Question

If the locus of the point, whose distances from the point $(2, 1)$ and $(1, 3)$ are in the ratio $5 : 4$, is \[ ax^2 + by^2 + cxy + dx + ey + 170 = 0, \] then the value of $a^2 + 2b + 3c + 4d + e$ is equal to:

• A. 5
• B. -27
• C. 37
• D. 437

Answer 1

The Correct Option is C

Let \( P(x, y) \)

\[ \frac{(x - 2)^2 + (y - 1)^2}{(x - 1)^2 + (y - 3)^2} = \frac{25}{16} \]

Expanding and simplifying:

\[ 9x^2 + 9y^2 + 14x - 118y + 170 = 0 \]

From the equation:

\[ a^2 + 2b + 3c + 4d + e = 81 + 18 + 0 + 56 - 118 \]

Calculating:

\[ = 155 - 118 \]

\[ = 37 \]

Answer 2

To find the locus of a point whose distances from the points \((2, 1)\) and \((1, 3)\) are in the ratio \(5:4\), we need to set up an equation based on the ratio of distances.

The distance of a point \((x, y)\) from \((2, 1)\) is given by:

\(d_1 = \sqrt{(x - 2)^2 + (y - 1)^2}\)

The distance of \((x, y)\) from \((1, 3)\) is:

\(d_2 = \sqrt{(x - 1)^2 + (y - 3)^2}\)

Given the ratio:

\(\frac{d_1}{d_2} = \frac{5}{4}\)

Squaring both sides, we have:

\(\left(\frac{d_1}{d_2}\right)^2 = \left(\frac{5}{4}\right)^2\) \(\Rightarrow \frac{(x - 2)^2 + (y - 1)^2}{(x - 1)^2 + (y - 3)^2} = \frac{25}{16}\)

Cross-multiplying gives:

\(16 \left[(x - 2)^2 + (y - 1)^2\right] = 25 \left[(x - 1)^2 + (y - 3)^2\right]\)

Expanding both sides:

\(16 \left[x^2 - 4x + 4 + y^2 - 2y + 1\right] = 25 \left[x^2 - 2x + 1 + y^2 - 6y + 9\right]\)

Simplifying further:

\(16x^2 - 64x + 16y^2 - 32y + 80 = 25x^2 - 50x + 25y^2 - 150y + 250\)

Bringing all terms to one side gives:

\(\Rightarrow -9x^2 - 9y^2 - 14x + 118y - 170 = 0\)

Thus, the equation is:

\(9x^2 + 9y^2 + 14x - 118y + 170 = 0\)

Identifying coefficients, we have: \(a = 9\), \(b = 9\), \(c = 0\), \(d = 14\), \(e = -118\).

We need to find the value of:

\(a^2 + 2b + 3c + 4d + e\)

Substitute the values:

\(= 9^2 + 2 \times 9 + 3 \times 0 + 4 \times 14 - 118\) \(= 81 + 18 + 56 - 118\) \(= 37\)

Therefore, the value is \(37\).