Where Two Lines Cross Is the Answer to Both Equations
Draw two simultaneous equations on one pair of axes and read the solution from their crossing point, find where each line meets the axes, compute the area of the triangle they enclose, and check the answer algebraically.
Why does the crossing point of two lines solve both equations at once?
A line is the set of all the points whose coordinates satisfy its equation — that is what drawing a graph means. Every point on the line has coordinates adding to , and no point off it does.
So if a point lies on two lines, its coordinates satisfy two equations. And that is precisely what solving a pair of simultaneous equations asks for.
Worked example. Solve and graphically.
Tabulate each line, including its intercepts:
- passes through , and
- passes through , and
Draw both on one pair of axes. They cross at , so
Check in both equations: and , as required.
This is the same answer elimination would have given, reached by a completely different route — and the picture tells you something the algebra does not. You can see that there is exactly one crossing point, which is why there is exactly one solution.
This page covers the ICSE Class 9 Mathematics chapter on graphical solution: drawing two lines together, finding axis intercepts, the area of the triangle they enclose with an axis, and checking the result algebraically.
So if a point lies on two lines, its coordinates satisfy two equations. And that is precisely what solving a pair of simultaneous equations asks for.
Worked example. Solve and graphically.
Tabulate each line, including its intercepts:
- passes through , and
- passes through , and
Draw both on one pair of axes. They cross at , so
Check in both equations: and , as required.
This is the same answer elimination would have given, reached by a completely different route — and the picture tells you something the algebra does not. You can see that there is exactly one crossing point, which is why there is exactly one solution.
This page covers the ICSE Class 9 Mathematics chapter on graphical solution: drawing two lines together, finding axis intercepts, the area of the triangle they enclose with an axis, and checking the result algebraically.
How do you find where a line meets the two axes?
**Put to find where it crosses the x-axis, and to find where it crosses the y-axis. Those two points are all you need to draw the line, and questions ask for them by name.
Worked example 1.** Find the intercepts of .
Worked example 2. Find the intercepts of .
A negative intercept is perfectly normal — it simply means the line crosses that axis below the origin, and your axes must be drawn long enough in the negative direction to show it. Deciding how far the axes must run is part of the question, and a graph that cuts off a required intercept loses the mark.
Worked example 3 — a line through the origin. Find the intercepts of .
Putting gives , and putting gives . Both intercepts are the same point, the origin — so this line needs a second tabulated point, such as , before it can be drawn at all.
Worked example 4 — reading intercepts from the equation. For , the intercepts are and .
Notice the shortcut: dividing the constant by each coefficient gives the two intercepts directly, and . **That works for any equation in the form , and it is much faster than substituting twice.
Check it on example 1**: and , which are the intercepts found above. The larger coefficient gives the smaller intercept, since a steeper contribution needs less distance to reach the same total.
Worked example 1.** Find the intercepts of .
Worked example 2. Find the intercepts of .
A negative intercept is perfectly normal — it simply means the line crosses that axis below the origin, and your axes must be drawn long enough in the negative direction to show it. Deciding how far the axes must run is part of the question, and a graph that cuts off a required intercept loses the mark.
Worked example 3 — a line through the origin. Find the intercepts of .
Putting gives , and putting gives . Both intercepts are the same point, the origin — so this line needs a second tabulated point, such as , before it can be drawn at all.
Worked example 4 — reading intercepts from the equation. For , the intercepts are and .
Notice the shortcut: dividing the constant by each coefficient gives the two intercepts directly, and . **That works for any equation in the form , and it is much faster than substituting twice.
Check it on example 1**: and , which are the intercepts found above. The larger coefficient gives the smaller intercept, since a steeper contribution needs less distance to reach the same total.
How do you find the area of the triangle two lines make with an axis?
The three vertices are the two axis intercepts and the point of intersection. The base lies on the axis and the height is the distance of the crossing point from that axis.
Worked example 1 — with the x-axis. Draw and , and find the area of the triangle they form with the x-axis.
First the intercepts on the x-axis:
- meets it at
- meets it at
Now the intersection. Adding the two equations gives , so and — the point .
So the triangle has vertices , and :
The height is the y-coordinate of the crossing point, because the base lies along the x-axis and height is measured perpendicular to the base. Do not use the distance between the intercepts and the crossing point — those are slant sides, not heights.
Worked example 2 — with the y-axis. Find the area of the triangle formed by and with the y-axis.
The y-intercepts are and . For the intersection, substitute into the first equation:
So the vertices are , and :
Now the height is the x-coordinate, since the base is on the y-axis. Which coordinate is the height depends on which axis the question names — and that is the one decision this question type is testing.
Worked example 3 — subtracting a negative. Notice that the base came out as , not . The two intercepts are on opposite sides of the origin, so the distance between them is . Distance is the difference of the coordinates, and with a negative value that difference is a sum. Sketching the axis with both points marked makes this impossible to get wrong.
One line of a triangle can also be a single line with both axes. The line together with both axes forms a triangle with vertices , and , whose area is
Here the two intercepts are the base and the height, because they lie along perpendicular axes — the quickest area calculation in the whole chapter.
Worked example 1 — with the x-axis. Draw and , and find the area of the triangle they form with the x-axis.
First the intercepts on the x-axis:
- meets it at
- meets it at
Now the intersection. Adding the two equations gives , so and — the point .
So the triangle has vertices , and :
The height is the y-coordinate of the crossing point, because the base lies along the x-axis and height is measured perpendicular to the base. Do not use the distance between the intercepts and the crossing point — those are slant sides, not heights.
Worked example 2 — with the y-axis. Find the area of the triangle formed by and with the y-axis.
The y-intercepts are and . For the intersection, substitute into the first equation:
So the vertices are , and :
Now the height is the x-coordinate, since the base is on the y-axis. Which coordinate is the height depends on which axis the question names — and that is the one decision this question type is testing.
Worked example 3 — subtracting a negative. Notice that the base came out as , not . The two intercepts are on opposite sides of the origin, so the distance between them is . Distance is the difference of the coordinates, and with a negative value that difference is a sum. Sketching the axis with both points marked makes this impossible to get wrong.
One line of a triangle can also be a single line with both axes. The line together with both axes forms a triangle with vertices , and , whose area is
Here the two intercepts are the base and the height, because they lie along perpendicular axes — the quickest area calculation in the whole chapter.
How do you verify a graphical solution algebraically?
Substitute the coordinates you read off into both original equations. A graph is a measurement and a measurement can be a little wrong; substitution is exact.
Worked example 1. A graph suggests that and cross at . Verify it.
So the reading was exact. Solving algebraically confirms it: from , , so , giving and .
Worked example 2 — a misread graph. Suppose the same pair was read as . Substituting:
The first equation fails, so the reading is wrong. This is exactly why the check is compulsory: a point read to the nearest grid line can satisfy one equation and miss the other.
Worked example 3 — when the lines never meet. Draw and .
The first passes through and ; the second, which simplifies to , passes through and . The two lines are parallel and never cross, so the pair has no solution.
The algebra says the same thing. The coefficient ratios match, , but the constant ratio does not, . Equal coefficient ratios with a different constant ratio means parallel lines — the condition from the simultaneous-equations chapter, now visible on the page.
Worked example 4 — when they are the same line. Draw and .
The second is just the first multiplied by , so both graphs are the same line. Every point on it satisfies both equations, so there are infinitely many solutions — and the graph shows why: there is no single crossing point to read.
So the picture distinguishes all three cases at a glance.
- Lines crossing once — exactly one solution
- Lines parallel — no solution, the equations are inconsistent
- Lines coinciding — infinitely many solutions, the equations are dependent
And the graph's honest limitation. If the solution is something like , no graph will give it to you exactly. The graphical method shows you what is happening; the algebra gives you the number — which is why an exam question asking for an exact fractional answer will always be algebraic.
Worked example 1. A graph suggests that and cross at . Verify it.
So the reading was exact. Solving algebraically confirms it: from , , so , giving and .
Worked example 2 — a misread graph. Suppose the same pair was read as . Substituting:
The first equation fails, so the reading is wrong. This is exactly why the check is compulsory: a point read to the nearest grid line can satisfy one equation and miss the other.
Worked example 3 — when the lines never meet. Draw and .
The first passes through and ; the second, which simplifies to , passes through and . The two lines are parallel and never cross, so the pair has no solution.
The algebra says the same thing. The coefficient ratios match, , but the constant ratio does not, . Equal coefficient ratios with a different constant ratio means parallel lines — the condition from the simultaneous-equations chapter, now visible on the page.
Worked example 4 — when they are the same line. Draw and .
The second is just the first multiplied by , so both graphs are the same line. Every point on it satisfies both equations, so there are infinitely many solutions — and the graph shows why: there is no single crossing point to read.
So the picture distinguishes all three cases at a glance.
- Lines crossing once — exactly one solution
- Lines parallel — no solution, the equations are inconsistent
- Lines coinciding — infinitely many solutions, the equations are dependent
And the graph's honest limitation. If the solution is something like , no graph will give it to you exactly. The graphical method shows you what is happening; the algebra gives you the number — which is why an exam question asking for an exact fractional answer will always be algebraic.
Exam tip
What layout earns full marks on a graphical solution?
Show both tables, state the scale, label each line with its equation, and write the solution as a coordinate pair. Then verify it. Every one of those is separately creditable.
- Write a table for each equation, with at least three points including both intercepts. Two tables side by side, clearly headed with their equations
- State the scale: * cm unit on both axes*. Use the same scale on both axes, or the lines will be distorted and an area will come out wrong
- Choose the range of the axes after tabulating, so that every intercept and the crossing point fit on the page
- Label each line with its equation next to it on the graph. With two unlabelled lines, the examiner cannot tell which is which
- Mark the point of intersection with a cross and write its coordinates beside it
- Verify by substituting into both original equations, and show that working
- For an area, name the three vertices first, then state the base and the height as separate lines with their units
- Give the area in square units, and lengths in units
The trap in area questions. The height is the coordinate perpendicular to the base: use the y-coordinate of the crossing point for a base on the x-axis, and the x-coordinate for a base on the y-axis. And when the two intercepts are on opposite sides of the origin, the base is the sum of their distances, not the difference of the numbers as written. Mark both points on a quick sketch of the axis and count.
- Write a table for each equation, with at least three points including both intercepts. Two tables side by side, clearly headed with their equations
- State the scale: * cm unit on both axes*. Use the same scale on both axes, or the lines will be distorted and an area will come out wrong
- Choose the range of the axes after tabulating, so that every intercept and the crossing point fit on the page
- Label each line with its equation next to it on the graph. With two unlabelled lines, the examiner cannot tell which is which
- Mark the point of intersection with a cross and write its coordinates beside it
- Verify by substituting into both original equations, and show that working
- For an area, name the three vertices first, then state the base and the height as separate lines with their units
- Give the area in square units, and lengths in units
The trap in area questions. The height is the coordinate perpendicular to the base: use the y-coordinate of the crossing point for a base on the x-axis, and the x-coordinate for a base on the y-axis. And when the two intercepts are on opposite sides of the origin, the base is the sum of their distances, not the difference of the numbers as written. Mark both points on a quick sketch of the axis and count.
Did you know
How do two mobile plans decide which is cheaper?
Two taxi services quote differently. The first charges a fixed plus per kilometre; the second charges plus per kilometre. Which should you take?
It depends on the distance, and the graph answers it in one picture. Writing for the fare and for the kilometres:
Draw both on one pair of axes. The first starts higher on the y-axis — a bigger fixed charge — but climbs more gently. The second starts lower and climbs faster. So they must cross, and where they cross the two services cost the same:
and the common fare there is
**So below km the second service is cheaper, and above km the first one is. The crossing point is called the break-even point, and reading which line is lower on which side of it is the whole decision.
The graph carries information the single number does not.** It shows at a glance how much cheaper each option is at any distance — the vertical gap between the lines — and it makes the trade-off visible: a high fixed charge with a low rate suits long journeys, and the reverse suits short ones.
And the two failure cases have meanings here too. If the two services had the same rate per kilometre but different fixed charges, the lines would be parallel and one would be cheaper at every distance — no break-even point exists. If both the fixed charge and the rate matched, the lines would coincide and the two would cost the same always. The three graphical cases from the previous section are exactly the three answers this comparison can have.
It depends on the distance, and the graph answers it in one picture. Writing for the fare and for the kilometres:
Draw both on one pair of axes. The first starts higher on the y-axis — a bigger fixed charge — but climbs more gently. The second starts lower and climbs faster. So they must cross, and where they cross the two services cost the same:
and the common fare there is
**So below km the second service is cheaper, and above km the first one is. The crossing point is called the break-even point, and reading which line is lower on which side of it is the whole decision.
The graph carries information the single number does not.** It shows at a glance how much cheaper each option is at any distance — the vertical gap between the lines — and it makes the trade-off visible: a high fixed charge with a low rate suits long journeys, and the reverse suits short ones.
And the two failure cases have meanings here too. If the two services had the same rate per kilometre but different fixed charges, the lines would be parallel and one would be cheaper at every distance — no break-even point exists. If both the fixed charge and the rate matched, the lines would coincide and the two would cost the same always. The three graphical cases from the previous section are exactly the three answers this comparison can have.
Exam relevance
Where does the graphical method reappear in JEE preparation?
This is foundation work whose direct successor is examined in the Class 10 board paper and whose ideas resurface throughout Class 11 and 12.
Where it leads. Class 10 examines the graphical method for a pair of linear equations explicitly, together with the consistency conditions you saw as parallel and coincident lines. Class 11 Straight Lines turns the same picture into algebra — the intersection of two lines, the area of a triangle from three vertices, the angle between two lines — all JEE Main material. In Class 12, Linear Programming is this chapter at scale: several lines, a shaded feasible region, and the answer at one of its corner points.
Where the consistency conditions go. The three cases — one solution, none, infinitely many — become conditions on the determinant of the coefficient matrix in Class 12 Matrices and Determinants. A non-zero determinant is lines crossing; a zero determinant is parallel or coincident. JEE Main asks directly for the value of a parameter that makes a system inconsistent, and the graph is the picture behind that question.
Where it appears in Physics and Chemistry. Reading an intersection is how a graphical solution to a physics problem is found — the meeting point of two position-time graphs is when and where two bodies meet, and the intercept-and-slope reading of a straight-line graph is the standard method of analysing experimental data. The break-even example above has the same shape as two bodies starting at different points with different speeds.
Question types to expect. At this level: draw, read, find intercepts, find an area, verify. In competitive papers: the point of intersection computed algebraically, the area of the triangle a line makes with the axes, and the parameter condition for a system to be inconsistent.
One result worth carrying forward. The triangle a line makes with both axes has area , which is the last worked example of the area section written in general. That formula is a standard JEE Main one-liner, and you have already derived its only case.
The single trap that costs marks. Reading a solution off a graph and reporting it as exact. If the crossing point is not on a grid intersection, the read value is an estimate — and in a competitive paper the options will include the nearby wrong value. Always finish with the algebra.
Board versus competitive emphasis. ICSE marks the tables, the scale, the labelled lines and the verification; a competitive paper marks a coordinate or an area. **The transferable habit is reading the picture for the kind of answer and using algebra for the value itself.**
Where it leads. Class 10 examines the graphical method for a pair of linear equations explicitly, together with the consistency conditions you saw as parallel and coincident lines. Class 11 Straight Lines turns the same picture into algebra — the intersection of two lines, the area of a triangle from three vertices, the angle between two lines — all JEE Main material. In Class 12, Linear Programming is this chapter at scale: several lines, a shaded feasible region, and the answer at one of its corner points.
Where the consistency conditions go. The three cases — one solution, none, infinitely many — become conditions on the determinant of the coefficient matrix in Class 12 Matrices and Determinants. A non-zero determinant is lines crossing; a zero determinant is parallel or coincident. JEE Main asks directly for the value of a parameter that makes a system inconsistent, and the graph is the picture behind that question.
Where it appears in Physics and Chemistry. Reading an intersection is how a graphical solution to a physics problem is found — the meeting point of two position-time graphs is when and where two bodies meet, and the intercept-and-slope reading of a straight-line graph is the standard method of analysing experimental data. The break-even example above has the same shape as two bodies starting at different points with different speeds.
Question types to expect. At this level: draw, read, find intercepts, find an area, verify. In competitive papers: the point of intersection computed algebraically, the area of the triangle a line makes with the axes, and the parameter condition for a system to be inconsistent.
One result worth carrying forward. The triangle a line makes with both axes has area , which is the last worked example of the area section written in general. That formula is a standard JEE Main one-liner, and you have already derived its only case.
The single trap that costs marks. Reading a solution off a graph and reporting it as exact. If the crossing point is not on a grid intersection, the read value is an estimate — and in a competitive paper the options will include the nearby wrong value. Always finish with the algebra.
Board versus competitive emphasis. ICSE marks the tables, the scale, the labelled lines and the verification; a competitive paper marks a coordinate or an area. **The transferable habit is reading the picture for the kind of answer and using algebra for the value itself.**
Key takeaways
What should you be able to do with graphs of two equations?
One picture, three possible answers, and one compulsory check.
- A line is the set of all points satisfying its equation, so a point on two lines satisfies both — that is the solution
- Tabulate three points per line, including both intercepts, and draw with a ruler on the same scale on both axes
- Intercepts: put for the x-axis and for the y-axis; for they are and
- The triangle with an axis has the two intercepts and the crossing point as vertices; the height is the coordinate perpendicular to the base
- When intercepts straddle the origin, the base is the sum of their distances
- A line with both axes makes a triangle of area x-intercept y-intercept
- Crossing once means one solution; parallel means none; coinciding means infinitely many
- Parallel shows as equal coefficient ratios with a different constant ratio
- Always verify by substituting the read coordinates into both original equations
The sharpest self-test is the area question. Take and , find the triangle they form with the y-axis, and check whether you used the x-coordinate of the crossing point as the height.
- A line is the set of all points satisfying its equation, so a point on two lines satisfies both — that is the solution
- Tabulate three points per line, including both intercepts, and draw with a ruler on the same scale on both axes
- Intercepts: put for the x-axis and for the y-axis; for they are and
- The triangle with an axis has the two intercepts and the crossing point as vertices; the height is the coordinate perpendicular to the base
- When intercepts straddle the origin, the base is the sum of their distances
- A line with both axes makes a triangle of area x-intercept y-intercept
- Crossing once means one solution; parallel means none; coinciding means infinitely many
- Parallel shows as equal coefficient ratios with a different constant ratio
- Always verify by substituting the read coordinates into both original equations
The sharpest self-test is the area question. Take and , find the triangle they form with the y-axis, and check whether you used the x-coordinate of the crossing point as the height.