The image of a point 3 t -1-1-4 t - - t - R -0 in a line- lies on 3 x -4 y -1-0- Then the slope of lines is areA- 7B- -7 1D- 1-71

The correct options are A 7 D 17(3t+1,1 4t) represents L_1:4x+3y 7=0 nbsp; and L_2:3x 4y+1=0. nbsp; So, the image of every point on L_1 about a line lies o ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{1}{7}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $7$
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 $x + 7 y - 8 = 0$

Suggest Corrections

Similar questions

Q. 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

Q. 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)

Q. The coordinates of a point on the line

x + y + 1 = 0

which is at a distance

\frac{1}{5}

unit from the line

3 x + 4 y + 2 = 0

are

Q. If the point

(1, 1)

lies on the line passing through the points

(a, 0)

and

(0, b)

, then :

\frac{1}{a} + \frac{1}{b} =

Q. The circle

C

passing through the origin, has the line

3 x + 4 y = 0

as its tangent at the origin. If the image of the centre of

C

w.r.t the line

\frac{x}{\frac{25}{3}} + \frac{y}{\frac{25}{4}} = 1

lies on the line

3 x + 4 y = 0,

then the equation of

C

Join BYJU'S Learning Program

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 options are A 7 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)