If the line-s 1-tx-2ty-3t2-0 and t2x-3-ty-6-0- t-0 are perpendicular to each other- then the number of possible values of t is

Slope of both the lines are m_1=1+t2t, m_2=t^23 t nbsp; Given lines are perpendicular nbsp; m_1× m_2 = 1 nbsp; ⇒1+t2t×t^23 t= 1 ⇒ (1+t)t= 2(3 t) ∵ ...

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Standard XIII

Mathematics

Condition for Two Lines to Be Parallel

If the line's...

Question

If the line's $(1 + t) x - 2 t y + 3 t^{2} = 0$ and $t^{2} x - (3 - t) y + 6 = 0, t \neq 0$ are perpendicular to each other, then the number of possible value(s) of $t$ is

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Solution

The correct option is A $0$
Slope of both the lines are
$m_{1} = \frac{1 + t}{2 t}, m_{2} = \frac{t^{2}}{3 - t}$

Given lines are perpendicular
$m_{1} \times m_{2} = - 1 \Rightarrow \frac{1 + t}{2 t} \times \frac{t^{2}}{3 - t} = - 1 \Rightarrow (1 + t) t = - 2 (3 - t) ∵ t \neq 0 \Rightarrow t + t^{2} = - 6 + 2 t \Rightarrow t^{2} - t + 6 = 0 D = 1 - 4 \times 6 = - 23 < 0$

So, no real value of $t$ exists.

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If the line's (1+t)x−2ty+3t2=0 and t2x−(3−t)y+6=0, t≠0 are perpendicular to each other, then the number of possible value(s) of t is

The correct option is A 0Slope of both the lines are m1=1+t2t, m2=t23−t Given lines are perpendicular m1×m2=−1⇒1+t2t×t23−t=−1⇒(1+t)t=−2(3−t)∵t≠0⇒t+t2=−6+2t⇒t2−t+6=0D=1−4×6=−23<0 So, no real value of t exists.

If the line's $(1 + t) x - 2 t y + 3 t^{2} = 0$ and $t^{2} x - (3 - t) y + 6 = 0, t \neq 0$ are perpendicular to each other, then the number of possible value(s) of $t$ is