The equation of perpendicular bisectors of the sides AB and AC of a triangle ABC are x - y + 5 = 0 and x + 2y = 0 respectively. If the point A is (1, -2), then the equation of line BC is

23x + 14 y - 40 = 0

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

14x - 23y + 40 = 0

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

23x - 14y + 40 = 0

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

14x + 23y - 40 = 0

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

Open in App

Solution

The correct option is D 14x + 23y - 40 = 0

The middle point F of AB is $(\frac{x_{1} + 1}{2}, \frac{y_{1} - 2}{2})$ lies on line (i). Therefore $x_{1} - y_{1} = - 13$
Also AB is perpendicular to FN. So the product of their slopes is – 1.
i.e., $\frac{y_{1} + 2}{x_{1} - 1} \times 1 = - 1 o r x_{1} + y_{1} = - 1$
On solving (ii) and (iii), we get B(-7, 6)
Similarly $C (\frac{11}{5}, \frac{2}{5})$
Hence the equation of BC is 14x + 23y - 40 = 0.

Suggest Corrections

Similar questions

The equations of the perpendicular bisectors of the sides $A B$ and $A C$ of a triangle $A B C$ are $x - y + 5 = 0$ and $x + 2 y = 0$ respectively. If the co-ordinates of $A$ are $(1, - 2)$ , then the equation of $B C$ is

Q. A line through A(-5, -4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 0 and x - y - 5 = 0 at B, C and D respectively. If

{(\frac{15}{A B})}^{2} + {(\frac{10}{A C})}^{2} = {(\frac{6}{A D})}^{2}

then the equation of the line is

The equation of tangent to the curve y = x³ + 2x + 6 which is perpendicular to the line x + 14y + 4 = 0 is :

Q. The equation of the perpendicular bisectors of the sides AB and AC of a

Δ

ABC are

x - y + 5 = 0

and

x + 2 y = 0

respectively. If the point A is

(1, - 2)

, then the equation of the line BC is?

Q. The equation of the perpendicular bisector of the sides

A B

and

A C

of a

△ A B C

are

x - y + 5 = 0

and

x + 2 y = 0,

respectively. If the point

A

(1, - 2)

then the equation of the line

B C

Join BYJU'S Learning Program