Class 11 Maths · Original EZT practice
Straight Lines (Class 11 Chapter 9): summary and practice
Straight Lines turns coordinate geometry into algebra: the slope m = tan theta = (y2 - y1)/(x2 - x1), the parallel and perpendicular tests, the five forms of a line's equation in this reprint, and the distance of a point from a line. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Introduction
Coordinate geometry describes a line with numbers: where it sits and how steeply it climbs. Every question in this chapter comes back to one quantity - the slope.
2. Slope of a line
The slope m is tan theta, where theta is the angle the line makes with the positive x-axis, and through two points it is (y2 - y1)/(x2 - x1). Parallel lines share a slope (m1 = m2); perpendicular lines satisfy m1 m2 = -1; the angle between two lines comes from tan of the difference formula.
3. The five forms of the equation
Section 9.3 builds five ways to write a line in this reprint: horizontal or vertical lines, point-slope form, two-point form, slope-intercept form and intercept form. Each form packages different given information - pick the one that matches what the problem hands you.
4. Distance of a point from a line
The distance of (x1, y1) from ax + by + c = 0 is |a x1 + b y1 + c| / sqrt(a^2 + b^2), and the distance between the parallel lines y = mx + c1 and y = mx + c2 follows by putting any point of one line into the other's formula.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Find the slope through two points
Choose the method: m = (y2 - y1)/(x2 - x1) - rise over run.
EZT practice question
Find the slope of the line through (0, 1) and (3, 4).
Show the worked solution
- Rise: 4 - 1 = 3.
- Run: 3 - 0 = 3.
- m = 3/3.
Answer: m = 1 (the line sits at 45 degrees to the x-axis).
Check: m = tan theta = 1 gives theta = 45 degrees, which matches equal rise and run.
Watch for this: Keep the same point order in top and bottom - (4 - 1)/(0 - 3) flips the sign.
Now try it yourself
Slope through (1, 2) and (5, 10)?
Check your answer
m = 2.
(10 - 2)/(5 - 1) = 8/4.
Test perpendicular lines with slopes
Choose the method: Lines are perpendicular exactly when m1 m2 = -1.
EZT practice question
Are y = 2x + 1 and y = -x/2 + 3 perpendicular?
Show the worked solution
- Slopes are 2 and -1/2.
- Product: 2 x (-1/2).
- = -1.
Answer: Yes - the slopes multiply to -1, so the lines are perpendicular.
Check: A perpendicular slope is the negative reciprocal: for 2 that is -1/2, which is what the second line has.
Watch for this: The test is the PRODUCT equal to -1, not the sum equal to 0.
Now try it yourself
Is y = 3x - 2 perpendicular to y = -x/3 + 7?
Check your answer
Yes.
3 x (-1/3) = -1.
Write a line in point-slope form
Choose the method: y - y1 = m(x - x1), then simplify to ax + by + c = 0.
EZT practice question
Equation of the line through (2, 3) with slope 4?
Show the worked solution
- Point-slope: y - 3 = 4(x - 2).
- Expand: y - 3 = 4x - 8.
- Bring everything to one side.
Answer: 4x - y - 5 = 0.
Check: Substitute (2, 3): 8 - 3 - 5 = 0 - the point lies on the line, and the slope read from -a/b is 4.
Watch for this: Distribute the 4 to BOTH terms: 4(x - 2) is 4x - 8, not 4x - 2.
Now try it yourself
Line through (1, -1) with slope 2?
Check your answer
2x - y - 3 = 0.
y + 1 = 2(x - 1), then simplify.
Distance of a point from a line
Choose the method: d = |a x1 + b y1 + c| / sqrt(a^2 + b^2).
EZT practice question
Distance of the origin from 3x + 4y - 10 = 0?
Show the worked solution
- Numerator: |0 + 0 - 10| = 10.
- Denominator: sqrt(9 + 16) = 5.
- Divide.
Answer: d = 10/5 = 2.
Check: The foot of the perpendicular from (0, 0) is (6/5, 8/5), whose distance from the origin is sqrt(36 + 64)/5 = 10/5 = 2.
Watch for this: The absolute value is not optional - distance is never negative.
Now try it yourself
Distance of (1, 1) from x + y - 4 = 0?
Check your answer
sqrt(2), about 1.41.
|1 + 1 - 4|/sqrt(2) = 2/sqrt(2) = sqrt(2).
Distance between parallel lines
Choose the method: Pick any point on one line, apply the point-to-line formula on the other.
EZT practice question
Distance between y = 2x + 1 and y = 2x + 5?
Show the worked solution
- Point (0, 1) lies on the first line.
- Second line as 2x - y + 5 = 0.
- d = |0 - 1 + 5|/sqrt(4 + 1) = 4/sqrt(5).
Answer: 4/sqrt(5), about 1.79.
Check: Parallel lines keep a constant gap - any other point of the first line, like (1, 3), gives the same 4/sqrt(5).
Watch for this: Write BOTH lines with the same coefficients before reading off c1 and c2.
Now try it yourself
Distance between y = 3x and y = 3x + 10?
Check your answer
sqrt(10), about 3.16.
From (0, 0) to 3x - y + 10 = 0: |10|/sqrt(9 + 1).
Handle horizontal and vertical lines
Choose the method: Horizontal: y = a (slope 0). Vertical: x = b (slope undefined).
EZT practice question
What is the slope of the line x = 5, and its distance from the origin?
Show the worked solution
- x = 5 is vertical - every point has x-coordinate 5.
- Slope: run is zero, so (y2 - y1)/(x2 - x1) divides by zero.
- Distance from origin: |0 - 5|/sqrt(1) using a = 1, b = 0, c = -5.
Answer: Slope undefined; distance 5.
Check: Geometrically the line is 5 units right of the y-axis, so the perpendicular gap from the origin is exactly 5.
Watch for this: Vertical lines have NO slope value - undefined is the correct statement, not zero.
Now try it yourself
Slope and origin-distance of y = -3?
Check your answer
Slope 0; distance 3.
Horizontal lines have tan theta = 0, and the gap is the absolute ordinate.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Slope formula?
Show answer and reason
m = (y2 - y1)/(x2 - x1).
Rise over run between two points.
2. Parallel vs perpendicular?
Show answer and reason
m1 = m2 vs m1 m2 = -1.
Equal slopes never meet; negative reciprocals cross at 90 degrees.
3. Point-slope form?
Show answer and reason
y - y1 = m(x - x1).
One point plus slope fixes the line.
4. Point-to-line distance?
Show answer and reason
|a x1 + b y1 + c|/sqrt(a^2 + b^2).
Perpendicular distance is the shortest one.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find the slope through (1, 2) and (5, 10).
- Distance of (1, 1) from x + y - 4 = 0.
- Write the line through (2, 3) with slope 4 in the form ax + by + c = 0.
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Questions about this practice page
- What does the slope of a line actually measure?
- Steepness and direction: how much y changes per unit of x. Positive slope climbs left to right, negative slope falls, zero is horizontal, and a vertical line has no defined slope because the run is zero.
- How many forms does the equation of a line have in this chapter?
- Five in the Reprint 2026-27: horizontal/vertical lines, point-slope, two-point, slope-intercept and intercept form (9.3.1 to 9.3.5). The older normal form is not part of this reprint.
- When do I use the two-point form instead of point-slope?
- When you know two points but not the slope. The two-point form folds the slope calculation into the equation; computing m first and using point-slope reaches the same line.
- Why the absolute value in the distance formula?
- The numerator a x1 + b y1 + c is positive on one side of the line and negative on the other. Distance is a length, so the formula takes the magnitude and drops the side information.
- Are these the NCERT exercise questions for Straight Lines (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.