The image of a point 3t-1-1-4t- - t-0 in a line- lies on 3x-4y-1-0- Then the slope of lines is are

(3t+1,1 4t) represents L_1:4x+3y 7=0. nbsp; L_2:3x 4y+1=0 nbsp; The image of every point on L_1 about a line lies on the line L_2. ⇒ The line must be the an ...

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Byju's Answer

Standard XIII

Mathematics

Bisectors of Angle between Two Lines

The image of ...

Question

The image of a point $(3 t + 1, 1 - 4 t), \forall t \in R - {0}$ in a line, lies on $3 x - 4 y + 1 = 0.$ Then the slope of line(s) is (are)

7

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\frac{1}{7}

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- \frac{1}{7}

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Solution

The correct option is D $- \frac{1}{7}$
$(3 t + 1, 1 - 4 t)$ represents $L_{1} : 4 x + 3 y - 7 = 0$ and $L_{2} : 3 x - 4 y + 1 = 0$ .
So, the image of every point on $L_{1}$ about a line lies on the line $L_{2}$
$\Rightarrow$ The line must be the angle bisector of $L_{1}$ and $L_{2} .$

The angle bisectors of $L_{1}$ and $L_{2}$ are $7 x - y - 6 = 0$ and $x + 7 y - 8 = 0$

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The image of a point (3t+1,1−4t),∀ t∈R−{0} in a line, lies on 3x−4y+1=0. Then the slope of line(s) is (are)

The correct option is D −17(3t+1,1−4t) represents L1:4x+3y−7=0 and L2:3x−4y+1=0. So, the image of every point on L1 about a line lies on the line L2 ⇒ The line must be the angle bisector of L1 and L2. The angle bisectors of L1 and L2 are 7x−y−6=0 and x+7y−8=0

The image of a point $(3 t + 1, 1 - 4 t), \forall t \in R - {0}$ in a line, lies on $3 x - 4 y + 1 = 0.$ Then the slope of line(s) is (are)