JEE Mathematics 2024 Question with Solution

Given:

• $A(a,b)$ is the centroid, $B(3,4)$ is the circumcenter, and $C(-6,-8)$ is the orthocenter.
• $P(2a+3, 7b+5)$
• Required distance is from $P$ to the line $2x+3y-4=0$, measured parallel to $x-2y-1=0$.

Find: The required oblique distance and the correct option.

From the solution,

$$a = 0, \quad b = 0$$

So,

$$P(2a+3, 7b+5) = (3,5)$$

Now measure distance from $P(3,5)$ to the line $2x+3y-4=0$ along a line parallel to $x-2y-1=0$.

Let a point on this directed line be written as

$$x = 3 + r\cos\theta, \quad y = 5 + r\sin\theta$$

Since the line is parallel to $x-2y-1=0$, the slope gives

$$\tan\theta = \frac{1}{2}$$

Hence,

$$\cos\theta = \frac{2}{\sqrt{5}}, \quad \sin\theta = \frac{1}{\sqrt{5}}$$

Substitute $x$ and $y$ into $2x+3y-4=0$:

$$2(3+r\cos\theta) + 3(5+r\sin\theta) - 4 = 0$$ $$17 + r(2\cos\theta + 3\sin\theta) = 0$$

So,

$$r(2\cos\theta + 3\sin\theta) = -17$$

Now,

$$2\cos\theta + 3\sin\theta = 2\cdot \frac{2}{\sqrt{5}} + 3\cdot \frac{1}{\sqrt{5}} = \frac{7}{\sqrt{5}}$$

Therefore,

$$r \cdot \frac{7}{\sqrt{5}} = -17$$ $$r = -\frac{17\sqrt{5}}{7}$$

Distance is the magnitude of $r$:

$$|r| = \frac{17\sqrt{5}}{7}$$

Therefore, the required distance is $\frac{17\sqrt{5}}{7}$, so the correct option is C.

Given: The extracted solution explicitly states the correct option is C and uses $a=0, b=0$.

Find: The distance measured parallel to $x-2y-1=0$.

Using

$$a=0, \quad b=0$$

we get

$$P=(2a+3, 7b+5)=(3,5)$$

A line parallel to $x-2y-1=0$ has direction ratio $(2,1)$, so a point at directed distance $r$ from $P$ can be represented as

$$(3,5) + r\left(\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$

Thus,

$$x = 3 + \frac{2r}{\sqrt{5}}, \quad y = 5 + \frac{r}{\sqrt{5}}$$

Now impose the condition that this point lies on

$$2x+3y-4=0$$

So,

$$2\left(3 + \frac{2r}{\sqrt{5}}\right) + 3\left(5 + \frac{r}{\sqrt{5}}\right) - 4 = 0$$ $$6 + \frac{4r}{\sqrt{5}} + 15 + \frac{3r}{\sqrt{5}} - 4 = 0$$ $$17 + \frac{7r}{\sqrt{5}} = 0$$ $$r = -\frac{17\sqrt{5}}{7}$$

Hence the required distance is

$$\left|r\right| = \frac{17\sqrt{5}}{7}$$

So the answer is C.