JEE Mathematics 2025 Question with Solution

Given: The triangle is formed by the lines $y = x + 1$, $y = 4x - 8$, and $y = mx + c$. Its orthocentre is at $(3, -1)$.

Find: The value of $m - c$.

From the intersection of $y = x + 1$ and $y = 4x - 8$,

$$x + 1 = 4x - 8$$

so

$$3x = 9 \Rightarrow x = 3$$

and then

$$y = 3 + 1 = 4$$

Hence one vertex is $A = (3, 4)$.

Since the orthocentre is $(3, -1)$, the altitude from $A$ is the vertical line $x = 3$. Therefore the opposite side must be horizontal. The opposite side lies on the line $y = mx + c$ together with the other intersection points, so the condition used in the extracted solution leads to

$$m = 1 + c$$

Therefore,

$$m - c = 0$$

So, the correct option is A.

Given: The orthocentre of the triangle formed by the three given lines is $(3, -1)$.

Find: $m - c$.

The extracted solution first finds the known vertex by intersecting the first two lines:

$$y = x + 1$$

and

$$y = 4x - 8$$

Equating,

$$x + 1 = 4x - 8$$

which gives

$$3x = 9$$

and hence

$$x = 3$$

Substituting into $y = x + 1$,

$$y = 4$$

So the vertex is $A = (3, 4)$.

The solution then uses the orthocentre property with slopes of the three lines:

• slope of $y = x + 1$ is $1$
• slope of $y = 4x - 8$ is $4$
• slope of $y = mx + c$ is $m$

Using the orthocentre condition as stated in the provided solution, it concludes that

$$m = 1 + c$$

Therefore,

$$m - c = 0$$

Thus, the correct option is A, that is, $0$.

Note: The solution does not include the full algebraic derivation of $m = 1 + c$; the final conclusion is taken from the solution itself.