ICSE Class 10 Mathematics • Chapter 09
Distance between $A(x_1, y_1)$ and $B(x_2, y_2)$:
$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Example 1: Find distance between A(3, 4) and B(7, 1).
$AB = \sqrt{(7-3)^2 + (1-4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ units
Applications:
If P divides line joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in ratio $m:n$:
Internal Division: $P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$
External Division: $P = \left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}\right)$
Special Case - Midpoint Formula: (when m:n = 1:1)
$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
Example 2: Find point P dividing A(2, 3) and B(7, 8) in ratio 2:3.
$P = \left(\frac{2(7) + 3(2)}{2+3}, \frac{2(8) + 3(3)}{2+3}\right) = \left(\frac{14+6}{5}, \frac{16+9}{5}\right) = (4, 5)$
Centroid of triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$:
$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$
Slope (gradient) of line through $(x_1, y_1)$ and $(x_2, y_2)$:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta$
where θ is angle line makes with positive x-axis.
| Line Type | Slope | Angle |
|---|---|---|
| Horizontal | m = 0 | θ = 0° |
| Vertical | m = undefined | θ = 90° |
| Rising (↗) | m > 0 | 0° < θ < 90° |
| Falling (↘) | m < 0 | 90° < θ < 180° |
| Condition | Slope Relation |
|---|---|
| Parallel Lines | $m_1 = m_2$ (slopes are equal) |
| Perpendicular Lines | $m_1 \times m_2 = -1$ (product = −1) |
Example 3: Line AB passes through A(2, 3) and B(4, 7). Line CD is perpendicular to AB. Find slope of CD.
Slope of AB = $\frac{7-3}{4-2} = \frac{4}{2} = 2$
For perpendicular: $m_{CD} = -\frac{1}{m_{AB}} = -\frac{1}{2}$
$y = mx + c$
where m = slope, c = y-intercept
Line with slope m passing through $(x_1, y_1)$:
$y - y_1 = m(x - x_1)$
Line through $(x_1, y_1)$ and $(x_2, y_2)$:
$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$
Line with x-intercept a and y-intercept b:
$\frac{x}{a} + \frac{y}{b} = 1$
Example 4: Find equation of line passing through (2, 5) with slope 3.
Using point-slope form: $y - 5 = 3(x - 2)$
$y - 5 = 3x - 6$
$y = 3x - 1$ or $3x - y - 1 = 0$
Example 5: Find equation of line with x-intercept 4 and y-intercept −3.
Using intercept form: $\frac{x}{4} + \frac{y}{-3} = 1$
$\frac{x}{4} - \frac{y}{3} = 1$
$3x - 4y = 12$
From equation ax + by + c = 0:
| Formula | Expression |
|---|---|
| Distance | $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ |
| Midpoint | $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ |
| Section (m:n) | $\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)$ |
| Centroid | $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$ |
| Slope | $\frac{y_2-y_1}{x_2-x_1}$ |
| Parallel | $m_1 = m_2$ |
| Perpendicular | $m_1 \times m_2 = -1$ |
BOARD Find the equation of line passing through (1, 2) and perpendicular to line 3x + 4y = 12.
BOARD The line joining A(−2, 9) and B(6, 3) is trisected at P and Q. Find coordinates of P and Q.