CUET UG Mathematics Coordinate Geometry Straight Lines Slope Forms of Equation Distance from a Point

Must‑Know (15–20 detailed bullets)

• The slope ( m ) of a line passing through two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is given by ( m = \frac{y_2 - y_1}{x_2 - x_1} ); for points (2, 3) and (5, 9), slope = ( \frac{9 - 3}{5 - 2} = 2 ).

• A line parallel to the x-axis has slope 0; for example, the line ( y = 4 ) has slope 0.

• A line parallel to the y-axis has undefined slope; for example, the line ( x = -3 ) has no defined slope.

• The slope of a line making an angle ( \theta ) with the positive x-axis is ( m = \tan\theta ); if ( \theta = 45^\circ ), then ( m = \tan 45^\circ = 1 ).

• Two lines are parallel if and only if their slopes are equal; lines ( y = 2x + 1 ) and ( y = 2x - 3 ) are parallel because both have slope 2.

• Two lines are perpendicular if the product of their slopes is ( -1 ); lines with slopes 3 and ( -\frac{1}{3} ) are perpendicular since ( 3 \times -\frac{1}{3} = -1 ).

• The equation of a line in slope-intercept form is ( y = mx + c ), where ( m ) is slope and ( c ) is y-intercept; for a line with slope 4 and y-intercept –2, the equation is ( y = 4x - 2 ).

• The point-slope form of a line is ( y - y_1 = m(x - x_1) ); for a line with slope 3 passing through (1, –2), the equation is ( y + 2 = 3(x - 1) ).

• The two-point form of a line passing through ( (x_1, y_1) ) and ( (x_2, y_2) ) is ( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} ); for points (1, 2) and (3, 6), the equation becomes ( \frac{y - 2}{4} = \frac{x - 1}{2} ).

• The intercept form of a line is ( \frac{x}{a} + \frac{y}{b} = 1 ), where ( a ) and ( b ) are x- and y-intercepts; a line cutting x-axis at 4 and y-axis at –3 has equation ( \frac{x}{4} + \frac{y}{-3} = 1 ).

• The general equation of a straight line is ( Ax + By + C = 0 ), where ( A, B, C ) are constants and ( A^2 + B^2 \neq 0 ); for ( 2x - 3y + 6 = 0 ), ( A = 2, B = -3, C = 6 ).

• The slope of the line ( Ax + By + C = 0 ) is ( m = -\frac{A}{B} ), provided ( B \neq 0 ); for ( 3x - 4y + 12 = 0 ), slope = ( -\frac{3}{-4} = \frac{3}{4} ).

• The distance of a point ( (x_1, y_1) ) from the line ( Ax + By + C = 0 ) is ( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} ); distance of (1, 2) from ( 3x - 4y + 5 = 0 ) is ( \frac{|3(1) - 4(2) + 5|}{\sqrt{3^2 + (-4)^2}} = \frac{|0|}{5} = 0 ).

• The perpendicular distance between two parallel lines ( Ax + By + C_1 = 0 ) and ( Ax + By + C_2 = 0 ) is ( \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} ); distance between ( 2x + 3y - 4 = 0 ) and ( 2x + 3y + 6 = 0 ) is ( \frac{|-4 - 6|}{\sqrt{4 + 9}} = \frac{10}{\sqrt{13}} ).

• The angle ( \theta ) between two lines with slopes ( m_1 ) and ( m_2 ) is given by ( \tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| ); for lines with slopes 1 and 2, ( \tan\theta = \left| \frac{1 - 2}{1 + (1)(2)} \right| = \frac{1}{3} ).

• If a line makes intercepts ( a ) and ( b ) on the axes, its area with the coordinate axes is ( \frac{1}{2}|ab| ); for ( \frac{x}{3} + \frac{y}{-4} = 1 ), area = ( \frac{1}{2} \times 3 \times 4 = 6 ) sq units.

• The foot of the perpendicular from a point ( (x_1, y_1) ) to the line ( Ax + By + C = 0 ) can be found using parametric form; verify from NCERT.

• Three points ( (x_1, y_1), (x_2, y_2), (x_3, y_3) ) are collinear if the area of the triangle formed is zero, or slope between any two pairs is same; (0, 0), (2, 2), (4, 4) are collinear since slope = 1 in each case.

• The equation of a line parallel to ( Ax + By + C = 0 ) is ( Ax + By + k = 0 ); a line parallel to ( 2x + 3y + 5 = 0 ) through (1, 1) is ( 2x + 3y - 5 = 0 ).

• The equation of a line perpendicular to ( Ax + By + C = 0 ) is ( Bx - Ay + k = 0 ); perpendicular to ( 3x - 4y + 7 = 0 ) is ( 4x + 3y + k = 0 ).

Difficulty Level

Intermediate — requires understanding of multiple forms and application in problems involving distance and angles, but formulas are direct from NCERT.

Common CUET Traps

• Trap: Assuming the slope of a vertical line is zero.
  Avoid: Vertical lines have undefined slope; only horizontal lines have slope zero.

• Trap: Using distance formula without absolute value or square root in denominator.
  Avoid: Always use ( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} ); missing modulus or denominator leads to wrong answer.

• Trap: Confusing intercept form ( \frac{x}{a} + \frac{y}{b} = 1 ) with signs when intercepts are negative.
  Avoid: If y-intercept is –3, write ( \frac{y}{-3} ), not ( \frac{y}{3} ); sign matters.

Practice MCQs

Q1. What is the slope of the line passing through (–1, 2) and (3, –2)?
A. 1
B. –1
C. 0
D. 2

Answer: B
Explanation: Slope = ( \frac{-2 - 2}{3 - (-1)} = \frac{-4}{4} = -1 )
Why others fail: Option A is chosen by reversing numerator or denominator.

Q2. Which of the following is the equation of a line with slope 2 and y-intercept 5?
A. ( y = 2x - 5 )
B. ( y = 5x + 2 )
C. ( y = 2x + 5 )
D. ( y = -2x + 5 )

Answer: C
Explanation: Slope-intercept form is ( y = mx + c ), so ( y = 2x + 5 )
Why others fail: Option A has correct slope but wrong sign for intercept.

Q3. The perpendicular distance of the point (3, 4) from the line ( 3x - 4y + 10 = 0 ) is:
A. 0
B. 1
C. ( \frac{3}{5} )
D. ( \frac{1}{5} )

Answer: B
Explanation: Distance = ( \frac{|3(3) - 4(4) + 10|}{\sqrt{9 + 16}} = \frac{|9 - 16 + 10|}{5} = \frac{3}{5} )? Wait: ( |3| = 3 ), so ( \frac{3}{5} ) → correction: ( |9 - 16 + 10| = |3| = 3 ), so ( \frac{3}{5} ) → answer should be C.
Wait: Recalculate: 3×3 = 9, –4×4 = –16, +10 → 9 –16 = –7 +10 = 3 → |3| = 3, √(9+16)=5 → 3/5.
But options include ( \frac{3}{5} ) as C.

Answer: C
Explanation: ( \frac{|3(3) - 4(4) + 10|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 - 16 + 10|}{5} = \frac{3}{5} )
Why others fail: Option B (1) is chosen by miscalculating numerator as 5 or ignoring signs.

Q4. The equation of a line perpendicular to ( 2x + 3y + 5 = 0 ) will have the form:
A. ( 2x - 3y + k = 0 )
B. ( 3x - 2y + k = 0 )
C. ( 3x + 2y + k = 0 )
D. ( 2x + 3y + k = 0 )

Answer: B
Explanation: Perpendicular to ( Ax + By + C = 0 ) is ( Bx - Ay + k = 0 ), so ( 3x - 2y + k = 0 )
Why others fail: Option A is chosen by swapping coefficients without sign change.

Q5. If the line ( 3x - 4y + 7 = 0 ) makes an angle ( \theta ) with the positive x-axis, then ( \tan\theta = )?
A. ( \frac{3}{4} )
B. ( \frac{4}{3} )
C. ( -\frac{3}{4} )
D. ( -\frac{4}{3} )

Answer: A
Explanation: Slope = ( -\frac{A}{B} = -\frac{3}{-4} = \frac{3}{4} ), and ( m = \tan\theta ), so ( \tan\theta = \frac{3}{4} )
Why others fail: Option C is chosen by forgetting negative sign in formula ( -A/B ).

Last‑Minute Revision (15–20 one‑liners)

• ⚠️ Slope of horizontal line = 0; vertical line = undefined.
• ⚠️ ( m = \tan\theta ), where ( \theta ) is angle with positive x-axis.
• ⚠️ Parallel lines: ( m_1 = m_2 ); perpendicular: ( m_1 m_2 = -1 ).
• ⚠️ Slope from general form: ( m = -\frac{A}{B} ), if ( B \neq 0 ).
• ⚠️ Point-slope form: ( y - y_1 = m(x - x_1) ).
• ⚠️ Two-point form: ( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} ).
• ⚠️ Intercept form: ( \frac{x}{a} + \frac{y}{b} = 1 ); a = x-intercept, b = y-intercept.
• ⚠️ General form: ( Ax + By + C = 0 ); ( A^2 + B^2 \neq 0 ).
• ⚠️ Distance from point to line: ( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} ).
• ⚠️ Distance between parallel lines: ( \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} ) for same ( A, B ).
• ⚠️ Angle between two lines: ( \tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| ).
• ⚠️ Equation of line parallel to ( Ax + By + C = 0 ) is ( Ax + By + k = 0 ).
• ⚠️ Equation of line perpendicular to ( Ax + By + C = 0 ) is ( Bx - Ay + k = 0 ).
• ⚠️ Three points collinear if slope between each pair is equal.
• ⚠️ Foot of perpendicular: use parametric method; verify from NCERT.
• ⚠️ Area of triangle formed by line with axes: ( \frac{1}{2}|ab| ).
• ⚠️ Line with slope 1 makes 45° with x-axis.
• ⚠️ If slope is –1, angle with x-axis is 135° (not 45°).
• ⚠️ Distance is always non-negative — modulus is crucial.
• ⚠️ Mnemonic: "Perpendicular swap and negate" → for ( Ax + By ), perpendicular is ( Bx - Ay ).