ICSE Class 10 Maths Equation of a Line: direct answer
ICSE Class 10 Maths Equation of a Line questions test whether you can connect a straight-line equation with coordinates, slope, intercepts, parallel lines and perpendicular lines. In Concise Mathematics Selina Solutions Class 10 ICSE Chapter 14 Equation of a Line, most answers come from four ideas: substitute a point in the line, use the slope formula, choose the correct form of a straight line, and simplify the equation carefully.
This replacement page is written as a working sheet for Chapter 14. It keeps the useful Selina-style question types from the existing page, corrects weak presentation, and adds teacher notes, formula reference, worked examples, a quick answer index and common mistake corrections. Use it after attempting the textbook problems so that you can check both the final answer and the method.
Formula reference for Chapter 14
Before solving the exercises, revise the standard formulas. The official CISCE Mathematics syllabus is available from CISCE. For related coordinate geometry practice, you can also compare the school-level treatment with NCERT Class 10 Mathematics, but follow your ICSE textbook method in the examination.
| Use case | Formula or rule | Teacher note |
|---|---|---|
| Point on a line | Substitute x and y in the equation. If L.H.S. = R.H.S., the point lies on the line. | This is the shortest method for Exercise \(14(A)\) verification questions. |
| Slope through two points | m=\dfrac{y_2-y_1}{x_2-x_1}, when x_2\ne x_1 | If x_2=x_1, the line is vertical and slope is not defined. |
| Inclination | m=\tan\theta | For a usual ICSE question, use standard angles such as 30^\circ, 45^\circ, 60^\circ. |
| Parallel lines | m_1=m_2 | Equal slopes mean equal inclination. |
| Perpendicular lines | m_1m_2=-1, when both slopes are defined | A horizontal line is perpendicular to a vertical line; the product rule does not apply when one slope is not defined. |
| Point-slope form | \(y-y_1=m(x-x_1)\) | Use it when one point and slope are known. |
| Two-point form | \(y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\) | Use it when two points are known. |
| Slope-intercept form | y=mx+c | m is slope and c is the y-intercept. |
| Intercept form | \dfrac{x}{a}+\dfrac{y}{b}=1 | a is the x-intercept and b is the y-intercept. Do not use it if an intercept is 0. |
| Midpoint | \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\) | If a line bisects a segment, it passes through this midpoint. |
| Section formula | \(\left(\dfrac{m x_2+n x_1}{m+n},\dfrac{m y_2+n y_1}{m+n}\right)\) | For internal division in the ratio m:n, keep the order of the two given points fixed. |
Concept snapshot: how to read a straight line
Think of a line equation as an address rule. A point such as \((2,3)\) either follows the rule or breaks it. For example, in x-2y+5=0, the point \((1,3)\) follows the rule because \(1-2(3)+5=0\). Once a point follows the rule, it lies on the line.
The slope is the line’s tilt. If the slope is positive, the line rises as x increases. If the slope is negative, it falls. If the slope is 0, the line is horizontal. If the line is vertical, its slope is not defined because the denominator in \dfrac{y_2-y_1}{x_2-x_1} becomes 0.
Exercise 14(A): points, bisection and concurrency
Exercise \(14(A)\) mainly checks substitution, midpoint, section formula and simultaneous equations. The important habit is to write the condition first: a point lies on a line only when it satisfies the equation of the line.
Question 1(a): The point \((m,-4)\) lies on x+y=4. Find m.
Step 1: Substitute x=m and y=-4 in the line x+y=4.
m+(-4)=4
Step 2: Solve the linear equation.
m-4=4
m=8
Final answer: m=8. The correct option is 8.
Question 1(b): The line kx-y=9 passes through \((6,3)\). Find k.
Step 1: Substitute x=6 and y=3.
6k-3=9
Step 2: Isolate k.
6k=12
k=2
Final answer: k=2. The correct option is 2.
Question 1(c): Does \((2,-4)\) lie on 3x-\dfrac{y}{2}=10?
Step 1: Substitute x=2 and y=-4 in the left side.
3(2)-\dfrac{-4}{2}=6-(-2)=8
Step 2: Compare with the right side.
8\ne 10
Final answer: No, \((2,-4)\) does not lie on the line. The correct option is no.
Question 1(d): Lines x+y=8 and x-y=0 meet at a point on mx-2y=0. Find m.
Step 1: From x-y=0, get x=y.
Step 2: Substitute x=y in x+y=8.
y+y=8
2y=8
y=4
Step 3: Since x=y, x=4. The intersection point is \((4,4)\).
Step 4: Substitute \((4,4)\) in mx-2y=0.
4m-2(4)=0
4m-8=0
m=2
Final answer: m=2. The correct option is 2.
Question 1(e): x+y=4 bisects the join of \((0,k)\) and \((4,0)\). Find k.
Step 1: Find the midpoint of \((0,k)\) and \((4,0)\).
\left(\dfrac{0+4}{2},\dfrac{k+0}{2}\right)=\left(2,\dfrac{k}{2}\right)
Step 2: Since the line bisects the segment, the midpoint lies on x+y=4.
2+\dfrac{k}{2}=4
\dfrac{k}{2}=2
k=4
Final answer: k=4. The correct option is 4.
Question 2: Which points lie on x-2y+5=0?
Step 1: Test \((1,3)\).
1-2(3)+5=1-6+5=0
Step 2: Test \((0,5)\).
0-2(5)+5=-5\ne 0
Step 3: Test \((-5,0)\).
-5-2(0)+5=0
Step 4: Test \((5,5)\).
5-2(5)+5=0
Final answer: The points \((1,3)\), \((-5,0)\) and \((5,5)\) lie on the line. The point \((0,5)\) does not lie on the line.
Question 3(i): Is it true that \dfrac{x}{2}+\dfrac{y}{3}=0 passes through \((2,3)\)?
Step 1: Substitute \((2,3)\) in the left side.
\dfrac{2}{2}+\dfrac{3}{3}=1+1=2
Step 2: Compare with 0.
2\ne 0
Final answer: The statement is false.
Question 3(ii): If \((2,a)\) lies on 2x-y=3, is a=5?
Step 1: Substitute x=2 and y=a.
2(2)-a=3
4-a=3
a=1
Final answer: The statement is false because a=1, not 5.
Question 4: For what value of k will \((3,-k)\) lie on 9x+4y=3?
Step 1: Substitute x=3 and y=-k.
9(3)+4(-k)=3
27-4k=3
Step 2: Solve for k.
24=4k
k=6
Final answer: k=6.
Question 5: The line \dfrac{3x}{5}-\dfrac{2y}{3}+1=0 contains \((m,2m-1)\). Find m.
Step 1: Substitute x=m and y=2m-1.
\dfrac{3m}{5}-\dfrac{2(2m-1)}{3}+1=0
Step 2: Multiply throughout by 15.
9m-10(2m-1)+15=0
Step 3: Expand and simplify.
9m-20m+10+15=0
-11m+25=0
m=\dfrac{25}{11}
Final answer: m=\dfrac{25}{11}.
Question 6: Does 3x-5y=6 bisect the join of \((5,-2)\) and \((-1,2)\)?
Step 1: Find the midpoint.
\left(\dfrac{5+(-1)}{2},\dfrac{-2+2}{2}\right)=(2,0)
Step 2: Test the midpoint in 3x-5y=6.
3(2)-5(0)=6
Final answer: Yes, the line bisects the join because the midpoint \((2,0)\) lies on the line.
Question 7(i): y=3x-2 bisects the join of \((a,3)\) and \((2,-5)\). Find a.
Step 1: Write the midpoint.
\left(\dfrac{a+2}{2},\dfrac{3+(-5)}{2}\right)=\left(\dfrac{a+2}{2},-1\right)
Step 2: Substitute the midpoint in y=3x-2.
-1=3\left(\dfrac{a+2}{2}\right)-2
1=\dfrac{3(a+2)}{2}
2=3a+6
a=-\dfrac{4}{3}
Final answer: a=-\dfrac{4}{3}.
Question 7(ii): x-6y+11=0 bisects the join of \((8,-1)\) and \((0,k)\). Find k.
Step 1: Find the midpoint.
\left(\dfrac{8+0}{2},\dfrac{-1+k}{2}\right)=\left(4,\dfrac{k-1}{2}\right)
Step 2: Substitute in x-6y+11=0.
4-6\left(\dfrac{k-1}{2}\right)+11=0
4-3(k-1)+11=0
18-3k=0
k=6
Final answer: k=6.
Question 8(i): \((-3,2)\) lies on ax+3y+6=0. Find a.
Step 1: Substitute x=-3 and y=2.
-3a+3(2)+6=0
-3a+12=0
a=4
Final answer: a=4.
Question 8(ii): y=mx+8 contains \((-4,4)\). Find m.
Step 1: Substitute x=-4 and y=4.
4=m(-4)+8
4=-4m+8
-4=-4m
m=1
Final answer: m=1.
Question 9: P divides the join of \((2,1)\) and \((-3,6)\) in the ratio 2:3. Does P lie on x-5y+15=0?
Step 1: Use the section formula for internal division in the ratio 2:3.
P=\left(\dfrac{2(-3)+3(2)}{2+3},\dfrac{2(6)+3(1)}{2+3}\right)
P=\left(\dfrac{-6+6}{5},\dfrac{12+3}{5}\right)=(0,3)
Step 2: Test \((0,3)\) in x-5y+15=0.
0-5(3)+15=0
Final answer: \(P=(0,3)\), and it lies on the line x-5y+15=0.
Question 10: Q divides the join of \((5,-4)\) and \((2,2)\) in the ratio 1:2. Does x-2y=0 contain Q?
Step 1: Find Q by section formula.
Q=\left(\dfrac{1(2)+2(5)}{1+2},\dfrac{1(2)+2(-4)}{1+2}\right)
Q=\left(\dfrac{12}{3},\dfrac{-6}{3}\right)=(4,-2)
Step 2: Test \((4,-2)\) in x-2y=0.
4-2(-2)=8\ne 0
Final answer: \(Q=(4,-2)\), and it does not lie on x-2y=0.
Question 11: Find the intersection of 4x+3y=1 and 3x-y+9=0. Then find k if it lies on \((2k-1)x-2y=4\).
Step 1: From 3x-y+9=0, write y=3x+9.
Step 2: Substitute this in 4x+3y=1.
4x+3(3x+9)=1
13x+27=1
x=-2
Step 3: Find y.
y=3(-2)+9=3
Step 4: Substitute \((-2,3)\) in \((2k-1)x-2y=4\).
(2k-1)(-2)-2(3)=4
-4k+2-6=4
-4k=8
k=-2
Final answer: The intersection point is \((-2,3)\), and k=-2.
Question 12: Show that 2x+5y=1, x-3y=6 and x+5y+2=0 are concurrent.
Step 1: Solve the first two lines. From x-3y=6, write x=6+3y.
Step 2: Substitute in 2x+5y=1.
2(6+3y)+5y=1
12+11y=1
11y=-11
y=-1
Step 3: Find x.
x=6+3(-1)=3
Step 4: Check whether \((3,-1)\) lies on the third line.
3+5(-1)+2=0
Hence proved: All three lines pass through \((3,-1)\), so they are concurrent.
Exercise 14(B): slope and inclination
Exercise \(14(B)\) usually moves from substitution to slope. The key connection is m=\tan\theta, where m is the slope and \theta is the angle made with the positive x-axis.
Worked example 1: Find inclination when slope is \sqrt{3}
Step 1: Use m=\tan\theta.
\tan\theta=\sqrt{3}
Step 2: Use the standard value \tan60^\circ=\sqrt{3}.
\theta=60^\circ
Final answer: The inclination is 60^\circ.
Question type: Slope of a perpendicular line
Step 1: If one line has slope 5, let the slope of its perpendicular be m.
Step 2: Use m_1m_2=-1.
5m=-1
m=-\dfrac{1}{5}
Final answer: The slope of the perpendicular line is -\dfrac{1}{5}.
Question type: Slope of the line through the origin and \((-3,4)\)
Step 1: The origin is \((0,0)\). Use the slope formula.
m=\dfrac{4-0}{-3-0}
m=-\dfrac{4}{3}
Final answer: The slope is -\dfrac{4}{3}.
Exercise 14(C): parallel and perpendicular lines
Exercise \(14(C)\) connects slope with geometry. Equal slopes show parallel lines. Slopes with product -1 show perpendicular lines. If you are finding a median, first find the midpoint of the opposite side.
Question type: Line through \((-7,4)\) and \((5,4)\)
Step 1: Find the slope.
m=\dfrac{4-4}{5-(-7)}=\dfrac{0}{12}=0
Step 2: A line with slope 0 is horizontal.
Final answer: The line is parallel to the x-axis.
Worked example 2: Median and a parallel line in a triangle
In \triangle ABC, let \(A=(4,7)\), \(B=(-2,3)\) and \(C=(0,1)\). Find the equation of the median through A, and find the equation of the line through B parallel to AC.
Step 1: The median through A goes to the midpoint of BC. Let this midpoint be D.
D=\left(\dfrac{-2+0}{2},\dfrac{3+1}{2}\right)=(-1,2)
Step 2: Find the slope of AD.
m_{AD}=\dfrac{2-7}{-1-4}=\dfrac{-5}{-5}=1
Step 3: Use point-slope form through \(A=(4,7)\).
y-7=1(x-4)
y=x+3
Step 4: Find the slope of AC.
m_{AC}=\dfrac{1-7}{0-4}=\dfrac{-6}{-4}=\dfrac{3}{2}
Step 5: A line parallel to AC has the same slope \dfrac{3}{2}. Through \(B=(-2,3)\):
y-3=\dfrac{3}{2}(x+2)
2y-6=3x+6
3x-2y+12=0
Final answer: Median through A: y=x+3. Line through B parallel to AC: 3x-2y+12=0.
Question type: Perpendicular from \((-1,2)\) to the line through \((1,4)\) and \((2,3)\)
Step 1: Find the slope of the given line.
m=\dfrac{3-4}{2-1}=-1
Step 2: The perpendicular slope is 1, because \((-1)(1)=-1\).
Step 3: Use point-slope form through \((-1,2)\).
y-2=1(x+1)
y=x+3
Final answer: The perpendicular line is y=x+3.
Exercise 14(D): forms of the equation of a line
Exercise \(14(D)\) usually asks you to choose the most efficient form of a line. If the intercepts are given, use \dfrac{x}{a}+\dfrac{y}{b}=1. If one point and a slope are given, use \(y-y_1=m(x-x_1)\). If two points are given, first find the slope or use the two-point form.
Question type: x-intercept 5 and y-intercept 3
Step 1: Use intercept form with a=5 and b=3.
\dfrac{x}{5}+\dfrac{y}{3}=1
Step 2: Multiply by 15.
3x+5y=15
Final answer: 3x+5y=15.
Question type: x-intercept -4 and y-intercept 6
Step 1: Use a=-4 and b=6.
\dfrac{x}{-4}+\dfrac{y}{6}=1
Step 2: Multiply by 12.
-3x+2y=12
Final answer: -3x+2y=12.
Worked example 3: Line through \((2,3)\) with x-intercept 4
Step 1: An x-intercept of 4 means the line passes through \((4,0)\).
Step 2: Find the slope through \((4,0)\) and \((2,3)\).
m=\dfrac{3-0}{2-4}=-\dfrac{3}{2}
Step 3: Use point-slope form through \((4,0)\).
y-0=-\dfrac{3}{2}(x-4)
2y=-3x+12
3x+2y=12
Final answer: The required line is 3x+2y=12.
Question type: Lines through \((-2,0)\) equally inclined to the coordinate axes
Step 1: Lines equally inclined to the coordinate axes make angles 45^\circ or 135^\circ with the positive x-axis.
Step 2: Their slopes are 1 and -1.
Step 3: For slope 1 through \((-2,0)\):
y-0=1(x+2)
x-y+2=0
Step 4: For slope -1 through \((-2,0)\):
y-0=-1(x+2)
x+y+2=0
Final answer: The two lines are x-y+2=0 and x+y+2=0.
Question type: Angle between y=x+1 and y=\sqrt{3}x-1
Step 1: Compare y=x+1 with y=mx+c. The slope is 1.
\tan\theta_1=1\Rightarrow \theta_1=45^\circ
Step 2: Compare y=\sqrt{3}x-1 with y=mx+c. The slope is \sqrt{3}.
\tan\theta_2=\sqrt{3}\Rightarrow \theta_2=60^\circ
Step 3: Find the acute angle between the two lines.
\theta=60^\circ-45^\circ=15^\circ
Final answer: The angle between the two lines is 15^\circ.
Test Yourself: mixed coordinate geometry practice
The Test Yourself questions generally combine two or more skills. Read the question and identify the first fixed fact: a midpoint, an intercept, a slope, a point on a line, or a parallel/perpendicular condition.
Worked example 4: Midpoint on axes and equation of a line
A point \(P(2,-3)\) is the midpoint of AB, where A lies on the x-axis and B lies on the y-axis. Find A, B, the slope of AB, and the equation of AB.
Step 1: Let \(A=(a,0)\) and \(B=(0,b)\).
Step 2: Use the midpoint formula.
\left(\dfrac{a+0}{2},\dfrac{0+b}{2}\right)=(2,-3)
Step 3: Compare coordinates.
\dfrac{a}{2}=2\Rightarrow a=4
\dfrac{b}{2}=-3\Rightarrow b=-6
Step 4: Therefore, \(A=(4,0)\) and \(B=(0,-6)\).
Step 5: Find the slope of AB.
m=\dfrac{-6-0}{0-4}=\dfrac{3}{2}
Step 6: Use point-slope form through \(A=(4,0)\).
y-0=\dfrac{3}{2}(x-4)
2y=3x-12
3x-2y-12=0
Final answer: \(A=(4,0)\), \(B=(0,-6)\), slope =\dfrac{3}{2}, and the equation is 3x-2y-12=0.
Examiner’s mindset for ICSE Class 10 Maths
In coordinate geometry, the working is as important as the final equation. A neat answer usually shows the formula, substitutes the correct coordinates, simplifies signs carefully, and gives the final equation in a standard form. Avoid jumping directly to the answer, because sign errors in slope and intercept questions are easy to miss.
For Chapter 14, a strong solution usually has this order: identify the known condition, choose the formula, substitute coordinates, simplify, and check the final equation when possible. For example, after finding 3x+2y=12, you can test \((4,0)\) and \((2,3)\) quickly to confirm the line is correct.
For wider revision, use this page with ICSE Class 10 Maths resources, the Selina Maths Class 10 Solutions index, and related ML Aggarwal Class 10 Maths solutions for extra practice.
Common mistakes students make
- Using the wrong sign in slope: In m=\dfrac{y_2-y_1}{x_2-x_1}, use the same order in numerator and denominator. Do not write \dfrac{y_2-y_1}{x_1-x_2}.
- Forgetting that a vertical line has no defined slope: If two points have the same x-coordinate, the denominator becomes 0. Do not call the slope 0; slope 0 belongs to a horizontal line.
- Confusing intercepts with coordinates: An x-intercept 5 means the point \((5,0)\). A y-intercept 3 means the point \((0,3)\).
- Using the perpendicular rule blindly: The rule m_1m_2=-1 is valid when both slopes are defined. For horizontal and vertical lines, reason from the graph.
- Not checking the final equation: If a line is found through two points, substitute both points back into the final equation. This catches many arithmetic errors.
Quick answer index
This index lists the worked items covered on this page. It is meant for quick checking after you have studied the steps above.
| Section | Question type | Answer |
|---|---|---|
| Exercise \(14(A)\) | \((m,-4)\) on x+y=4 | m=8 |
| Exercise \(14(A)\) | kx-y=9 through \((6,3)\) | k=2 |
| Exercise \(14(A)\) | \((2,-4)\) on 3x-\dfrac{y}{2}=10 | No |
| Exercise \(14(A)\) | Intersection of x+y=8, x-y=0 on mx-2y=0 | m=2 |
| Exercise \(14(A)\) | x+y=4 bisects \((0,k)\), \((4,0)\) | k=4 |
| Exercise \(14(A)\) | Points on x-2y+5=0 | \((1,3)\), \((-5,0)\), \((5,5)\) |
| Exercise \(14(A)\) | \dfrac{x}{2}+\dfrac{y}{3}=0 through \((2,3)\) | False |
| Exercise \(14(A)\) | \((2,a)\) on 2x-y=3 | a=1; statement a=5 is false |
| Exercise \(14(A)\) | \((3,-k)\) on 9x+4y=3 | k=6 |
| Exercise \(14(A)\) | \((m,2m-1)\) on \dfrac{3x}{5}-\dfrac{2y}{3}+1=0 | m=\dfrac{25}{11} |
| Exercise \(14(A)\) | 3x-5y=6 bisects given join | Yes |
| Exercise \(14(A)\) | y=3x-2 bisection | a=-\dfrac{4}{3} |
| Exercise \(14(A)\) | x-6y+11=0 bisection | k=6 |
| Exercise \(14(A)\) | \((-3,2)\) on ax+3y+6=0 | a=4 |
| Exercise \(14(A)\) | \((-4,4)\) on y=mx+8 | m=1 |
| Exercise \(14(A)\) | Section formula point P | \(P=(0,3)\), lies on x-5y+15=0 |
| Exercise \(14(A)\) | Section formula point Q | \(Q=(4,-2)\), does not lie on x-2y=0 |
| Exercise \(14(A)\) | Intersection and parameter | Intersection \((-2,3)\), k=-2 |
| Exercise \(14(A)\) | Concurrency | Concurrent at \((3,-1)\) |
| Exercise \(14(B)\) | Slope \sqrt{3} | Inclination 60^\circ |
| Exercise \(14(B)\) | Perpendicular to slope 5 | -\dfrac{1}{5} |
| Exercise \(14(B)\) | Through \((0,0)\) and \((-3,4)\) | Slope -\dfrac{4}{3} |
| Exercise \(14(C)\) | Through \((-7,4)\) and \((5,4)\) | Parallel to x-axis |
| Exercise \(14(C)\) | Median through A | y=x+3 |
| Exercise \(14(C)\) | Line through B parallel to AC | 3x-2y+12=0 |
| Exercise \(14(C)\) | Perpendicular through \((-1,2)\) | y=x+3 |
| Exercise \(14(D)\) | Intercepts 5 and 3 | 3x+5y=15 |
| Exercise \(14(D)\) | Intercepts -4 and 6 | -3x+2y=12 |
| Exercise \(14(D)\) | Through \((2,3)\), x-intercept 4 | 3x+2y=12 |
| Exercise \(14(D)\) | Equally inclined to axes through \((-2,0)\) | x-y+2=0, x+y+2=0 |
| Exercise \(14(D)\) | Angle between y=x+1 and y=\sqrt{3}x-1 | 15^\circ |
| Test Yourself | Midpoint on axes | \(A=(4,0)\), \(B=(0,-6)\), slope =\dfrac{3}{2}, line 3x-2y-12=0 |
How to use this page with the textbook
First, solve the Selina exercise on paper without looking at the answer. Then compare your method with the worked steps above. If your final equation differs, substitute the given point or intercepts into both equations. The correct equation must satisfy the data in the question.
For more chapter practice, use ICSE Books PDF resources only as support; do not replace your class notebook and textbook working. Coordinate geometry improves when you draw a small rough graph before writing algebra.
Frequently Asked Questions
What is the main formula in ICSE Class 10 Maths Equation of a Line?
The main formula is the slope formula m=\dfrac{y_2-y_1}{x_2-x_1}, along with the line forms \(y-y_1=m(x-x_1)\), y=mx+c, and \dfrac{x}{a}+\dfrac{y}{b}=1. Use the formula that matches the data given in the question.
How do I know if a point lies on a line in Selina Chapter 14?
Substitute the point’s x-coordinate and y-coordinate into the equation of the line. If the left side equals the right side, the point lies on the line; otherwise, it does not.
When should I use the intercept form in Equation of a Line questions?
Use intercept form when the x-intercept and y-intercept are both given and neither is 0. The form is \dfrac{x}{a}+\dfrac{y}{b}=1, where a is the x-intercept and b is the y-intercept.
What is the difference between parallel and perpendicular lines in slope questions?
Parallel lines have equal slopes, so m_1=m_2. Perpendicular lines have slopes whose product is -1, so m_1m_2=-1, provided both slopes are defined.
How can I check my final equation of a line?
Substitute the given point, intercept or midpoint back into your final equation. For example, if your answer is 3x+2y=12 and the line should pass through \((2,3)\), then \(3(2)+2(3)=12\), so the check works.
