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:
- 5
- -27
- 37
- 437
The Correct Option is C
Approach Solution - 1
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:
\[ = 155 - 118 \]
\[ = 37 \]
Approach Solution -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 \(37\).
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