Coordinate Geometry · Topic 5 of 6

✖️ Intersection of Lines

2 exam question types, fully solved

The point where two lines cross satisfies both equations at once — so finding it is exactly solving a pair of simultaneous equations. Once you have that point, any expression in it is easy.

Solve the two equations together

• Intersection point. Treat the two line equations as simultaneous equations and solve — elimination (add/subtract to kill a variable) is usually fastest. The solution is the crossing point.

• Expressions at the point. Once is known, substitute it into whatever the question asks for — don't carry the variables further than you must.

• Concurrency. Three lines are concurrent when the crossing point of any two also lies on the third — find the point from two, then check it in the third.

📐Coordinate Plane ExplorerLive calculations as you move points

Line 1: a₁x + b₁y = c₁

a₁

b₁

c₁

Line 2: a₂x + b₂y = c₂

a₂

b₂

c₂

Step 1 — Eliminate y(a₁b₂−a₂b₁)x = c₁b₂−c₂b₁ → (1·-1 − 1·1)x = 4·-1 − 2·1-2x = -6 → x = 3

Step 2 — Back-substitutey = (c₁ − a₁x)/b₁ = (4 − 1·3)/1y = 1

Intersection pointSatisfies BOTH: 1·3 + 1·1 = 4 ✓ and 1·3 + -1·1 = 2 ✓(3, 1)

Solving x + y = 7 and x − y = 1 gives x = (then y = 3).

Q1The lines x + y = 5 and x − y = 1 intersect at:

Q2The lines x = 2 and y = 3 meet at:

Q3The lines y = x and y = 4 − x meet at:

📚

Real exam questions — Intersection of Lines

2 question types · 8 solved examples from real SSC papers

The point where two lines cross is simply the one pair $(x,y)$ that obeys both equations — so every “intersection” question is really a pair of linear equations in disguise. Solve by elimination, plug the point into whatever is asked, and for three lines just force them through one shared point.

How to solve this type

The point of intersection is the common solution of the two line equations, so solve them simultaneously. The fastest tool is elimination: make one variable's coefficients equal (multiply an equation by a number), then add or subtract to cancel it and get one unknown; back-substitute for the other. If a coefficient is already

1

, substitution is even quicker. In an MCQ you can also plug the option coordinates into both equations — the correct point fits BOTH, never just one.

Lines $7x+11y=3$ and $8x+y=15$ intersect at $P$ . $P$ also lies on which line?

2x-y=1

3x+2y=3

2x+y=2

3x+5y=1

Lines $2x+y=8$ and $x+2y=7$ intersect at:

(3,2)

(2,3)

(4,0)

(1,5)

Solve $3x+y=5$ and $2x-y=5$ , then find $x+y$ .

3

2

1

0

Lines $4x-2y=10$ and $4x+ky=2$ intersect at $(a,4)$ . Find $k$ .

4

-3

-4

3