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Standard X

Mathematics

Slope of a Line

The line L1≡ ...

Question

The line $L_{1} \equiv 4 x + 3 y - 12 = 0$ intersects the x - axis and y - axis at A and B respectively. A variable line perpendicular to $L_{1}$ intersect the x and y - axes at P and Q respectively. The locus of the circumcentre of triangle ABQ is

6x - 8y + 7 =0

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6x + 8y - 25 =0

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8x - 6y + 7 =0

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14x - 12y + 3 =0

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Solution

The correct option is A 6x - 8y + 7 =0
clearly circumcentre of triangle ABQ will lie on perpendicular bisector of line AB, which is 6x - 8y + 7 =0

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Similar questions

Q. The line

L_{1} \equiv 4 x + 3 y - 12 = 0

intersects the

x -

and

y -

axes at

A

and

B

, respectively. A variable line perpendicular to

L_{1}

intersects the

x -

and the

y -

axes at

P

and

Q

, respectively. Then the locus of the circumcentre of triangle

A B Q

Q. A straight line

L_{1} : \frac{x}{a} + \frac{y}{b} = 1

intersects the

x

-axis and

y

-axis at

P

and

Q

respectively and a straight line

L_{2}

perpendicular to

L_{1}

cuts the

x

-axis and

y

-axis at

R

and

S

respectively. The locus of the point of intersection of the lines

P S

and

Q R

Q. A straight line

L_{1} : \frac{x}{a} + \frac{y}{b} = 1

intersects the

x

-axis and

y

-axis at

P

and

Q

respectively and a straight line

L_{2}

perpendicular to

L_{1}

cuts the

x

-axis and

y

-axis at

R

and

S

respectively. The locus of the point of intersection of the lines

P S

and

Q R

Q. A line

L_{1}

intersects

x

and

y

axes at

P

and

Q

respectively. Another line

L_{2},

perpendicular to

L_{1}

cuts

x

and

y

axes at

R

and

S

respectively. The locus of the point of intersection of the lines

P S

and

Q R

is a circle passing through the

Q. Let a given line

L_{1}

intersect the

x

and

y -

axis at

P

and

Q

, respectively. Let another line

L_{2}

, perpendicular to

L_{1}

, cut the

x -

axis and

y -

axis at

R

and

S

, respectively. Show that the locus of the point of intersection of the lines

P S

and

Q R

, is a circle passing through the origin ?

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The line L1≡4x+3y−12=0 intersects the x - axis and y - axis at A and B respectively. A variable line perpendicular to L1 intersect the x and y - axes at P and Q respectively. The locus of the circumcentre of triangle ABQ is

The correct option is A 6x - 8y + 7 =0clearly circumcentre of triangle ABQ will lie on perpendicular bisector of line AB, which is 6x - 8y + 7 =0

The line $L_{1} \equiv 4 x + 3 y - 12 = 0$ intersects the x - axis and y - axis at A and B respectively. A variable line perpendicular to $L_{1}$ intersect the x and y - axes at P and Q respectively. The locus of the circumcentre of triangle ABQ is

The correct option is A 6x - 8y + 7 =0
clearly circumcentre of triangle ABQ will lie on perpendicular bisector of line AB, which is 6x - 8y + 7 =0