Quick Revision

Points and Quadrants

Signs of $(x, y)$ : first quadrant $(+, +)$ , second $(-, +)$ , third $(-, -)$ , fourth $(+, -)$ .
x is the abscissa and y the ordinate.
A point on the x-axis is $(a, 0)$ and a point on the y-axis is $(0, b)$ ; neither lies in a quadrant.
Distance from the x-axis $= |y|$ ; distance from the y-axis $= |x|$ .

For example: $(6, -11)$ is in the fourth quadrant, 11 units from the x-axis and 6 units from the y-axis.

x-axis: $(x, -y)$ . y-axis: $(-x, y)$ . Origin: $(-x, -y)$ .
$y = x$ : $(y, x)$ . $y = -x$ : $(-y, -x)$ .
$x = a$ : $(2a - x, y)$ . $y = b$ : $(x, 2b - y)$ .
A point and its image are the same distance from the mirror, on opposite sides.
In any other line: find the foot F of the perpendicular, then the image is $2F - P$ .

For example: in $x = 3$ , $(1, 5)$ goes to $(5, 5)$ ; in $y = -1$ it goes to $(1, -7)$ .

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ ; from the origin, $\sqrt{x^2 + y^2}$ .
Triplets save the square root: 3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29.
A distance with an unknown coordinate usually gives two answers.
Shapes: find every side and both diagonals. Four equal sides: rhombus; add equal diagonals: square. Opposite sides equal: parallelogram; add equal diagonals: rectangle.
Collinear: the two shorter distances add up to the longest.

Midpoint $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ ; the other end is $B = 2M - A$ .
Internal division in $m : n$ : $(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n})$ .
The ratio from one coordinate: $(x - x_1) : (x_2 - x)$ .
The x-axis divides AB in the ratio $-y_1 : y_2$ ; the y-axis in $-x_1 : x_2$ .
[Tier 2] External division: $\frac{mx_2 - nx_1}{m - n}$ , and the same for y.

$Point P dividing AB in the ratio m to n$

Centroid $G(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})$ ; a missing vertex is $3G - A - B$ .
Parallelogram ABCD: $D = A + C - B$ , since the diagonals bisect each other.
Area $= \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ .
With one vertex at O: $\frac{1}{2}|x_1y_2 - x_2y_1|$ . Otherwise move a vertex to O first.
Area 0 means the points are collinear. A quadrilateral is two triangles.

[SSC Tip] An answer without the $\frac{1}{2}$ is twice the area, a likely wrong option.

$m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta$ , where $\theta$ is the angle with the positive x-axis.
For $ax + by + c = 0$ , $m = -\frac{a}{b}$ .
Slope 1 means $45^\circ$ , $-1$ means $135^\circ$ , $\sqrt{3}$ means $60^\circ$ , $\frac{1}{\sqrt{3}}$ means $30^\circ$ .
A line parallel to the x-axis has slope 0; parallel to the y-axis, undefined.
Collinear points: equal slopes.

$Slope as rise over run with angle theta$

Slope-intercept: $y = mx + c$ , where c is the y-intercept.
Point-slope: $y - y_1 = m(x - x_1)$ . Through two points: find the slope, then use this form.
Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$ .
$x = a$ is upright and $y = b$ is level; $x = 0$ is the y-axis and $y = 0$ the x-axis.
A point lies on a line when it satisfies the equation. To find k, put the point in.

Parallel: $m_1 = m_2$ ; the line through a point is $ax + by + k = 0$ .
Perpendicular: $m_1m_2 = -1$ , or $a_1a_2 + b_1b_2 = 0$ ; the line is $bx - ay + k = 0$ .
Perpendicular bisector of AB: through the midpoint, slope $-\frac{1}{m}$ .
Distance from $(x_1, y_1)$ to $ax + by + c = 0$ : $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$ ; between parallel lines, $\frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$ .
[Tier 2] Angle: $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$ .

$ax + by = c$ cuts the x-axis at $(\frac{c}{a}, 0)$ and the y-axis at $(0, \frac{c}{b})$ .
Area with the axes: $\frac{c^2}{2|ab|}$ .
Two lines and the x-axis: base between the two x-intercepts, height $= |y|$ of the meeting point.
$x = 0$ , $y = 0$ , $x = p$ and $y = q$ enclose a rectangle of area $|pq|$ .

$Line cutting the axes at c over a and c over b$

For $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ :

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ : the lines cross, one solution.
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ : parallel lines, no solution.
All three ratios equal: the same line, infinitely many solutions.
Where two lines meet: solve the pair, or test the options in both equations.

The algebra is in Chapter 11.

For example: $x + 2y = 3$ and $2x + 4y = 9$ are parallel: $\frac{1}{2} = \frac{2}{4}$ , but $\frac{3}{9}$ is different.

Sketch the axes and the points every time; a wrong sign shows up at once.
Mistakes to watch: slope as $\frac{x_2 - x_1}{y_2 - y_1}$ , slope $\frac{a}{b}$ for $-\frac{a}{b}$ , area without the $\frac{1}{2}$ , a perpendicular slope of $-m$ or $\frac{1}{m}$ , the intercept a for $\frac{c}{a}$ , a section ratio taken the wrong way round, and a reflection in the wrong axis.

If the PYQ Practice Sets or the Chapter Test are still left, do them now, against the clock.