The equation $(y - 1) = m (x - 2)$ represents a family of lines passing through $(2, 1) .$ But this equation does not include the line $x - 2 = 0,$ since the slope of the line is not defined. A line $y = m x$ through $(0, 0)$ cuts the lines $x + y = 3$ and $x + y = 5$ at two distinct points A and B respectively. $A B = d$

(4 - d^{2}) m^{2} - 2 m d^{2} + 4 - d^{2} = 0

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(4 - d^{2}) m^{2} - 2 m d^{2} + 6 - d^{2} = 0

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(4 - d 2) m^{2} + 2 m d^{2} + 4 - d^{2} = 0

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(4 - d^{2}) m^{2} + 2 m d^{2} + d^{2} - 4 = 0

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Solution

The correct option is A $(4 - d^{2}) m^{2} - 2 m d^{2} + 4 - d^{2} = 0$
$A \equiv (\frac{3}{1 + m}, \frac{3 m}{1 + m}), B \equiv (\frac{5}{1 + m}, \frac{5 m}{1 + m})$

Since $A B^{2} = d^{2} \Rightarrow (4 - d^{2}) - 2 m d^{2} + 4 - d^{2} = 0$

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The equation (y–1)=m(x–2) represents a family of lines passing through (2,1). But this equation does not include the line x–2=0, since the slope of the line is not defined. A line y=mx through (0,0) cuts the lines x+y=3 and x+y=5 at two distinct points A and B respectively. AB=d

The correct option is A (4–d2)m2–2md2+4–d2=0A≡(31+m,3m1+m),B≡(51+m,5m1+m) Since AB2=d2⇒(4−d2)−2md2+4−d2=0

The correct option is A $(4 - d^{2}) m^{2} - 2 m d^{2} + 4 - d^{2} = 0$
$A \equiv (\frac{3}{1 + m}, \frac{3 m}{1 + m}), B \equiv (\frac{5}{1 + m}, \frac{5 m}{1 + m})$

Since $A B^{2} = d^{2} \Rightarrow (4 - d^{2}) - 2 m d^{2} + 4 - d^{2} = 0$