If y = mx + 1 passes through (2,5), find m.

Correct — A.
Substitute the point (x = 2, y = 5) into y = mx + 1:
5 = 2m + 1
Subtract 1 from both sides: 2m = 4

Divide by 2: m = 2
Check: y = 2x + 1 at x = 2 gives 4 + 1 = 5 → option (a)

Why the others are wrong

• (b)
  1 — With m = 1 the line is y = x + 1, which gives y = 3 at x = 2, not 5. The point (2, 5) is not on it.
• (c)
  0 — m = 0 makes y = 1 a horizontal line. y is 1 at every x, so it never reaches y = 5 at x = 2.
• (d)
  3 — With m = 3, y = 3 × 2 + 1 = 7 at x = 2. The line passes above (2, 5), not through it.

A point lies on a line exactly when its coordinates satisfy the equation. So a line with one unknown, like y = mx + 1, is fixed by one point: substitute x and y, then solve for the unknown.

In y = mx + c, m is the slope and c is the y-intercept. Here c = 1, so the line meets the y-axis at (0, 1). From there to (2, 5), y rises 4 while x rises 2, and 4 ⁄ 2 = 2.

Key facts

• A point (x₁, y₁) is on a line when substituting x₁ and y₁ makes the equation true
• In y = mx + c, m is the slope and the line meets the y-axis at (0, c)
• Slope through two points = (y₂ − y₁) ⁄ (x₂ − x₁)

Common traps

• Dropping the intercept: 5 = 2m gives m = 2.5, which is not an option. That is the sign the + 1 was lost.
• Swapping the coordinates: putting x = 5, y = 2 gives 2 = 5m + 1 and m = 1⁄5.

No directly related past PYQ was found.