First slide

Equation of line in Straight lines

Question

$The line L_{1} \equiv 4 x + 3 y - 12 = 0 intersects the x - and y -axis at A and B, respectively. A varıable ine$
$perpendicular to L_{1} intersects the x - and the y -axis at P and Q, respectively. Then the locus of the circumcenter of triangle A B Q is$

Moderate

A

$3 x - 4 y + 2 = 0$
B

$4 x + 3 y + 7 = 0$
C

$6 x - 8 y + 7 = 0$
D

None of theses

Solution

Clearly, the circumcenter of triangle ABQ will lie on the perpendicular bisector of line AB.
If A(3,0) and B(0,4) then midpoint of AB = $(\frac{3}{2}, 2)$
Let the equation of perpendicular bisector be 3x-4y+k=0, which passes through $(\frac{3}{2}, 2)$
Now, the equation of perpendicular bisector of line $A B is 3 x - 4 y + \frac{7}{2} = 0 . Hence,$
$the locus of circumcenter is 6 x - 8 y + 7 = 0$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App