# Coordinate Geometry Maths Notes

## Coordinate Geometry Maths Notes

In the previous standard, we understood, how to find the distance between any two points on a number line.

Here coordinate of point P is 3 and coordinate of point Q is 5.
∴ d(P,Q) = 5 – 3 = 2
If coordinate of point Q is x2 and coordinate of point P is x1 and x2 > x1, then d(P, Q) = x2 – x1.
Using the same concept, we will understand to find the distance between two points in the XY-plane.

(i) To find the distance between any two points on any axis:

In the figure, points P(x1, 0) and Q(x2, 0) lie on X-axis. Q lies to the right side of P.
∴ X2 > X1
∴ d(P, Q) = x2 – x1.

(ii) To find distance between any two po1nts on Y-axis.
In the figure, points R(0, y1) and S(0, y2) lie on Y-axis.
The points S lie above point R.
∴ y2 > y1

∴ d(R, S) = y2 – y1

To find the distance between the two points if the segment joining these points is parallel to any axis in the XY-plane:

(i) In the figure, seg KT is parallel to X-axis.
∴ y-coordinates of points K and T are equal.
We draw seg KP and seg TQ perpendicular to X-axis.
∴ ▢ KPQT is a rectangle.
∴ KT = PQ … (Opposite sides of a rectangle)
as PQ = x2 – x1,
KT = x2 – x1

(ii)

In the figure, seg CD is parallel to Y-axis.
∴ x-coordinates of points C and D are equal.
Draw seg CS and segDR perpeidicuIar to Y-axis.
∴ ▢CSRD is a rectangle.
∴ CD = SR … (Opposite sides of a rectangle)
as SR = y2 – y1, CD = y2 – y1

Distance formula:

In the figure, P(x1, y1) and Q(x2, y2) are any two points in the XY-plane.
From point Q draw perpendicular QS on X-axis.
Similarly from point P draw perpendicular PT on seg QS.
seg QS is parallel to Y-axis.
∴ the x-coordinate of point T is x2.
seg PT is parallel to X-axis.
∴ the y-coordinate of point T is y1.
∴ PT = d(P,T) = x2 – x1;
QT = d(Q, T) = y2 – y1
In right angled △ PTQ.
PQ2 + QT2 … (Pythagoras theorem)
= (x2 – x1)2 + (y2 – y1)2
∴ PQ = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
This is known as distance formula.
Note that,
$$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ = $$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$$

Remember this!

• Coordinates of origin are (0, 0). Hence if coordinates of point A are (x, y), then d(O, A) = $$\sqrt{x^{2}+y^{2}}$$
• If points A(x1, y1), B(x2, y2) 1i in the XY-plane, then d(A, B) = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
that is,
AB2 = (x2 – x1)2 + (y2 – y1)2 = (x1 – x2)2 + (y1 – y2)2.

Division of a line segment:

In the figure, XZ = 4 and ZY = 6
∴ $$\frac{\mathrm{XZ}}{\mathrm{ZY}}=\frac{4}{6}=\frac{2}{3}$$
∴ we can say that, ‘Point Z divides the line segment XY in the ratio 2 : 3’.

Section Formula:

In the figure, point P on the seg RS in XY-plane, divides seg RS in the ratio m : n.
Assume R(x1, y1), S(x2, y2) and P(x, y). Draw seg RT, seg PQ and seg SM perpendicular to X-axis.
∴ T(x1, 0); Q(x, 0) and M(x2, 0).
∴ TQ = x – x1 and QM = x2 – x ……… (1)
seg RT || seg PQ seg SM.
∴ by the property of intercepts made by three parallel lines, $$\frac{\mathrm{RP}}{\mathrm{PS}}=\frac{\mathrm{TQ}}{\mathrm{QM}}=\frac{m}{n}$$
Now TQ = x – x1 and QM = x2 – x … [From (1)1
∴ $$\frac{x-x_{1}}{x_{2}-x}=\frac{m}{n}$$
∴ n(x – x1) = m(x2 – x)
∴ nx – nx1 = mx2 – mx
∴ mx + nx = mx2 + nx1
∴ x(m + n) = mx2 + nx1
∴ x = $$\frac{m x_{2}+n x_{1}}{m+n}$$
Similarly drawing perpendiculars from points R, P and S to Y-axis, we get, y = $$\frac{m y_{2}+n y_{1}}{m+n}$$
∴ coordinates of the point, which divides the line
segment joining the points R(x1, y1) and S(x2, y2) in the ratio m : n are given by
$$\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$$

Coordinates of the midpoint of a segment:

If P(x1, y1) and Q(x2, y2) are two points and M(x, y) is the midpoint of seg PQ then m = n.

Let’s recall

• The medians of a triangle are concurrent.
• The point of concurrence of the medians is called centroid.
• Centroid divides the median in the ratio 2: 1.

Centroid Formula:

Suppose, in ∆PQR, P(x1, y1), Q(x2, y2) and R(x3, y3) are the vertices. Seg PM is a median and G(x, y) is the centroid.
By definition of the median, M is the midpoint of seg QR.
Coordinates of M, by midpoint formula is
$$\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$$
G(x, y) is the centroid of ∆ PQR
∴ PG : GM = 2 : 1
By section formula,

Thus, if (x1, y1), (x2, y2) and (x3, y3) are the vertices of a triangle then the coordinates of the centroid are
$$\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$$.
This is called centroid formula.

Remember this!

Section formula:

The coordinates of a point which divides the line segment joined by two distinct points (x1, y1) and (x2, y2) in the ratio m : n are
$$\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$$

Midpoint formula:

The coordinates of midpoint of a line segment
joining two distinct points (x1, y1) and (x2, y2) are
$$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$$

Centroid formula:

If (x1, y1), (x2, y2) and (x3, y3) are the vertices of a triangle then coordinates of the centroid are
$$\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$$

Slope of a Line:

In the figure P(x1, y1) and Q(x2, y2) are any two paints in XY-plane.
the ratio $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$ is constant.
It is true for any two points on line l.
The ratio $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$ is called the slope of the line l.
Usually slope is denoted by letter m.
∴ m = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

Remember this!

• Slope of X-axis and any line parallel to X-axis is O.
• Slope of Y-axis and any line parallel to Y-axis cannot be determined.
• Parallel lines have equal slopes.
• When two distinct fines have sanie slope, then the two lines are parallel.

Slope of line-using ratio in trigonometry

The tangent ratio of an angle made by the line with the positive direction of X-axis is called the slope of that line.

In the figure, line l makes an angle θ with the positive direction of X-axis.
∴ slope of line l = m = tan θ.