Introduction

Coordinate geometry beautifully blends algebra and geometry. Introduced by René Descartes in 1637, it allows us to represent geometric figures algebraically. In this chapter, we focus on the simplest yet most powerful figure – the straight line. You’ll learn slope, conditions for parallelism and perpendicularity, equations of lines in different forms, distance formulas, and solved NCERT examples.

Key Formulas Used

Distance between two points:

PQ=(x2−x1)2+(y2−y1)2

Section formula (internal division):

(mx2+nx1m+n,my2+ny1m+n)

Midpoint formula:

(x1+x22,y1+y22)

Slope of line through two points:

m=y2−y1x2−x1

Condition for parallel lines: Slopes equal → $m_{1} = m_{2}$
Condition for perpendicular lines: Slopes negative reciprocals → $m_{1} \cdot m_{2} = - 1$
Equation of line (point-slope form):

y−y1=m(x−x1)

Equation of line (two-point form):

y−y1=y2−y1x2−x1(x−x1)

Slope-intercept form:

y=mx+c

Intercept form:

xa+yb=1

Distance of point from line:

d=∣Ax1+By1+C∣A2+B2

✅ Solved NCERT Examples (Step by Step)

Example 1

Find slope of lines: (a) Through (3, -2) and (-1, 4) (b) Through (3, -2) and (7, -2) (c) Through (3, -2) and (3, 4) (d) Inclination 60° with x-axis

Solution: (a) $m = \frac{4 - (- 2)}{- 1 - 3} = \frac{6}{- 4} = - \frac{3}{2}$ (b) $m = \frac{- 2 - (- 2)}{7 - 3} = 0$ (c) $m = \frac{4 - (- 2)}{3 - 3} = \frac{6}{0}$ → Not defined (d) $m = \tan 60 ° = \sqrt{3}$

Example 2

If angle between two lines is 45° and slope of one line is 1, find slope of other line.

Solution: Formula: $\tan θ = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}$ Here, $θ = 45 °$ , $m_{1} = 1$ . $\tan 45 ° = 1 = \frac{m_{2} - 1}{1 + m_{2}}$ Cross multiplying: $1 + m_{2} = m_{2} - 1$ → Contradiction. Alternate case: $\frac{m_{2} - 1}{1 + m_{2}} = - 1$ . Solving: $m_{2} - 1 = - (1 + m_{2})$ → $m_{2} = - 3$ . So slope of other line = -3.

Example 3

Line through (-2, 6) and (4, 8) is perpendicular to line through (8, 12) and (x, 24). Find x.

Solution: Slope of first line: $m_{1} = \frac{8 - 6}{4 - (- 2)} = \frac{2}{6} = \frac{1}{3}$ Slope of second line: $m_{2} = \frac{24 - 12}{x - 8} = \frac{12}{x - 8}$ Condition: $m_{1} \cdot m_{2} = - 1$ $\frac{1}{3} \cdot \frac{12}{x - 8} = - 1$ → $\frac{12}{3 (x - 8)} = - 1$ $\frac{4}{x - 8} = - 1$ → $x - 8 = - 4$ → $x = 4$ .