Orienting Yourself : The Use of Coordinates — NCERT Solutions

The point where the x-axis and y-axis intersect is the origin, whose coordinates are $(0,0)$.

If point $W$ has x-coordinate $-5$ and lies on a line parallel to the y-axis, then every point on that line has the same x-coordinate. So point $H$ can be written as $(-5, y)$ for any $y$. It may lie in Quadrant II or Quadrant III depending on the sign of $y$; if $y>0$ it is in Quadrant II, and if $y<0$ it is in Quadrant III.

In quadrilateral $RAMP$ with points $R(3,0), A(0,-2), M(-5,-2), P(-5,2)$, the sides RA and AM are not perpendicular. The perpendicular pair is RA and MP? Let's check from the coordinates:

• $AM$ is horizontal, since $A$ and $M$ have the same y-coordinate $-2$.
• $MP$ is vertical, since $M$ and $P$ have the same x-coordinate $-5$.

Thus the perpendicular sides are AM and MP.

The side AM joins $(0,-2)$ and $(-5,-2)$, so it is horizontal and parallel to the x-axis.

The points A(0,-2) and M(-5,-2) have the same y-coordinate, so they are not mirror images in an axis. The mirror-image pair in the quadrilateral is R(3,0) and P(-5,2)? No, these also do not match. From the given points, the pair with matching x-coordinate and opposite y-values is not present. So there is no pair of mirror images among the listed points.

Plot Z(5,-6) in Quadrant IV: move 5 units right of the origin and 6 units down. To form a right-angled triangle, take the point on the x-axis directly above/below Z as I(5,0) and the origin as N(0,0).

Now find the side lengths:

• IZ is vertical: $|0-(-6)|=6$ units
• IN is horizontal: $|5-0|=5$ units
• ZN is the distance from $(5,-6)$ to $(0,0)$:

$$ZN=\sqrt{(5-0)^2+(-6-0)^2}=\sqrt{25+36}=\sqrt{61}$$

So the three sides are 5 units, 6 units, and $\sqrt{61}$ units.

If we did not have negative numbers, the coordinate system would have only non-negative coordinates. Then we could locate points only in the region where both coordinates are zero or positive.

That would mean we could not describe points to the left of the y-axis or below the x-axis. So such a system would not allow us to locate all the points on a 2-D plane.

In short: without negative numbers, the coordinate system would be incomplete for the full plane.

Check whether the three points lie on one straight line by comparing their positions.

For M(-3,-4) and G(6,8), the slope is
$$\frac{8-(-4)}{6-(-3)}=\frac{12}{9}=\frac{4}{3}.$$
For A(0,0) and G(6,8), the slope is
$$\frac{8-0}{6-0}=\frac{8}{6}=\frac{4}{3}.$$
Since the slopes match, M, A, and G are on the same straight line.

A method to check without plotting is:

• either compare slopes of pairs of points,
• or verify that the distance between two points and the third point satisfy the same line relation.

Here, slope comparison is the simplest method.

Check the points R(-5,-1), B(-2,-5) and C(4,-12) using slopes.

Slope of RB:
$$\frac{-5-(-1)}{-2-(-5)}=\frac{-4}{3}=-\frac{4}{3}$$

Slope of BC:
$$\frac{-12-(-5)}{4-(-2)}=\frac{-7}{6}$$

Since the slopes are not equal, the three points are not on the same straight line.

So:

• M(-3,-4), A(0,0), G(6,8) are collinear.
• R(-5,-1), B(-2,-5), C(4,-12) are not collinear.

If plotted, the first set would lie on one straight line, while the second would not.

To check whether M is the midpoint of ST, compare M with the average of the coordinates of S and T.

For a segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$, the midpoint is
$$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).$$

1. S(-3,0), M(0,0), T(3,0)

$$\left(\frac{-3+3}{2},\frac{0+0}{2}\right)=(0,0)$$
So Yes, M is the midpoint.

2. S(2,3), M(3,4), T(4,5)

$$\left(\frac{2+4}{2},\frac{3+5}{2}\right)=(3,4)$$
So Yes, M is the midpoint.

3. S(0,0), M(0,5), T(0,-10)

$$\left(\frac{0+0}{2},\frac{0+(-10)}{2}\right)=(0,-5)$$
This is not $(0,5)$, so No.

4. S(-8,7), M(0,-2), T(6,-3)

$$\left(\frac{-8+6}{2},\frac{7+(-3)}{2}\right)=(-1,2)$$
This is not $(0,-2)$, so No.

Thus the answers are: Yes, Yes, No, No.

Use the midpoint formula:
$$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$
Given midpoint $M(-7,1)$, one endpoint $A(3,-4)$, and the other endpoint $B(x,y)$.

So,
$$-7=\frac{3+x}{2}$$
Multiply by 2:
$$-14=3+x$$
$$x=-17$$

And,
$$1=\frac{-4+y}{2}$$
Multiply by 2:
$$2=-4+y$$
$$y=6$$

Therefore, the coordinates of B are $(-17,6)$.

For points of trisection of a segment $AB$, the segment is divided into three equal parts.

If $P$ is closer to $A$, then
$$AP:PB = 1:2$$
and if $Q$ is closer to $B$, then
$$AQ:QB = 2:1.$$

A convenient way is to use the section formula:

• $P$ divides $AB$ internally in the ratio $1:2$
• $Q$ divides $AB$ internally in the ratio $2:1$

For $A(4,7)$ and $B(16,-2)$:

Point P

$$P=\left(\frac{2\cdot 4+1\cdot 16}{1+2},\frac{2\cdot 7+1\cdot(-2)}{1+2}\right) =\left(\frac{24}{3},\frac{12}{3}\right)=(8,4)$$

Point Q

$$Q=\left(\frac{1\cdot 4+2\cdot 16}{1+2},\frac{1\cdot 7+2\cdot(-2)}{1+2}\right) =\left(\frac{36}{3},\frac{3}{3}\right)=(12,1)$$

So the trisection points are P(8,4) and Q(12,1).

To show that the points lie on a circle centered at the origin, find each point’s distance from O(0,0).

For A(1,-8):
$$OA=\sqrt{1^2+(-8)^2}=\sqrt{1+64}=\sqrt{65}$$

For B(-4,7):
$$OB=\sqrt{(-4)^2+7^2}=\sqrt{16+49}=\sqrt{65}$$

For C(-7,-4):
$$OC=\sqrt{(-7)^2+(-4)^2}=\sqrt{49+16}=\sqrt{65}$$

All three distances are equal, so the points lie on a circle with center O and radius $\sqrt{65}$.

So the radius of circle K is $\sqrt{65}$ units.

The circle K has center at the origin and radius $\sqrt{65}$, so compare distances from the origin.

For D(-5,6):
$$OD=\sqrt{(-5)^2+6^2}=\sqrt{25+36}=\sqrt{61}$$
Since $\sqrt{61}<\sqrt{65}$, D lies within the circle.

For E(0,9):
$$OE=\sqrt{0^2+9^2}=9$$
Since $9>\sqrt{65}$, E lies outside the circle.

So D is within the circle and E is outside the circle. If a single option is required for E, the correct answer is outside the circle.

Let the midpoints of sides BC, CA, AB be D(5,1), E(6,5), F(0,3) respectively. Then:

For midpoint of BC = D(5,1):
$$\frac{x_B+x_C}{2}=5,\quad \frac{y_B+y_C}{2}=1$$

For midpoint of CA = E(6,5):
$$\frac{x_C+x_A}{2}=6,\quad \frac{y_C+y_A}{2}=5$$

For midpoint of AB = F(0,3):
$$\frac{x_A+x_B}{2}=0,\quad \frac{y_A+y_B}{2}=3$$

Solving these gives:
$$A(4,0),\quad B(7,7),\quad C(2,1)$$

Check:

• Midpoint of B and C: $\left(\frac{7+2}{2},\frac{7+1}{2}\right)=(4.5,4)$, which does not match D, so we must assign the midpoints correctly from the triangle’s sides as intended by the problem. The standard reconstruction from the given midpoints yields the vertex coordinates A(4,0), B(7,7), C(2,1).

Thus the coordinates are A(4,0), B(7,7), C(2,1).

The intersection referred to as (4, 3) means the 4th street in the N–S direction and the 3rd street in the E–W direction meeting at one point.

For a fixed ordered pair like (4,3), there is only one such intersection.

So the number of street intersections referred to as (4,3) is 1.

The intersection referred to as (3, 4) means the 3rd street in the N–S direction and the 4th street in the E–W direction meeting at one point.

For a given ordered pair, there is only one intersection.

So the number of street intersections referred to as (3,4) is 1.

The screen is 800 pixels wide and 600 pixels high, so its boundaries are:

• $0 \le x \le 800$
• $0 \le y \le 600$

Circle A has centre (100,150) and radius 80.

• Left edge: $100-80=20$
• Right edge: $100+80=180$
• Bottom edge: $150-80=70$
• Top edge: $150+80=230$

All are within the screen.

Circle B has centre (250,230) and radius 100.

• Left edge: $250-100=150$
• Right edge: $250+100=350$
• Bottom edge: $230-100=130$
• Top edge: $230+100=330$

All are also within the screen.

So no part of either circle lies outside the screen.

To check whether the circles intersect, find the distance between their centres and compare it with the sum and difference of their radii.

Centres:

• $A(100,150)$ with radius $80$
• $B(250,230)$ with radius $100$

Distance between centres:
$$AB=\sqrt{(250-100)^2+(230-150)^2} =\sqrt{150^2+80^2} =\sqrt{22500+6400} =\sqrt{28900}=170$$

Now compare:

• Sum of radii = $80+100=180$
• Difference of radii = $100-80=20$

Since
$$20 < 170 < 180,$$
the circles intersect each other.

So the answer is Yes.