Straight Lines Class 11 One Shot | NCERT 11th Maths Complete Chapter-9 Revision | CBSE 2026-27 Exam

Straight Lines — Class 11 Revision

The lesson reviews the coordinate-geometry foundations needed to understand equations of straight lines, then develops slope, angle relationships, equation forms, and distance calculations. The teacher emphasizes that a clear understanding of the basics helps with both routine and more challenging problems.

1. Coordinate-Geometry Foundations

• A point on a line satisfies its equation. To test whether a point lies on a line, substitute its coordinates for (x) and (y). If the equation holds, the point lies on the line.

• Distance between two points: For (A(x_1,y_1)) and (B(x_2,y_2)), [ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. ]

• Section formula: If (P) divides (AB) internally in the ratio (AP:PB=m:n), then [ P\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right). ] The midpoint is the special case (m=n=1): [ \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right). ]

• Area of a triangle: For vertices ((x_1,y_1)), ((x_2,y_2)), and ((x_3,y_3)), [ \text{Area}=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|. ] The absolute value is needed because area cannot be negative.

2. Inclination and Slope

• The inclination of a line is the angle (\theta) it makes with the positive (x)-axis, measured anticlockwise.

• The slope, or gradient, of a line is [ m=\tan\theta. ]

• Given two points on a line, [ m=\frac{y_2-y_1}{x_2-x_1}, ] provided (x_2\ne x_1).

• A horizontal line has slope (0); a vertical line has an undefined slope.

• The sign and value of the slope indicate how steeply, and in which direction, the line rises or falls.

3. Parallel and Perpendicular Lines

• Two non-vertical lines are parallel when their slopes are equal: [ m_1=m_2. ]

• Two lines are perpendicular when [ m_1m_2=-1. ]

• Slopes can be used to verify whether lines through given pairs of points are parallel or perpendicular.

4. Angle Between Two Lines

For lines with slopes (m_1) and (m_2), the acute angle (\theta) between them satisfies [ \tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|. ] If (1+m_1m_2=0), the lines are perpendicular, so the angle is (90^\circ).

When a problem gives an angle and a relation between the slopes—for example, one slope being a multiple of the other—substitute into this formula and solve. Because of the absolute value, consider both sign cases; this can produce multiple valid slope pairs.

5. Common Forms of a Line’s Equation

• Vertical line: (x=k), where (k) is constant.
• Horizontal line: (y=k).

• Point–slope form: Through ((x_1,y_1)) with slope (m), [ y-y_1=m(x-x_1). ]

• Two-point form: Through ((x_1,y_1)) and ((x_2,y_2)), [ y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1), ] when (x_2\ne x_1).

• Slope–intercept form: [ y=mx+c, ] where (m) is the slope and (c) is the (y)-intercept.

• Intercept form: If the line has (x)-intercept (a) and (y)-intercept (b), [ \frac{x}{a}+\frac{y}{b}=1. ]

• General form: [ Ax+By+C=0. ] When (B\ne0), rearranging gives a slope of (-A/B). Rearranging a general equation into slope–intercept form is a useful way to identify its slope and intercept.

6. Standard Problem-Solving Methods

• Find a line through a point with a given slope: Substitute the point and slope into point–slope form, then simplify.
• Find a line through two points: Use the two-point form, or first calculate the slope and then use point–slope form.
• Prove points are collinear: Form the equation of a line through two points and check whether the third point satisfies it. Alternatively, compare slopes.
• Find a perpendicular bisector:
  1. Find the midpoint of the segment.
  2. Calculate the segment’s slope.
  3. Use the negative reciprocal as the perpendicular slope.
  4. Use point–slope form through the midpoint.
• Find a line from its intercepts: Substitute the intercepts into intercept form and simplify.
• Handle an angle given as the line’s inclination: Convert the angle into a slope using (m=\tan\theta), taking the quadrant and sign into account.
• Reflection across the (x)-axis: Use the fact that the angle of incidence equals the angle of reflection. Relate the slopes of the incident and reflected paths, then use their given points to find where the ray meets the axis.
• Find a point’s image across a line: Use the perpendicular from the point to the line and the fact that the line of reflection bisects the segment joining the point and its image. The perpendicular foot is the midpoint of that segment.

7. Distance Formulas

• Perpendicular distance from ((x_1,y_1)) to (Ax+By+C=0): [ d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}. ]

• Distance between parallel lines: First write them with the same (A) and (B) coefficients: [ Ax+By+C_1=0,\qquad Ax+By+C_2=0. ] Then [ d=\frac{|C_2-C_1|}{\sqrt{A^2+B^2}}. ]

• For problems that specify a distance from a line, substitute the point into the distance formula. Since the formula contains an absolute value, consider both resulting sign cases.

8. Study Guidance

• Draw a diagram when working with line relationships, angles, intersections, or reflections; it can make the setup clearer.
• Focus on the meaning of slope and on how the different equation forms relate to one another.
• The teacher recommends starting with NCERT for school-level preparation, then using additional reference material for further practice where needed.
• As this is a revision session, students are encouraged to practise the NCERT exercises, including miscellaneous questions. The auto-generated subtitles contain some garbled expressions and arithmetic, so the formulas above present the intended standard methods.

Speakers and Sources Featured

• Kuldeep Bhandari — the teacher and main presenter.
• Brief background music and song snippets — audible in the lesson; the specific song or source is not identified in the subtitles.