the form $$\bigl(0 , y\bigr)$$, where $$y$$ can be any real number (positive, negative or zero).

Step 4 | Read the required coordinate
Comparing the general form $$\bigl(0 , y\bigr)$$ with the ordered pair $$\bigl(x , y\bigr)$$, we see that

\[x = 0\]

Hence, the x‑coordinate of every point on the y‑axis is $$0$$.

Let us recall what an ordered pair  $$ (x , y) $$  means.

• First component $$x$$ = perpendicular distance of the point from the y-axis; taken positive if the point is to the right of the y-axis and negative if it is to the left.
• Second component $$y$$ = perpendicular distance of the point from the x-axis; taken positive if the point is above the x-axis and negative if it is below.

Now consider a point $$P$$ that lies on the x-axis.

1. Since $$P$$ is on the x-axis, its perpendicular distance from the x-axis is zero.
     Therefore the ordinate of $$P$$ is $$0$$; that is, $$y = 0$$.
2. The point may be any real‐number distance away from the y-axis.
     Let that signed distance be $$x$$. It can take any real value: $$x \in \mathbb{R}$$.

Hence the coordinates of every point on the x-axis can be written in the general form

Conversely, if the ordinate of a point is zero, the point must lie on the x-axis. Thus we obtain the desired generalisation, entirely analogous to the statement "a point on the y-axis has coordinates $$(0,y)$$".

Let the two points be $$P\,(x,\,y)$$ and $$Q\,(y,\,x)$$.

For two points to coincide, their corresponding coordinates must be equal:

First-coordinate equality    $$x = y,$$
Second-coordinate equality $$y = x.$$

Both statements are the same, so a single condition emerges:

If this condition holds, put $$x = y = k$$. Then

$$P\,(k,\,k) \quad\text{and}\quad Q\,(k,\,k),$$
which is one and the same point — any point that lies on the line $$y = x$$.

If $$x \neq y$$, at least one pair of corresponding coordinates differs, so the points are distinct.

Therefore, point $$Q\,(y, x)$$ coincides with point $$P\,(x, y)$$ only when $$x = y$$.

Given statement
If $$x \neq y$$, then $$(x,y) \neq (y,x)$$; and $$(x,y)=(y,x)$$ iff $$x=y$$. We must decide whether it is true.

Key fact about ordered pairs
An ordered pair is a pair with order. Two ordered pairs are equal precisely when their corresponding components are equal; that is

\[(a,b)=(c,d) \iff a=c \text{ and } b=d\]

This definition is part of the Class 9 syllabus and may be taken as known.

Part 1 : Show that $$x \neq y \;\Rightarrow\;(x,y) \neq (y,x)$$

1. Assume $$x \neq y$$.
2. Compare $$(x,y)$$ and $$(y,x)$$.
     • In $$(x,y)$$ the first component is $$x$$.
     • In $$(y,x)$$ the first component is $$y$$.
3. Because $$x \neq y$$, the first components differ. By the definition above, two ordered pairs with different first components can never be equal.
4. Therefore $$(x,y) \neq (y,x)$$.

Part 2 : Show that $$(x,y)=(y,x) \;\Leftrightarrow\; x=y$$

(i) Necessity ("only if"): Suppose $$(x,y)=(y,x)$$.
By the equality rule, the first components must be equal, hence $$x=y$$. Thus equality of pairs forces equality of the numbers.

(ii) Sufficiency ("if"): Now assume $$x=y$$.
Then both ordered pairs become $$(x,x)$$, so of course $$(x,y)=(y,x)$$.

Combining (i) and (ii) gives the required biconditional:

\[(x,y)=(y,x) \iff x=y\]

Conclusion
Both parts have been proved, so the claim is true.

Step 1 : Understand what “hitting” means.
When the door is pulled open it sweeps a quarter-circle whose
• centre is the hinge $$B_1$$, and
• radius is the breadth (width) of the door.
If any part of the wardrobe lies on or inside this quarter-circle, the door will hit it.

Step 2 : Read the relevant lengths from the plan.
The plan is drawn on a square grid; one small square represents $$10\text{ cm}$$ of actual length. Counting squares from the figure we obtain

• breadth of the bathroom door = $$9$$ squares = $$9\times10 = 90\text{ cm}$$,
• co-ordinates of the hinge $$B_1 = (5,\,4)$$ (in squares),
• co-ordinates of the nearest corner of the wardrobe $$W = (15,\,9)$$ (in squares).

Step 3 : Distance from the hinge to the nearest corner of the wardrobe.

By the distance formula,

Each square is $$10\text{ cm}$$, so

Step 4 : Compare with the present door breadth.
Required clearance $$\approx 111.8\text{ cm}$$, present door breadth $$= 90\text{ cm}$$.

Since $$90\text{ cm} < 111.8\text{ cm}$$, the tip of the door, when swung through 90°, stays about $$111.8 - 90 \approx 21.8\text{ cm}$$ short of the wardrobe.

Therefore the present door will not hit the wardrobe.

Step 5 : What if the door is made wider?
Let the new breadth be $$x\text{ cm}$$. The door just grazes the wardrobe when its radius equals the hinge-to-wardrobe distance, i.e. when

Hence

• if $$x \le 111.8\text{ cm}$$ (roughly $$112\text{ cm}$$), the door still clears the wardrobe and no change is needed;
• if $$x > 111.8\text{ cm}$$, the door will strike the wardrobe. Two simple remedies are:
  • shift the hinge to the other jamb of the same doorway so that the door now opens against the opposite wall (away from the wardrobe), or
  • replace the hinged door by a sliding or folding door, which does not sweep an arc into the room.

Conclusion. With the present $$90\text{ cm}$$ door there is a clearance of about $$22\text{ cm}$$, so it does not touch the wardrobe. Any door wider than the critical value $$50\sqrt{5}\text{ cm} \approx 111.8\text{ cm}\;(\approx 112\text{ cm})$$ will collide with the wardrobe and one of the changes above is needed.

Given points

Point A has coordinates $$A\,(3,\,4)$$ and point D has coordinates $$D\,(7,\,1)$$.

1. Distance covered along the x-axis

1. The x-coordinate of A is $$x_A = 3$$.
2. The x-coordinate of D is $$x_D = 7$$.
3. Change in the x-coordinate: $$\Delta x = x_D - x_A = 7 - 3 = 4$$.
4. Since distance is always taken as a positive quantity, the distance moved along the x-axis is $$|\Delta x| = 4\text{ units}$$.

2. Distance covered along the y-axis

1. The y-coordinate of A is $$y_A = 4$$.
2. The y-coordinate of D is $$y_D = 1$$.
3. Change in the y-coordinate: $$\Delta y = y_D - y_A = 1 - 4 = -3$$.
4. The negative sign shows the movement is downward, but the distance (magnitude) is $$|\Delta y| = 3\text{ units}$$.

Result

Thus, the movement from A to D covers:

• $$4\text{ units}$$ along the x-axis, and
• $$3\text{ units}$$ along the y-axis.

Let us recall the construction done just before the question was asked.

• Point A represents 0 on the number-line and point B represents 2; hence $$AB = 2\text{ units}$$ lies on the number-line.
• At B a perpendicular of length 1 unit was erected to meet point C; thus $$BC = 1\text{ unit}$$ and $$\angle ABC = 90^{\circ}$$.

Thus △ABC is a right-angled triangle with the lengths of the two perpendicular sides already known.

By the Pythagoras Theorem applied to △ABC,

$$\text{(hypotenuse)}^{2}=\text{(base)}^{2}+\text{(height)}^{2}\;.$$

Therefore

$$AC^{2}=AB^{2}+BC^{2}=2^{2}+1^{2}=4+1=5,$$

so

$$AC = \sqrt{5}.\quad(1)$$

In the construction, the compass was then kept at centre A with radius $$AC$$ and an arc was drawn to cut the (produced) number-line at point D. As every point on the arc is at a distance $$AC$$ from A, we have

$$AD = AC.$$ Using (1), this gives

\[AD = \sqrt{5}.\]

Hence yes, the already known distances $$AB$$ and $$BC$$ do help, because together they determine $$AC$$ by the Pythagoras theorem, and the construction ensures $$AD = AC = \sqrt{5}.$$

Step 1 – Understand what “reflection in the origin” means on a number line
On a horizontal number line the origin is the point marked $$0$$. To reflect a point $$P$$ having coordinate $$x$$ in the origin, we have to plot a new point $$P'$$ such that the origin $$O$$ is exactly the midpoint of the segment $$PP'$$. Algebraically this means

$$OP = OP' \text{ and } O \text{ divides } PP' \text{ internally in the ratio }1:1.$$

Step 2 – Find the coordinate of the image
Let the coordinate of $$P'$$ be $$k$$. Using the midpoint formula for points on a straight line, the origin $$O$$ (coordinate $$0$$) satisfies
\[\frac{x + k}{2} = 0\]
Solving for $$k$$ gives
$$x + k = 0 \;\;\Rightarrow\;\; k = -x.$$
Thus, after the reflection the coordinate changes from $$x$$ to $$-x$$.

Step 3 – Analyse what is unchanged
The distance of a point from the origin on a number line is the absolute value $$|x|$$. For the image point $$P'$$ the distance is $$|-x|$$. But $$|-x| = |x|$$. Therefore the magnitude or absolute value of the number remains exactly the same.

Step 4 – Analyse what is changed
Although the distance from the origin is unchanged, the direction from the origin reverses. A positive number $$x$$ moves to the right of $$0$$, while its image $$-x$$ moves the same distance to the left, and vice-versa. In algebraic terms, the sign of the number changes (positive becomes negative, negative becomes positive), so the actual numerical value changes from $$x$$ to $$-x$$.

Conclusion
With a reflection in the origin:

• The distance from the origin (the absolute value $$|x|$$) stays the same.
• The sign/direction changes, so the number itself becomes its negative.

Let the coordinates of the given vertices of $$\triangle ADM$$ be

• $$A\, (x_1,\,y_1)$$
• $$D\, (x_2,\,y_2)$$
• $$M\, (x_3,\,y_3)$$

Earlier you reflected the triangle in the y-axis. In that case every point $$P(x,\,y)$$ went to $$P'( -x,\,y)$$ and you noted the following facts:

1. The image triangle $$\triangle A'D'M'$$ is congruent to $$\triangle ADM$$ because the distance formula gives

\[\,AD = A'D'\qquad DM = D'M'\qquad MA = M'A'\]

(all three pairs of corresponding sides are equal).

2. Each original point and its image are equidistant from the reflecting line; e.g.

\[\text{dist}(A,\,y\text{-axis}) = |x_1|\quad\text{and}\quad \text{dist}(A',\,y\text{-axis}) = |-x_1| = |x_1|.\]

The question now asks what happens if, instead, you reflect $$\triangle ADM$$ in the x-axis.

1. Coordinates after reflection in the x-axis

The rule for reflection in the x-axis is

\[P(x,\,y) \;\longrightarrow\; P''(x,\,-y).\]

Hence

• $$A''\,(x_1,\,-y_1)$$
• $$D''\,(x_2,\,-y_2)$$
• $$M''\,(x_3,\,-y_3)$$

2. Congruence of the two triangles

Using the distance formula between two generic points $$P(x_p,\,y_p)$$ and $$Q(x_q,\,y_q)$$,

\[PQ = \sqrt{(x_q-x_p)^2 + (y_q-y_p)^2}.\]

Compute, for example, $$AD$$ and $$A''D''$$:

\[\begin{aligned} AD &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2},\\ A''D'' &= \sqrt{\bigl(x_2 - x_1\bigr)^2 + \bigl((-y_2) - (-y_1)\bigr)^2}\\ &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = AD. \end{aligned}\]

The same calculation holds for the other two sides, so

\[\triangle ADM \cong \triangle A''D''M''.\]

3. Distance of a point and its image from the x-axis

The perpendicular distance of any point $$P(x,\,y)$$ from the x-axis is $$|y|$$. For $$A$$ and $$A''$$:

\[\text{dist}(A,\,x\text{-axis}) = |y_1|, \quad \text{dist}(A'',\,x\text{-axis}) = |-y_1| = |y_1|.\]

Thus each vertex and its image lie at the same distance from (but on opposite sides of) the x-axis. The corresponding statement clearly holds for $$D,\,D''$$ and $$M,\,M''$$.

4. Conclusion

All the facts you recorded for reflection in the y-axis — congruence of the two triangles and equal perpendicular distances of corresponding points from the reflecting line — hold unchanged when the triangle is reflected in the x-axis. The only difference is that reflected points now have the same x-coordinate and the opposite y-coordinate, instead of the other way round.

Hence the answer is Yes: the observations remain the same.

The Cartesian plane is formed by two number lines that are drawn at right angles to each other:

• the horizontal line is the $$x$$ axis,
• the vertical line is the $$y$$ axis.

The point where these two axes meet is called the origin.

At the origin there is no displacement along either axis, so both the $$x$$ and $$y$$ measurements are zero:

$$x\text{-coordinate}=0, \; y\text{-coordinate}=0$$

Consequently, the coordinates of the point of intersection of the two axes are $$\bigl(0,0\bigr)$$.

Given: The point W has x-coordinate equal to $$-5$$.

Step 1 — Equation of the line through W parallel to the y-axis.
A line parallel to the y-axis is vertical, so every point on it has the same x-coordinate. If such a line passes through a point whose x-coordinate is $$k$$, its equation is simply $$x = k$$.

Since W has x-coordinate $$-5$$, the required line is

Step 2 — Coordinates of any point H on that line.
Every point H lying on $$x = -5$$ has x-coordinate fixed at $$-5$$, while its y-coordinate can be any real number. Hence

$$H(-5,\,y),\; y \in \mathbb{R}.$$

Step 3 — In which quadrants can H lie?

• Quadrant II (x negative, y positive): when $$y > 0$$, the point $$(-5,\,y)$$ has $$x = -5 < 0$$ and $$y > 0$$, so H lies in Quadrant II.
• Quadrant III (x negative, y negative): when $$y < 0$$, the point $$(-5,\,y)$$ has $$x < 0$$ and $$y < 0$$, so H lies in Quadrant III.
• If $$y = 0$$, then $$H = (-5,\,0)$$ lies on the negative x-axis; points on an axis are not considered to be in any quadrant.
• Quadrants I and IV require $$x > 0$$, which is impossible here since $$x = -5$$.

Conclusion: Every point on the required line has coordinates $$(-5,\,y)$$. Such a point can lie only in Quadrant II (when $$y > 0$$) or Quadrant III (when $$y < 0$$); if $$y = 0$$ the point lies on the negative x-axis.

Step 1 : Draw the axes and plot the point

1. Draw a horizontal line and mark its centre O. Call it the x-axis.
2. Through O draw a vertical line; call it the y-axis.
3. Choose a convenient scale, e.g. 1 cm = 1 unit on both axes and label the positive and negative directions.
4. To plot $$Z\,(5,-6)$$ move 5 units to the right of the origin and 6 units down. Mark the point and label it $$Z(5,-6)$$.

Step 2 : Construct a right-angled triangle IZN

1. Through $$Z$$ draw a line perpendicular to the x-axis (i.e. a vertical line). It meets the x-axis at $$I(5,0)$$.
2. Through $$Z$$ draw a line perpendicular to the y-axis (i.e. a horizontal line). It meets the y-axis at $$N(0,-6)$$.
3. Join $$I$$ to $$N$$. The triangle $$IZN$$ is now complete.

Step 3 : Verify the right angle

The segment $$IZ$$ is vertical (parallel to the y-axis) and the segment $$ZN$$ is horizontal (parallel to the x-axis). Since the x-axis and the y-axis are perpendicular, $$IZ \perp ZN$$. Hence $$\angle IZN$$ is a right angle.

Step 4 : Find the lengths of the three sides

• Vertical side: $$IZ = |0 - (-6)| = 6\text{ units}$$.
• Horizontal side: $$ZN = |5 - 0| = 5\text{ units}$$.
• Hypotenuse: $$IN$$ can be found by the Distance Formula or by Pythagoras.
  Using Pythagoras, \[IN = \sqrt{IZ^{2} + ZN^{2}} = \sqrt{6^{2} + 5^{2}} = \sqrt{36 + 25} = \sqrt{61}\;\text{units}\] A decimal approximation is $$IN \approx 7.81\text{ units}$$.

Conclusion

The right-angled triangle $$IZN$$ has side lengths
$$IZ = 6\text{ units},\; ZN = 5\text{ units},\; IN = \sqrt{61}\text{ units}\,(\approx7.81\text{ units}).$$

Step 1 : What does “no negative numbers” mean?
In ordinary arithmetic the set of real numbers is $$\mathbb R = \{ \ldots , -3,-2,-1,0,1,2,3,\ldots \}$$. If we forbid negative numbers, the only numbers left are the non-negative real numbers

\[ \mathbb R_{\ge 0}=\{\,x\mid x\ge 0\,\}=\{0,1,2,3,\ldots\}\cup\{\text{all positive fractions and decimals}\}. \]

These are the numbers that would be available to use as coordinates.

Step 2 : Building the coordinate system with only non-negative numbers
A point in the Cartesian plane is written as an ordered pair $$\bigl(x,\,y\bigr)$$ where $$x$$ is read from the horizontal x-axis and $$y$$ from the vertical y-axis. If both $$x$$ and $$y$$ are forced to satisfy $$x\ge0$$ and $$y\ge0$$, then only the following positions are possible:

• The origin $$(0,0)$$.
• All points on the positive part of the x-axis, $$\bigl(x,0\bigr)$$ with $$x>0$$.
• All points on the positive part of the y-axis, $$\bigl(0,y\bigr)$$ with $$y>0$$.
• All interior points of the first quadrant, $$\bigl(x,y\bigr)$$ with $$x>0;,\,y>0;$$.

This restricted coordinate system therefore covers only the first quadrant together with the two positive semi-axes.

Step 3 : Which points are lost?
Any point whose $$x$$-coordinate, $$y$$-coordinate, or both are negative will be unreachable. Typical examples are

• $$(-4,\,2)$$ in the second quadrant, because $$x=-4<0;$$.
• $$(3,-1)$$ in the fourth quadrant, because $$y=-1<0;$$.
• $$( -5,-7)$$ in the third quadrant, because both coordinates are negative.

Since the complete Cartesian plane consists of four quadrants, the restricted system can describe only one of them. Hence three-quarters of the plane—in fact, infinitely many points—cannot be located.

Step 4 : Final conclusion
Without negative numbers the coordinate system reduces to a “one-quadrant” grid containing all points $$(x,y)$$ with $$x\ge0,\,y\ge0$$. Because points having a negative $$x$$ or $$y$$ coordinate cannot be expressed, such a system cannot locate every point in two-dimensional space.

Therefore, negative numbers are essential for a complete 2-D Cartesian coordinate system.

To decide whether $$M(-3,-4)$$, $$A(0,0)$$ and $$G(6,8)$$ lie on one straight line we can compare the slopes of the segments that would join them.

1. Slope of segment $$MA$$
For $$M(x_1,y_1)=(-3,-4)$$ and $$A(x_2,y_2)=(0,0)$$, the two–point formula $$m=\dfrac{y_2-y_1}{x_2-x_1}$$ gives
$$m_{MA}=\dfrac{0-(-4)}{0-(-3)}=\dfrac{4}{3}$$

2. Slope of segment $$AG$$
With $$A(0,0)$$ and $$G(6,8)$$ we get
$$m_{AG}=\dfrac{8-0}{6-0}=\dfrac{8}{6}=\dfrac{4}{3}$$

3. Compare

Since the slopes are equal, the two segments are parts of the same straight line; therefore the three points are collinear.

Other non-graphical check: Using the distance formula we find $$MA=5$$, $$AG=10$$ and $$MG=15$$; because $$MA+AG=MG$$, the points must lie on a single straight line.

Given points
R  : $$(-5,\,-1)$$   B  : $$(-2,\,-5)$$   C  : $$(4,\,-12)$$

In Problem 6 you learnt that three points lie on the same straight line (are collinear) iff the slope between any two of them is the same as the slope between the other two.

Step 1 — Find the slope of $$\overline{RB}$$

For points $$\bigl(x_1,\,y_1\bigr)$$ and $$\bigl(x_2,\,y_2\bigr)$$ the slope is $$m = \dfrac{y_2-y_1}{x_2-x_1}$$.

Take R as $$\bigl(x_1,\,y_1\bigr)$$ and B as $$\bigl(x_2,\,y_2\bigr)$$:

$$m_{RB}=\dfrac{y_B-y_R}{x_B-x_R}=\dfrac{-5-(-1)}{-2-(-5)}=\dfrac{-4}{3}=-\dfrac43$$

Step 2 — Find the slope of $$\overline{BC}$$

Now B is $$\bigl(x_1,\,y_1\bigr)$$ and C is $$\bigl(x_2,\,y_2\bigr)$$:

$$m_{BC}=\dfrac{y_C-y_B}{x_C-x_B}=\dfrac{-12-(-5)}{4-(-2)}=\dfrac{-7}{6}=-\dfrac76$$

Step 3 — Compare the two slopes

$$m_{RB}=-\dfrac43 \quad\text{and}\quad m_{BC}=-\dfrac76$$

Since $$-\dfrac43\neq-\dfrac76$$, the slopes are different. Therefore points R, B and C are not collinear.

(You may verify by also finding $$m_{RC} = \dfrac{-12-(-1)}{4-(-5)} = -\dfrac{11}{9}$$, which is again unequal.)

Step 4 — Plotting the two sets of points

• First set (from Problem 6): plot the three points you examined there. You would have found they do lie on the same straight line; join them with a ruler to check.
• Second set (R, B, C): mark R at $$(-5,-1)$$, B at $$(-2,-5)$$, C at $$(4,-12)$$. Join any two of them with a straight edge; you will see the third point does not fall on that line, confirming the algebraic result.

Hence, using the slope test and a quick graph, only the first trio of Problem 6 is collinear; the points R, B and C are not.

Mid-point formula
For two points $$S\bigl(x_1,\,y_1\bigr)$$ and $$T\bigl(x_2,\,y_2\bigr)$$, the coordinates of their mid-point M are \[ \bigl(\tfrac{x_1+x_2}{2},\,\tfrac{y_1+y_2}{2}\bigr). \] M will be the required mid-point iff its given coordinates satisfy the two equalities $$x_M = \tfrac{x_1+x_2}{2}\;\text{ and }\; y_M = \tfrac{y_1+y_2}{2}.$$

Connection observed
Whenever M is the mid-point of ST, each coordinate of M equals the arithmetic mean (average) of the corresponding coordinates of S and T, i.e. $$x_M=\tfrac{x_S+x_T}{2},\; y_M=\tfrac{y_S+y_T}{2}.$$

Given:

• Point $$A\,(3,\,-4)$$
• Point $$B\,(x,\,y)$$ (unknown)
• Midpoint $$M\,(-7,\,1)$$ of the line segment $$AB$$

Step 1 — Write the midpoint formula.

The midpoint of a segment joining $$A\,(x_1,\,y_1)$$ and $$B\,(x_2,\,y_2)$$ is

Step 2 — Substitute the known values.

Here $$x_1 = 3,\; y_1 = -4,\; x_2 = x,\; y_2 = y$$ and the midpoint is $$(-7,\,1)$$. Therefore

$$\frac{3 + x}{2} = -7 \qquad\text{and}\qquad \frac{-4 + y}{2} = 1.$$

Step 3 — Solve for x.

$$\frac{3 + x}{2} = -7 \;\Longrightarrow\; 3 + x = -14 \;\Longrightarrow\; x = -17.$$

Step 4 — Solve for y.

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

Step 5 — State the coordinates of B.

Hence $$B\,(-17,\,6).$$

Check: Midpoint of $$A(3,-4)$$ and $$B(-17,6)$$ is $$\left(\tfrac{3+(-17)}{2},\,\tfrac{-4+6}{2}\right) = (-7,\,1) = M.$$ ✓

Idea recalled from the midpoint formula

The midpoint M of a segment joining $$A\,(x_1,y_1)$$ and $$B\,(x_2,y_2)$$ is obtained by taking the average of the corresponding co-ordinates:

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

If instead of an equal (1 : 1) share we want a point that shares the segment in the ratio m : n, we keep the same “weighted-average” idea but give the x– and y–coordinates of the point the respective weights m and n:

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

This is called the section formula for internal division.

Step 1 : Work out the required ratios

For trisection we need two points:

• P is one-third of the way from A to B, so $$AP:PB=1:2$$ (P is nearer to A).
• Q is two-thirds of the way from A to B, so $$AQ:QB=2:1$$ (Q is nearer to B).

Step 2 : Insert the co-ordinates of A(4,7) and B(16,−2)

• For P (ratio 1 : 2):

$$\begin{aligned} P_x &= \frac{1\times16\; +\;2\times4}{1+2}=\frac{16+8}{3}=8,\\[2pt] P_y &= \frac{1\times(-2)+2\times7}{1+2}=\frac{-2+14}{3}=4. \end{aligned}$$

Hence $$P\,(8,4).$$

• For Q (ratio 2 : 1):

$$\begin{aligned} Q_x &= \frac{2\times16\; +\;1\times4}{2+1}=\frac{32+4}{3}=12,\\[2pt] Q_y &= \frac{2\times(-2)+1\times7}{2+1}=\frac{-4+7}{3}=1. \end{aligned}$$

Hence $$Q\,(12,1).$$

Step 3 : Quick vector check (optional)

The vector $$\overrightarrow{AB}=(16-4,\,-2-7)=(12,-9).$$ One third of this is $$(4,-3).$$ Adding once to A gives P and adding twice gives Q — the same results as above, confirming the calculation.

Therefore, the co-ordinates of the required trisection points are

$$P\,(8,4)\quad\text{and}\quad Q\,(12,1).$$

Step 1 : Name the unknown vertices
Let the vertices be $$A(x_1,\,y_1)$$, $$B(x_2,\,y_2)$$ and $$C(x_3,\,y_3)$$.

Step 2 : Write the midpoint relations
If a point is the midpoint of a segment, each coordinate of that point is the average of the corresponding coordinates of the segment’s end-points.

• D is the midpoint of $$BC$$:  $$D\bigl(\tfrac{x_2+x_3}{2},\,\tfrac{y_2+y_3}{2}\bigr)=(5,1)$$
• E is the midpoint of $$CA$$:  $$E\bigl(\tfrac{x_3+x_1}{2},\,\tfrac{y_3+y_1}{2}\bigr)=(6,5)$$
• F is the midpoint of $$AB$$:  $$F\bigl(\tfrac{x_1+x_2}{2},\,\tfrac{y_1+y_2}{2}\bigr)=(0,3)$$

Step 3 : Translate each midpoint into two linear equations

Step 4 : Solve for the x-coordinates

Add the three x-equations:

$$ (x_2+x_3)+(x_3+x_1)+(x_1+x_2)=10+12+0 $$

$$ 2(x_1+x_2+x_3)=22 \;\;\Longrightarrow\;\; x_1+x_2+x_3=11 $$

Now determine each unknown:

• From $$x_2+x_3=10$$ and the above sum, $$x_1=11-(x_2+x_3)=11-10=1$$.
• Substitute $$x_1=1$$ into $$x_1+x_2=0$$  ⇒  $$x_2=-1$$.
• Finally, $$x_3=11-(x_1+x_2)=11-(1+(-1))=11$$.

Thus $$x_A=1,\; x_B=-1,\; x_C=11$$.

Step 5 : Solve for the y-coordinates

Add the three y-equations:

$$ (y_2+y_3)+(y_3+y_1)+(y_1+y_2)=2+10+6 $$

$$ 2(y_1+y_2+y_3)=18 \;\;\Longrightarrow\;\; y_1+y_2+y_3=9 $$

Now determine each unknown:

• From $$y_2+y_3=2$$ and the above sum, $$y_1=9-(y_2+y_3)=9-2=7$$.
• Substitute $$y_1=7$$ into $$y_1+y_2=6$$  ⇒  $$y_2=-1$$.
• Finally, $$y_3=9-(y_1+y_2)=9-(7+(-1))=3$$.

Thus $$y_A=7,\; y_B=-1,\; y_C=3$$.

Step 6 : Write the coordinates of the vertices

Step 7 : Verification (optional but recommended)
Midpoint of $$BC$$: $$\bigl(\tfrac{-1+11}{2},\,\tfrac{-1+3}{2}\bigr)=(5,1)=D$$ ✔
Midpoint of $$CA$$: $$\bigl(\tfrac{11+1}{2},\,\tfrac{3+7}{2}\bigr)=(6,5)=E$$ ✔
Midpoint of $$AB$$: $$\bigl(\tfrac{1+(-1)}{2},\,\tfrac{7+(-1)}{2}\bigr)=(0,3)=F$$ ✔

Hence the obtained coordinates are correct.